Linear and Exponential Relationships

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Unit 2 Linear and Exponential Relationships Lesson 10 Rational Exponents

70

Lesson 11 Functions

76

Lesson 12 Key Features of Functions

82

Lesson 13 Average Rate of Change

92

Lesson 14 Graphing Functions

100

Lesson 15 QJ Solving Systems of Linear Equations

108

Lesson 16 QJ Using Functions to Solve Equations

116

Lesson 17 LJJ Graphing Inequalities

124

Lesson 18 Translating Functions

134

Lesson 19 Reflecting Functions

144

Lesson 20 Stretching and Shrinking Functions

150

Lesson 21 Eid Functions in Context

160

Lesson 22 Qj Arithmetic Sequences

166

Lesson 23 G3 Geometric Sequences

172

Unit 2 Review

178

Unit 2 Performance Task

182

69

Rational Exponents An exponential expression includes a base raised to an exponent, or power. The properties of exponents can help you simplify many exponential expressions and solve equations involving exponents. Some of those properties are listed below. Product of powers: Power of a product:

(ab)m = amb

Power of a power: Power of zero: Quotient of powers: Power of a quotient:

a -" = - a n d - - an for all a * 0

Negative powers:

An exponential expression can be evaluated for any rational exponent. Until now, you have worked primarily with integer powers, but sometimes you may need to simplify or evaluate an exponential expression for other powers. The properties of exponents can help you rewrite expressions with fractional exponents in a more familiar form. You know that 2 • -^ = 1, 3 =3, and \'9 = 3. By applying the substitution property of equality and the power of a power property, you can find an equivalent form of a fractional exponent.

3' = 3

Substitute 2 • ^for 1. Apply the power of a power property. Evaluate inside the parentheses. Substitute V^for 3.

Raising a number to the power j is equivalent to taking its square root. In general, an exponential expression with a fractional exponent involves a root. In converting between the exponential and radical forms, the base becomes the radicand, the denominator of the fraction becomes the index of the root, and the numerator of the fraction becomes an integer exponent for the expression. A base a with exponent „ is the same as the nth root of the number a. an = Vo rn .

A base o with exponent jj- is the same as the nth root of the number a raised to the mth power. a" = ('Vo)m = 'Vo'

70

Unit 2: Linear and Exponential Relationships

Connect Simplify the expression

v'X

,

h x2.

Rewrite each radical expression using exponents. In the expression \fx, the unwritten index i is 2, so Vx = x2. In the expression 3vx, the index is 3, Simplify the numerator of the fraction,

so 3Vx = x1. In the expression 6"\r~s, the index is 6 and 6 fT the exponent is 5, so \ = x 6 .

Use the product of powers property to multiply the terms. i x2 x5

Now, rewrite the expression with no radicals. 1

1

4- '

X2"3

1

xz-xj 5 XG

Now, rewrite the expression. x2 - x 3

J jr

vG yv

3 -

c2 = — + x 2 x"

Simplify the fraction. Notice that the numerator and denominator of the fraction are identical. Any fraction with the same numerator and denominator is equal to 1. vl

1

„ Rewrite the exponential expression as a radical.

1

^ 4- x2 = 1 + x 7

3

To rewrite the expression x2 as a radical, the base, x, will be the radicand, the numerator of the fraction, 3, will be an exponent, and the denominator, 2, will be the index of the radical. ^ 1 + x^ - 1 + (Vx) 3

Since the expression has a rational exponent that is an improper fraction, it can be written in another way. Simplify. I ild l 5

s

x2 = x1 ' 2 - ( 1 + x1 = 1 + xVx

Lesson 10: Rational Exponents

71

Use a table to graph the equation y = 4*. Then, use the graph to confirm values of y for fractional values of x.

Make a table of values. Graph the equation using the values from the table.

Make a table of values with fractional values of x. y — *i

y

y=4* = (2 2 )*=2< = V2 -1.414

1.414

y = 43

1.587

y

= ^4^-1,587

= 4^ = V4 - 2

2

Compare the values in the table to the graph. According to the equation, when x = ^ and x - 3, y is close to 1^. The graph passes

i/=:45 = (V4) 3 = 23 = 8

8

near 1^ for these values of x. When x = ^ y = 2. The graph has a y-value of 2 halfway between the x-values of 0 and 1.

Graph the equation on your graphing calculator and use the TRACE or TABLE function to confirm your answers. Are your calculations for x = ^ and x = ^ more or less accurate than those on the calculator?

72

Unit 2: Linearand Exponential Relationships

When x - 2- y ~ 8. The graph has a y-value of 8 halfway between the x-values of 1 and 2, ^ The values of the equation for fractional values of x match the graph above.

EXAMPLE B

-

I—

Solve the equation 32 = 9V3 for x.

Write all terms as exponential expressions with the same base. The left side of the equation has a base of 3, so rewrite the expression 9^3 as an exponential expression with a base of 3. Substitutes for9.

9V3 0.2 2

nr

3 • V3

Rewrite the radical as an exponential expression. Use the product of powers property. Find a common denominator and add.

32I> 5

Solve the equation. Substitute the expression you found above for the right side of the equation. 3! = 9V3 x_

5

32 = 3 ? x

5

Substitute 3^ for 9^/3 . Since the bases are equal, set the exponents equal to each other. Multiply both sides of the equation by 2.

- 5

The solution to the equation is x - 5.

Solve the equation 64y — 16fory.

Lesson 10: Rational Exponents

73

Practice Rewrite each radical expression as an exponential expression with a rational exponent.

2.

1.

3.

Vx

6.

y5

REMEMBER The index of the radical becomes the denominator of the fractional exponent.

Rewrite each exponential expression as a radical expression. 4.

5.

5'

12-

Simplify each expression by using the properties of exponents. 278. 7. (16*)-(16*) 2T

: REMEMBER To multiply

; exponential expressions with the i same base, add the exponents.

Choose the best answer. 10.

5

Which is equivalent to x3? A. x B

x'

C.

D. x Vx 11.

74

Which is equivalent to V27n.12,? A.

3n2

B.

nV3

C.

n2V3"

D.

V3/?

Unit 2: Linear and Exponential Relationships

9-

Simplify each expression. 12.

13.

14.

15.

(x^ + ( ^

3

g_\ Va

\b

7(\32c

15'

Solve each equation. 16. 2 y - V 8

19.

17.

3V3 = 3'

18. 125* = 5

JlifiVMjliiWi* Write the expression V32 in three different but equivalent ways.

20. fiinna;* Write each expression as a power of 2 with a rational exponent.

2V2:

V4:_

1. 2' Using the exponents, order the terms from least to greatest.

Lesson 10: Rational Exponents

75

Functions A relation is a set of ordered pairs of the form (x, y). The equation y = x + 4 describes a relation. It relates the value of y to the value of x. A relation can be represented as an equation, a graph, a table, a mapping diagram, or a list of ordered pairs. A function is a special kind of relation in which each input, the first value in the ordered pair, is mapped to one and only one output, the second value in the ordered pair. This relation is not a function: (1,6), (3 8), (3, 9)

This relation is a function: (1,6), (3, 8), (5, 10)

\ The input 3 is assigned to two different outputs.

t t t Each input is assigned to only one output.

The set of all possible inputs for a function is called the function's domain. The set of all possible outputs for a function is called its range. The domain and range are sets that consist of values called elements. Look at the function shown in the mapping diagram. The domain for that function is the numbers -4, -3, -2, -1, and 0, or the set (-4, -3, -2, -1, 0}. The range is {-5, -4, -3,2}.

Input H-

Output

u

0

o -J _

n--

* *"

^ A function can be written as an equation using function notation. In the equation f(x) = 2* + 1, the notation f(x) is read as "fofx." It takes the place of y and stands for the output of the function for the input x. So, when x = 2, f(x) becomes f(2), and f(2) = 22 + 1 = 5. This means that the function/includes the ordered pair (2, 5). This same function can be represented by a graph. By replacing f(x) with y, the equation can be graphed on thexy-coordinate plane. The set of all the points on that graph is the function.

Most often, a function is named by the letter fand has input x, but a function can be named by almost any letter or symbol. For example, a function might be named g(x) or o(x). A function in another situation might be named h(t), so that the variable representing the input is t.

76

Unit 2: Linear and Exponential Relationships

-t Connect Does this graph represent a function?

Perform the vertical line test by drawing a vertical line through the graph atx = 5.

Determine if the graph represents a function. If any vertical line drawn through a graph passes through more than one point, the graph does not represent a function. The - >- vertical line passes through two points: (5, 3) and (5, - 3). This means that the input value of 5 maps to two different outputs, 3 and - 3, so the graph does not show a function. ^ The graph does not pass the vertical line test. It does not represent a function.

tresent a function?

x

-2

-i :

f(x) I\I

-24 *-4

i -2-i '• *>2

I

0

1

-2

_oJ-

2

I

3

9_

!

-a

Look at the input values in the table. No input, or x-value, appears in the table more than once, so each input corresponds to only one output. ^ This means that the table represents a function.

Construct a mapping diagram by using the elements in the table. How can a mapping diagram help you determine if the relation is a function?

Lesson 11: Functions

77

A function in which the input variable is an exponent is an exponential function. EXAMPLE f!» Evaluate the exponential function h(t) - 31 + 1 for t = 4.

Perform the calculations to find the output.

Substitute 4 for t in the equation. Evaluating a function means finding the output for a given input. In this case, the input is 4, so find the value of h(4).

h(t) - 3' + 1

- 3 +1 -81+1 h(4) - 82

h{4) = 34 + 1

A function in which the input variable is raised to the first power is a linear function. l^ilflia^s* The linear function g(x) = 3x + 4 has the domain {-2, -1,0, 1, 2}. Find the range of g(x)

Create a table of values to find all the elements in the range. In order to find the elements of the range, evaluate the function for each value in the domain. x

g(x) = 3x + 4

g (x)

-2

g(-2) - 3(-2) + 4= -2

-1 0

g(--i) = 3(-l) + 4 = 1 g(0) = 3(0) + 4 - 4

i 4

1

gO) = 3(1) + 4 - 7

7

2

g(2) = 3(2) + 4 - 10

-2

10

Collect the values of g(x) into a set. The values of g(x) are -2,1,4, 7, and 10. The range of g(x) is (-2,1, 4, 7,10}.

The function P(t) = 10 • 2'can be used to represent the population of bacteria in a Petri dish after t hours. What are the values of P(0) and P{4)? What do these values represent?

78

Unit 2: Linear and Exponential Relationships

Look at the table. What values of a would make f(x) a

*

-1

function? What values of a would indicate that f(x) is not a function?

Review the definition of a function.

f(x) -1.5

3

6

o

14

6

17

Find possible values for a. If a = —1, 3, or 6, an x-value would map to more than one f{x)-value. For example, if a = — 1, the x-value -1 would map to both -1.5 and 14. This means that the relation would no longer be a function.

Recall that a function is a relation in which every input, orx-value, maps to only one output, or f(x)-value. So, if the table represents a function, nox-valuecan map to more than one f(x)-value.

For any real number values of o other than — 1,3, and 6, every x-value would map to only one f(x)-value. ^ The relation f(x) is not a function when o is equal to -1, 3, or 6. The relation is a function when the value of a is any real number except -1, 3, and 6.

EXAMPLE

Find the value of the function g(x) = 5x 3 whenx = 8.

Substitute the value of x into the function. g(x) = 5x5 g(8) = 5 - 8 ^

2 Evaluate the function. Rewrite the rational exponent as a root. Since the denominator is 3, that will be the index of the radicand.

5 • 8J

Write two ordered pairs, each with the same y-coordinate. Could these two ordered pairs belong to the same function?

Rewrite the exponent as a radical.

g(8) = 5 • V8 9(8) = 5 - 2 K 9(8) = 10.

Lesson 11: Functions

79

Practice Determine whether each relation is a function.

1.

Input

+

n

3.

2.

Output i

-»- 1R

REMEMBER In a function, each input maps to only one output,

Does a vertical line pass through more than one point?

Evaluate the function for the given value of x. 4.

5.

f(x) = lOx^forx- 27

= 7rforx = 81

Write true or false for each statement. If false, rewrite the statement so it is true. 6.

The range of a function is the set of all of its inputs.

7.

To graph the function f(x) = x + 2, you can draw the graph ofy - x + 2 because the graph of fis the graph of the equation y = f (x).

Choose the best answer. 8.

Which value could not be substituted for a if the table represents a function?

24 a

4

1

4

2 A. C.

80

1 -1

: -i i B. 0 D.

-3

Unit 2: Linearand Exponential Relationships

9.

The function f(x) = 2x3 has the domain {0, 1, 4, 9}. Which of the following is the range of the function? A.

{2,2,16,54}

B.

{0,2,16,54}

C.

{0,2,8,162}

D.

{2,2,16,27}

Evaluate the functions for the inputs given. Show your work in the tables.

10.

11.

x

f(x) = 10* + 5

f(x)

0 1

f(D -

2

. f(2) =

3

f(3) =

4

! f{4) =

Solve.

12.

A ball dropped onto a hard floor from a height of 16 inches bounces back up to -= its previous heighten each successive bounce. The function h(b) = 16 • (^ can be used to represent this situation. To what height, in inches, will the ball rise on its third bounce? {Hint: Evaluate / 1\ h(b) = 16 • bd for b = 3.) Show your work.

EVALUAT The total charge for a babysitting job that lasts t hours can be represented by the function c(t) = 2 + 9t. Evaluate this function for the domain {1, 2, 3}. Show your work and briefly explain what each pair of values means in this problem situation.

14.

gill w ii 3'^ |f you switch the domain and range of a function, will the relation that results sometimes be a function? Will it always be a function? Give examples to justify your answer.

Lesson 11: Functions

81

Key Features of Functions Intercepts and End Behavior UNDERSTAN The graphs and tables of functions contain various key features. These key features are often important for understanding functions and using them to solve problems. The x-intercept of a function is the point (a, 0) at which the graph intersects the x-axis. The y-intercept is the point (0, b) at which the graph intersects the y-axis. In the graph of [(x) = 3X - 3 shown, the x-intercept is (1, 0) and the y-intercept is (0, -2).

fM

I

.

^—y-intercept

0

1

0

2

6

^—x-intercept

You can locate the x-intercept in a table by finding the row whose y-value is 0. The y-intercept is in the row whose x-value is 0. Functions can also be described in terms of their end behavior. In the graph of f(x) = 3* - 3, look at the arrows on each end of the graph. The arrow on the right end of the curve shows that as x increases, y also continuously increases. Since the value of y is continuously increasing, this function has no maximum value. The arrow on the left end of the curve shows that as x decreases (becomes more negative), yapproaches but never reaches - 3. This line that the graph approaches but never touches is called the asymptote of the function. Since the graph asymptotically approaches the line y = -3 but never intersects it, the function has no minimum value. If enough values are listed in a table, you can estimate end behavior based on the values of f(x). Starting from the top of the second column and moving down, notice that the value of f(x) gets larger and larger. Starting from the bottom of the column and moving up, notice that f(x) gets smaller (more negative) but never passes —3.

82

Unit 2: Linear and Exponential Relationships

yapproaches x | as x appro dies _i—i.

j a s y nptoe

f(x) approaches as x approaches

x -3

f(x) f(x) decreases

*- 97

~T 1

-2

—9 — 29

-1

-2-

0

-2

1

o

2

g

3

24

f(x) increases I without bound

r

Connect The function f(x) =

+ 2 is graphed below.

Identify the function's intercepts and describe its end behavior.

Find thex-andy-intercepts of the function. Where does the graph intersect the x-axis? The graph never intersects the x-axis, so the function does not have an x-intercept. Where does the graph intersect the y-axis? The graph intersects the y-axis at (0, 3). The function's y-intercept is (0, 3).

Describe the end behavior of the function. What happens to the graph as x-values approach —o°? As x-values approach -«^,y-values increase toward oo. What happens to the graph as x-values approach <*>? As x-values approach o°, y-values decrease toward 2. This means that the function has an asymptote of y = 2.

Does the function have a minimum and a maximum? If so, what are they? If not, why not?

Lesson 12: Key Features of Functions

83

Intervals of Functions '1* Remember that a function's domain is the set of all possible inputs. For a function such as f(x) = 3* - 3, the domain is the interval on the x-axis on which the function is defined, in which the graph exists. The range of a function is the interval on they-axis containing all possible outputs. Interval notation can be used to represent an interval. In interval notation, the end values of an interval are listed as a pair separated by a comma. A bracket beside a value means that it is included in the interval, while a parenthesis means that it is not. For example, the domain [0, 5) is equivalent to 0 ^ x < 5. The domain can be broken up into smaller intervals that share a certain characteristic. For example, it can be useful to divide the domain into sections in which the value of f(x) is positive and sections where it is negative. Look again at a graph of the function f{x) = 3X - 3. Determine the intervals where y is positive and where y is negative. The value of y is negative when x < 1. The value of y is positive when x > 1. Using interval notation, y is negative on the interval (-co, l)and positive on the interval ("], «=).

The domain can also be divided into sections where the value of f (x) is increasing from left to right and where it is decreasing from left to right. Look at the graph. From left to right, the graph is always curving upward. The value of y is always increasing as the value of x increases. This is true across the entire domain, from negative infinity (-=&) to positive infinity (<«). The function is always increasing. In other words, the interval of increase J S ( — oc f oc).

Usually, these intervals can also be determined from tables by looking at the values in the f(x) column.

84

Connect The graph below represents an exponential function f. The table below represents a linear function g.

y

. }-

1

3x

X

*-x

_c

2

~~3

1

; 3

o -i

6

-2

Compare and contrast functions f and g by using these key features: domain; range; intervals of increase and decrease; and positive and negative intervals.

Identify the domain and range. The end behavior of the graph of fshows that it extends indefinitely both left and right. Thus, its domain is all real numbers, orthe interval Since f has an asymptote of y = -l,its range isy > —1, orthe interval (- 1, =e). The table for function g does not list all values of x or g(x), but it also does not give evidence of any boundaries (such as an asymptote). Since g is a linear function, without other information, you may assume that the domain and range are all real numbers.

Compare the intervals of increase and decrease. The graph of f continuously curves downward. So, f is always decreasing. The table for function g shows that as x-values increase, g(x)-values decrease, so g is also a decreasing function.

Compare positive intervals and negative intervals for the functions.

Both functions are decreasing across their entire domains. The interval of decrease is (—cc ( so) for both functions.

The graph of /"intercepts the x-axis at (0, 0). The third row of values in the table shows that g also has an x-intercept of (0, 0). The functions are always decreasing. Functions fand g are both positive when x < 0, on the interval ( — ^, 0), and negative when x > 0, on the interval (0, •*).

Graph function g on the same grid as f. Compare and contrast the two graphs to check the answers on this page.

Lesson 12: Key Features of Functions

85

The domain of a linear function is{-l < x ^ 2}. The function hasanx-interceptat (1, 0) EXAMPLE and a y-intercept at (0, -3). Graph the function. Then identify the maximum, minimum, range, and intervals of increase and decrease for the function.

Graph the function, paying attention to the restricted domain. Plot the intercepts. Draw a line through the intercepts, but do not extend it to the left of —1 or to the right of 2 on thex-axis. .

e

•i --J

-1*

:!.; -6

I f ( I, 3 ) -

.. j.._.

'

2

0) -4

do nain rest

to> i

domai ir estri te d : 'to X < 2

-2

C

, 0

T

2

-

ed , ((0.-3)

1

-

...

Describe the function's minimum, maximum, and range.

When the domain is restricted to {—1 < x < 2}, the lowest point on the graph is at (-1, -6). Thus, the minimum y-value is —6. The highest point on the graph is at (2, 3). Thus, the maximum y-value is 3. The range is all values of y greater than or equal to the minimum, -6, and less than or equal to the maximum, 3. This can be represented as {—6 < y < 3}. Identify intervals of increase or decrease. The line segment slants up from left to right, so the function is always increasing. The value of y increases across the entire domain. The function increases on the interval {-1 < x < 2 } .

Identify intervals where the linear function graphed above is positive and where it is negative.

86

Unit 2: Linear and Exponential Relationships

A piano is being lowered from an apartment that is 18 feet above the sidewalk. The piano descends at a constant rate. The piano's elevation over time is represented by the linear function graphed below. Identify and interpret the key features of the graph. y

Piano's Descent

20

18 O 16

E" c 10

LU

1 2 3 4 5 6 7

Time (in minutes)

Identify and interpret the domain. The graph is shown to exist on the domain [0, 6], This domain contains the minutes over which the piano is being lowered. Identify and interpret intervals of increase and decrease. The function is decreasing for the entire domain. This means that the piano's elevation is always decreasing. The function has no interval of increase. This makes sense because the piano is always being lowered and never being raised. Identify and interpret the intercepts. They-intercept, (0,18), represents the piano's initial elevation of 18 feet. Thex-intercept, (6, 0), shows that it takes 6 minutes for the piano to reach the sidewalk, at an elevation of 0 feet.

What is the range for this function? What does it represent in the problem?

Lesson 12; Key Features of Functions

87

Practice Rewrite each domain in interval notation. 1.

{5
2.

3.

{x

{all real numbers}

REMEMBER A bracket means include the value, and a parenthesis means exclude the value.

For each graph, determine whether the function is increasing or decreasing. Identify the interval of increase or decrease. 5.

4.

o

Does the graph curve (or slant) upward or downward?

Identify the intercepts of the given function. 6.

88

y

7.

x-intercept:

x-intercept:

y-intercept:

y-intercept:

Unit 2: Linear and Exponential Relationships

Identify the intercepts of the given function 8.

24

x

-12

0

12

24

9.

_2

X

_!

0

1

2 ,

f(x>

-8

-6

-2

0

g(jr)

-9.99

x-intercept:

x-int ^rrppt-

y-intercept:

y-int) ^rcent:

-9.9 1 -9

0

90

Fill in each blank with an appropriate word or words. 10.

A point at which a graph crosses the y-axis is a(n)

11.

Afunction's never intersects.

is a line that the graph of the function approaches but

12. The of a function describes how its f(x)-values change as x approaches positive infinity or negative infinity. 13.

The greatest y-value on the graph of a function is the function's

Choose the best answer. 14.

Which statement about this function is not true?

15.

The table below shows some ordered pairs for an exponential function.

rw

Jl 1

5 6

_

0

0

1

5

2

35

3

215

Which statement about this function is not true? A.

Its domain is { — 4 < x < 6}.

B.

Its range is {-1 < y < 4}.

C. D.

A.

Its x-intercept is the same as its y-intercept.

It has a y-intercept at (0, 2).

B.

It is positive on the interval (0, cc).

It has a maximum of 6.

C.

It is increasing on the interval (-M,OO).

D.

As x approaches —°°, f(x) approaches^.

Lessonl2: Key Features of Functions

89

Describe the end behavior of each function. 16.

17.

r

Use the graph and table below for questions 18-20. The graph represents exponential function f. The table represents some ordered pairs for linear function g.

2

.

' 6

X

g(x)

-1

-8

0

-4

1

0

2

4

T

8

3

18.

Compare and contrast the intercepts of the functions.

19.

Compare the increasing and decreasing intervals of the functions.

20.

Compare the intervals on which the functions are positive and those on which they are negative.

90

Unit 2: Linear and Exponential Relationships

For each graph, describe the intervals where the function is positive and where it is negative.

21.

y

22.

y

Solve. INTERPRET, A cylinder contains 20 milliliters of water. The water begins to leak out as represented by the linear function graphed on the right. Identify the intercepts and interpret what they mean in this situation.

Volume of Water in Cylinder --> 22

e 20

$> 18 j= 16

i;; .S 10

£ 1

2

3

4

5

6

Time {in seconds)

24.

The cost of a taxi ride includes a $3 fee plus $2 for each mile traveled. So, a 1-mile ride costs $5 and a 2-mile ride costs $7. Create a graph to represent this linear function. Identify the domain for your graph and explain why you chose it.

}, :

14



Cost of Taxi Ride i . i

13 • i

0)

cn o

o

'•

i

i

10 9

T3

':

.

.

8 7

5 4 3 2 1 0

i

i

!

..!

i 1

i

' 2

3

4

5

6

Miles Traveled

Lesson 12: Key Features of Functions

91

•'

Average Rate of Change Finding Average Rate of Change UNDERSTAN|i*

Rates allow us to relate quantities measured in different units. For example, the table and graph below show a linear function that compares the number of hours a cashier works to his total earnings, in dollars. Cashier's Earnings

Cashier's Earnings Time in hours, x

Earnings in $, y

0 2 4

0 15 30

•o c C/J

o> c 'c '—

CD ill

0

1

2

3

4

5

6

7

Time Worked (in hours) The cashier's earnings change, depending on the number of hours he works. His pay rate is an example of a rate of change. A rate of change shows how one quantity changes relative to another quantity. To calculate the average rate of change between two ordered pairs (x, y, and (x2/ y2), use this formula: change in y

y2 - y,

average rate of change = -r change :— mx ~ x*j . x„i For the function describing the cashier's earnings, choose two ordered pairs, such as (2,15) and (4, 30). 30-15 15 -. r ^ average rate of change = 4 - 2 = ~2~ = 7-5®

In this case, the rate of change compares dollars earned to hours worked. So, the cashier's rate of pay is $7.50 per hour.

92

Unit 2: Linear and Exponential Relationships

Connect A basketball championship begins with 64 teams. Every time a team wins a game, it goes on to the next round. Once a team loses a game, it is eliminated from competition and does not play any more games. The number of teams in each round of the championship is a function of the round. That function is represented on the graph to the right. Compare the rate of change between rounds 1 and 2 to the rate of change between rounds 2 and 3.

y

Basketball Championship

1

2

3

4

5

6

Round

Calculate the average rate of change between rounds 1 and 2. Find the rate of change from (1, 64) to (2, 32). 32 - 64 2-1

32 1

32 teams per round

Between rounds 1 and 2, the number of teams decreases at a rate of 32 teams per round.

Calculate the average rate of change between rounds 2 and 3. Find the rate of change from (2, 32) to (3,16). 16 - 32 _ 3-2

16 1

16 teams per round

Between rounds 2 and 3, the number of teams decreases at a rate of 16 teams per round.

3

Compare the rates of change. The rate of change between rounds 1 and 2 is different than it is between rounds 2 and 3. The rate between rounds 2 and 3 is half what it was between rounds 1 and 2.

Choose a pair of points on the graph below and find the average rate of change between them. Compare your result with those of other students. Did they use the same two points?

Lesson 13: Average Rate of Change

93

Comparing Average Rates of Change UNDERSTAN t* The table below represents the linear function f(x) = 2x + 1. Notice that as x-values increase by 1, f(x)-values increase by a constant amount, 2. In other words, the function grows by an equal amount, 2, in each unit interval.

X

0

1

2

3

4

5

f(x)

1

3

5

7

9

11

A linear function has a constant rate of change. Its average rate of change is the same no matter what interval you are observing. The constant rate of change of a linear function is its slope. An exponential function has a graph that is a curve. An exponential growth function is always increasing, while an exponential decay function is always decreasing. The table below represents an exponential growth function.

+1 X

0

1

f(x)

1

2

Notice that as each x-value increases by 1, each f(x)-value is multiplied by 2. The value of the function, f (x), does not grow by constant amounts over equal intervals, so it does not have a constant rate of change. However, f (x) does grow by the same factor over equal intervals. This function increases by a factor of 2, or doubles, over each unit interval. The value of an exponential function grows by equal factors over equal intervals. If the factor by which the function changes is greater than 1, then the function represents exponential growth. If the factor is less than 1, then the function represents exponential decay. The average rates of change for an exponential function grow by the same factor as the values of the function. For the table above, the average rate of change doubled over each unit interval.

94

x

0

f (X) ;

1

~

Unit 2: Linear and Exponential Relationships

. 1 2

!

2 4

3

4

5

8

16

32

Connect Find and describe the average rate of change for four consecutive pairs of values in the table. What type of function is this?

x

-3

-2 I

f(x)

64

-1 j

16

0 !

4

1

1 Determine the average rate of change for consecutive pairs of values (x, f(x)). Be sure that the intervals are the same between each pair of points. The difference between each pair of x-values in the table is 1 unit, so the intervals are the same. between ( 3, 64} and 16-64 -48 _ o _ t — ^\

\

1

2,16):

48

between (-2,16) and (-1,4): 4-16 -12 - -12 1

between {-1, 4) and (0, 1): 1 -4

-3

o-(-i)

i

between (0, 1) and (l, 4):

Compare the average rates of change in order to classify the function. The average rates of change for the first four consecutive pairs of points are: -48,-12,-3,-f. These rates are different. The rate of change is not constant, so this is not a linear function. The value of f(x) decreases by a common factor over each interval. between {-3, 64) and (-2,16): _ 64

1 -0

between (-2,16) and (-1,4): 16

between (-1, 4) and (0,1):

between (0, l)and 1 1-1 1 4 Since the values of f(x) change by an equal factor over equal intervals and that factor is 4, this is an example of an exponential decay function. Why is it important to keep the intervals between each pair of values (x, f(x)} the same when comparing average rates of change?

Lesson 13: Average Rate of Change

95

Determine the average rate of change between several consecutive pairs of points for the function f(x) = -3x+ 2. Describe how the function is changing and classify it.

Create a table of ordered pairs for the function. x

f(x) = -3x

-2 ; f(-2) - -3(-2) +2

2

f(x)

6+2=8'

8

-1 • fl-1) == -3(-l) + 2 - 3 + 2 - 5 ;

5,

; 2; 0 , f{0) = -3(0) + 2 = 0 + 2 = 2 1 f(l) = -3(1) + 2 - -3 + 2 = -1 | -1 ;

2 f(2) = -3(2) + 2 = ~6

2 = -4 • -4

:

Determine the average rate of change for four consecutive pairs of values (x, f(x)). Be sure that the intervals are the same between each pair of points. between {-2, 8) and (-1, 5): 5-8 -1 -(-2)

between (-1, 5) and (0, 2): 2-5 -3 ^ 0-(- - 1 )

1 "

between (0,2) and (1, -1):

Compare the average rates of change.

-1-2

-3

~

1-0

1

J

between (1, -l)and(2, -4):

The average rates of change are all the same, -3. Since the rate of change is constant, f(x) = -3x + 2 must be a linear function. ^ The rate of change, or slope, is -3 for all pairs of values. The function is linear.

Does the equation y = —3x + 5 provide any clues about what the rate of change for the linear function is? Explain.

96

EXAMPLE Compare the rates of change for f(x) = 10* and function g, which is represented in the table.

i

x

9(x)

-1

]_ 8

0

1

1

8

2

64

3

512

Create a table of values for f. f(x

X

-1

f(-D

0 ! f (0) - 10° 1

= iox

-1 = 10

1 10 1

ifd) = 10 == 10 ]

2 f(2) = 102 = 100 3

f(3) = 103 = 1,000

f(x)

r

_ 10

1 10 100 1,000

Find the average rate of change for three consecutive intervals for function f. between (0, l)and 0,10): yf^ - ~ = 9 between (1,10) and (2,100): Find the average rate of change for three consecutive points for function g.

between (0, 1) and (1, 8): between (1,8) and (2, 64): 64-8 56 cc T^T: T'

100-10 _ 90 _ nn 2-1 T: between (2,100) and (3,1,000): 1,000-100

900

3-2

1

900

The rate of change of the function fis not constant. Each average rate of change is 10 times the previous rate of change.

between (2, 64} and (3, 512): 512-64 448 1 -448 3-2 The rate of change of the function g is not constant. Each average rate of change is 8 times the previous average rate of change. ^ The average rates of change for function fare growing more rapidly than the average rates of change for function g.

By what factor are the values of function f{x) growing? Does the equation f(x) ~ 10X help you determine that factor? How could you write an explicit expression for g(x)?

Lesson 13: Average Rate of Change

97

Practice Fill in the blanks by writing an operation sign and a number to show how the f (x)-values are changing in each unit interval. Then classify each function as linear or exponential. 1.

/ +1 >^ X

f(x)

-1

n \J

1

1

6

+1 i | .

V 1 6

+1\/+1

i * i

^ 3

36

216

L_^_

L

+131 4

1,296 Over each interval, does f(x) change by an equal amount or an equal factor?

2.

Fill in the blanks with an appropriate word or phrase. ,

..

between two ordered pairs (x, y) is the ratio

3.

The average

4.

In a linear function, the rate of change is also known as the

5.

The average rate of change for a

6.

The average rate of change for an exponential function grows by equal per unit interval.

change in y „ . in x

function is constant.

Use the graph for questions 7-10. 7.

Determine the average rate of change between —Inland (0, 2)

8.

Determine the average rate of change between (0, 2) and (1, 4).

9.

Determine the average rate of change between (1, 4) and {2,10).

10.

Write a sentence or two comparing the average rates of change you found. (If they vary, describe how they vary.)

98

Unit 2: Linear and Exponential Relationships

Use the information about function f (x), given as a table below, and function g(x) = 5" for questions 11-14. 11.

12.

Using the table on the right, find the average rate of change for three unit intervals for function f.

Complete the table to find four consecutive ordered pairs for the function

13.

fM

1 4

-1 0

1

1

4

2

16

Find the average rate of change for three unit intervals for function g.

g(x) = 5*.

,

x

-1 0

g(-D = g(0) =

1 2

g(2) =

14.

Compare the changes in the values of functions f and g.

15.

fJMisWsWa!* The graph shows how the total amount that a landscaper charges for a job changes depending on the number of hours she works. Identify the slope of the graph. Then interpret what this slope represents in this problem situation.

Cost of Landscaping Jobs

•m

•t

110 100 90 •o 80 70 0 GO O) 50 40 CD .n 30 O 20 "re *-< 10

.o

0

1

2

3

4

Length of Job (in hours)

Lesson 13: Average Rate of Change

99

_
Si Graphing Functions Graphing Linear Functions The various representations of a function give different details about the function. An equation in function notation explains the rule for generating an output from any given input. A table can list many, but usually not all, input/output pairs for the function. A graph is a visual representation of all the input/output pairs of the function. UNDERSTAN

The graph of a function is the graph ofy = f{x), soy takes on the value of the output. If you choose any point (x, y) on the graph of a function, they-coordinate is the output of the function when the x-coordinate is the input. Every point on the graph is a solution to the equation y - f(x). To understand the most about a function, it is often helpful to translate from one form to another. By examining the equation of the function, you can often identify key features that will help you construct the graph of the function. Examine the linear function represented symbolically as f(x) = ^x + 1. Its graph on the 3 xy-coordinate plane is y = f (x) or y = ^x + 1. The equation is in slope-intercept form, 3 y = mx + b, where m represents the slope and b represents the y-intercept. For y = ^x + 1, 3 the slope or rate of change, m, is ^, and the y-intercept, b, is 1. This is enough information to graph the function. The y-intercept of a linear equation 3 y = mx + bisat(0, b). Fory = ^x + 1, the y-intercept is at (0,1). Plotting this point starts your graph. Now you can use the slope to find another point on the graph. The slope is a rate of change that tells how to move from one point on the graph to another. It is the ratio -r =—. change mx Place your finger at the y-intercept and count 3 units up and 2 units to the right to find another point, (2, 4). Draw a straight line through those points. Every point on the line is a solution for y = fx + 1. So, (-4, -5), (-2, -2), (0,1), and (2, 4} are all solutions.

Sometimes, a linear equation will not be in slope-intercept form. In that case, you may need to put it in that form yourself before graphing it.

100

Unit2: Linear and Exponential Relationships

-«= Connect Graph the linear equation 6x + 3y = 12 on a coordinate plane. Identify at least three ordered pairs that are solutions for the equation.

Rewrite the equation in slope-intercept form and identify the y-intercept and the slope. 6x+3y=12 3y= -6x+12

Subtract 6x from both sides. Divide both sides by 3.

y= -2x + 4

Plot the y-intercept and use the slope to find a second point on the line. After plotting (0, 4), count down 2 units and 1 unit to the right. Plot a point there at (1,2).

y 2 uni s

(0, 4) I

Draw a line through the points. Locate a third point on the line.

You can "eyeball" another point on your line, or you can use the slope to find a third point. ^ The ordered pairs (0, 4), (1, 2), and (2, 0) are solutions for 6x + 3y = 12.

Use a graphing calculator to check your work. Press . Enter Y, - -2X+4. Press MiH;!!l. Does the graph on your calculator screen look like the graph drawn on the left? If you press f£ JJ M:^d!l, do the data in the table match the graph on the left?

Lesson 14: Graphing Functions

101

Graphing Exponential Functions You can also use key features to help you graph an exponential function. A general exponential function has the form f(x) = a • bx + c, where a ^ 0, b > Oand fa =£ 1, and c is a real number. UNDERSTAN

Examine the exponential function f(x) = 3 • 2" - 4. In this function, o = 3, b = 2, and c = -4. To graph this function on the xy-coordinate plane, graph y = 3 • 2* — 4. The simplest key feature to find from the graph is the horizontal asymptote. No matter what input x is entered, the term a • b x can never equal 0, sof(x) can never equal c. So the liney = c is a horizontal asymptote. Thus, the given function has a horizontal asymptote at y - -4. The parameter a tells where the graph lies in relation to the asymptote. • If o > 0, then the graph lies entirely above the asymptote. • If a < 0, then the graph lies entirely below the asymptote.

The y-intercept, (0, f(0)), of an exponential function is located at the point (0, o + c). f (0) - a - b° + c simplifies to f (0) = a - l + c = a + c since any number raised to the power of 0 is equal to 1. Fory = 3 -2" -4, the y-intercept is (0, -1). The parameter b describes how to move from one point to another on the graph. • If b > 1, the function curves away from the asymptote as x increases (as the graph moves to the right). • If 0 < b < 1, the function approaches the asymptote as x increases (as the graph moves to the right). For the example function, b — 2. This means the value of y will double (be multiplied by 2) as the graph moves 1 unit to the right. At x = 0 (the y-intercept), the graph is 3 units above the asymptote. Atx = 1, the graph will be twice as far from the asymptote, 6 units above it. Atx = 2, the graph will be twice that distance, or 12 units, above the asymptote.

102

Unit 2: Linear and Exponential Relationships

Connect / 1\ Use what you know about key features to graph f(x) = 3 ^

Identify the parameters. An exponential function has the form f(x) = a - b* + c.

HV 1 In y = 3hd , a = 3, b = ^, and c = 0.

»

Identify the asymptote. The asymptote is the line y = c, in this case, y = 0, or the x-axis.

Identify the y-intercept. Any number raised to the power of 0 equals 1. So, when x

0,y , = 3[i)° = 30)«:

The y-intercept will be at (0, 3).

y

Use b as a factor to find additional points on the graph. Since b = ^, moving along the x-axis 1 unit means dropping half the distance to the asymptote. Since the y-intercept is 3 units above the asymptote, the graph will be 1.5 units above the asymptote at x = 1 and 0.75 unit above it at x = 2.

y-intercep

asympto e

Since 0 < b < 1, connect these points with a smooth curve that approaches the asymptote.

How could you use the value of b to find points to the left of the y-intercept?

Lesson 14: Graphing Functions

103

Sean is at his grandmother's house, which is 60 miles from his home. He starts riding home at time f = 0. His distance from his home, d(t), after t hours can be modeled by the function d(t) = 60 - 15t. Graph the function for the domain 0< t<4. Explain why the domain must be restricted in that way and what the maximum and minimum values mean in this situation. EXAMPLE A

Identify the type of function. Is the function linear or exponential? The equation d(t) = 60 - 15t has a variable, f, raised to the power of 1. So, the function is linear.

Graph the function, choosing an appropriate scale and label for each axis.

Identify the slope and y-intercept from the equation. The graph of the function is the graph of the equation d(t) = 60 - 15t,or d(t}= -15t+60, for all values oft between 0 and 4, inclusive. The slope, m, of the line is -15, and its y-intercept, (0, fa), is (0, 60).

Plot the y-intercept. According to the slope, another point is 15 units down and 1 unit to the right, at (1, 45). Draw a line through those points. d(t)

Bike Ride Home

Identify and interpret the maximum and minimum. The maximum is 60. This is the farthest Sean is from home during his ride. The minimum is at 0. The point (4, 0) shows that he was 0 miles from his home— or at home—after 4 hours of riding. That means it took him 4 hours to reach his home.

1

2

3

4

5

Time (in hours)

The domain is restricted, so the graph is only the part of the line between t = 0 and t = 4. Why is the graph a line segment instead of an entire line? Do values outside the domain make sense?

104

Unit 2: Linearand Exponential Relationships

The equation f(x) = 3x + 1 represents a linear function f. EXAMPLE The table of values on the right represents an exponential function g. Graph functions f and g on the same coordinate plane. Then compare their properties.

X

0

1 3 1

1

3

2

9

3

27

-1

Graph the functions. To plot f (x) = 3x + 1, notice that it is in slope-intercept form. So, plot the y-intercept at (0,1) and then count 3 units up and 1 unit to the right. Draw a straight line through the points. To graph function g, plot and connect the coordinate pairs from the table.

Use the graphs to compare the functions. Both functions have the same domain: the set of all real numbers. Function f has a range that includes all real numbers. Function g approaches but never touches the x-axis (the line y = 0), so its range is y > 0. Both functions have the same y-intercept at (0,1). Both are increasing functions. However, shortly after x = 1, the graph of function g starts to increase at a much more rapid rate than the graph of function f, which continues to increase at a constant rate. Notice that around x - 1.5, the graph of function g overtakes the graph of function f.

Will an exponential function always overtake a linear function? Explain and give or sketch an example.

Lesson 14: Graphing Functions

105

Practice Circle the ordered pairs that are solutions for the graphed function. 1.

(-2,3) 0,2) (0, -2)

REMEMBER Each point on the graph is a solution for the equation.

0, -3)

(2,1) (4,0)

Choose the best answer. 3.

Which graph represents the function f(x) = 2(3X)? A.

r

B.

106

Unit 2: Linear and Exponential Relationships

C

D.

Graph each function. 4.

f(x) = -3x + 5

5.

"1 -ill

6.

The graph of f(x) = 2 \ -F is shown on the coordinate I plane. Graph g(x) = 2(5*) on the same coordinate plane. Then compare the end behavior of the two functions.

7.

f* * A hurricane is located off the coast when scientists begin tracking its distance from land. Its distance from land, d(t), after t hours can be modeled by the function d(t) = 120 - 20t. Graph the function for the domain 0 s t < 6. Identify the maximum and minimum values. Explain what each represents in this situation.

COMPAR

d(t)

:

120 110 100 on 3U

80 70

60 m ou 40

"

._

_

30 20 10 0

i j <

£

(

(

t 1D

Lesson 14: Graphing Functions

107

Solving Systems of Linear Equations A system of linear equations consists of two or more linear equations that use the same variables. UNDERSTAN

Recall that a linear equation in two variables generally has an infinite number of solutions: all of the(x, y) pairs that make the equation true. The solution to a system of equations is the point or points that make both or all of the equations true. Typically, a system of linear equations has one solution. If there is no coordinate pair that satisfies every equation in the system, then the system has no solution. When the equations have the same graph (because they are equivalent equations), the system has an infinite number of solutions: every (x, y) pair on that graph. You can use graphs to approximate the solution to a system of equations. To solve a system of equations graphically, graph each equation on the same coordinate plane. The solution is the point or points where the graphs of the equations intersect. Because those points are solutions to every equation in the system, they are the solutions for the system. The system shown on the graph on the right has one solution: (3, 2).

UNDERSTAN One way to solve a system of equations algebraically is to use the elimination method. In this method, equations are added and subtracted in order to eliminate all but one variable. This results in an equation in one variable, which can be solved. The value for that variable is then used to solve for the other variables.

Knowing the properties of equality is crucial to understanding how the elimination method works. For example, one step involves multiplying both sides of an equation by a constant factor. The multiplication property of equality assures that doing that will not change the solution of that equation. Another way to solve a system algebraically is the substitution method. In this method, a variable in one equation is replaced by an equivalent expression from another equation. This results in a new equation that has fewer variables. This can be repeated until only one variable remains in the equation. The value of the variable can be found from that equation and then used to find the values of the other variables. The substitution method is especially useful when a system of equations includes an equation with an isolated variable, such as y = 3x + 7. If the system does not include an equation in this form, you can take the necessary steps to isolate a variable in one of the system's equations.

108

Unit 2: Linear and Exponential Relationships

Connect Solve the system of equations by graphing. fy-2x-6

\x-2y =6 « Write the equations in slope-intercept form. The first equation is already in slope-intercept form. Isolate y in the second equation. x

Graph the first equation.

— 2y = 6 -2y= -x+ 6 y - jx ~ 3

Plot a point at the y-intercept, (0, —6). Then use the slope, 2, to plot a second point at (1, —4). Draw a line to connect the points.

Graph the second equation. Plot a point at the y-intercept, (0, —3). Then use the slope, -•/ to plot a second point at (2, —2). Draw a line to connect the points.

V

T Find the solution.

To find the solution to the system, find the point where the lines intersect. ^ The point of intersection appears to be {2,-2).

Substitute x = 2 and y = -2 into both equations in the system and confirm that true statements result.

Lesson 15: Solving Systems of Linear Equations

109

EXAMPLE I Solve the system by using the elimination method. -3x-2y = -10 2x + y = 7

Choose which variable to eliminate. Look at the coefficients of the y-terms. The y-term in the first equation has a coefficient of—2, and the y-term in the second equation has a coefficient of 1. Use the multiplication property of equality to multiply both sides of the second equation by 2. 2(2x + y) = 2(7)

Combine equations to eliminate one variable.

4x + 2y = 14 The addition property of equality allows you to add equivalent values to both sides of an equation. Add the new equation to the first equation from the original system to eliminate y.

This new equation has the same set of solutions as 2x + y = 7, because they are equivalent equations.

- 3 x - 2 y = -10 + 4 x + 2 y = 14 x+0 =

4 4

Use the value of x to solve for y. The substitution property of equality allows you to substitute 4 for x in the original second equation in order to solve for y. 2x + y '-= 7

2(4} + y = 7 8 +y=7 y = —1

Substitute x - 4 into the equation. Simplify. Subtract 8 from both sides of the equation.

^ The solution to the system is the ordered pair (4, -1). • 11 ^vm~f*i^m^^m*^m+~i^—* mi i i •• •

110

, ....• n. .»,. ,. • ,

Unit 2: Linear and Exponential Relationships

, , , .^i^

Substitute the x- and y-values of (4, -1) into both equations in the system and verify that the solution is correct.

EXAMPLE B

A system of equations and its graph are shown.

2y = 3 y=2 Use elimination to find the solution to the system. Show that the elimination method produces a new and simpler system of equations with the same solution as the original system.

1 Replace the first equation in the system. Multiply both sides of the second equation by -1, and then add the result to the first equation. x 4- 2y = 3 + - x - y = -2 Y=

1

Replace the first equation with this equation to produce a new system.

fy=l

Replace the second equation in the new system.

lx+y=2

Multiply both sides of the new first equation by -1, and then add the result to the second equation. x+ + Graph this new system, y

x

y- 2 -y=-1 - 1

Replace the second equation with this equation to produce a new system.

The systems have the same solution, (1,1). ^ Combining one equation in a system with a multiple of another equation yields a different system of equations with the same solution as the original system.

Compare the original system of equations to the final system of equations. In which system is the answer more obvious?

Lesson 15: Solving Systems of Linear Equations

111

EXAMPLE

Solve the system by using the substitution method.

'2y-3x=19

4y ~ -4

Isolate a variable in one equation. In the second equation, the coefficient of x is 1. So, the easiest course of action is to solve the second equation for x. Subtract 4y from both sides of the equation.

x + 4y = -4 x = -4 - 4y Perform the substitution and solve for the other variable. The substitution property of equality allows you to replace x with the expression -4 - 4y in the first equation from the system. Doing so allows you to solve fory. 2y-3x=19 2 y - 3 ( - 4 - 4 y ) = 19 Use the value of one variable to solve for the other variable.

2y + 12

14y+ 12 = 19 Apply the substitution property of equality

14y=7

again. Substitute^ for y in one of the equations and solve for x.

x + 4y = -4

x + 2 = -4 x = -6 The solution to the system is - 6,

TRX

Solve the system by substitution.

13x

112

Unit 2: Linear and Exponential Retationships

+ y-9

Problem Solving Bonnie has a jewelry-making business. She rents a studio space for $400 per month, and each necklace she makes costs her $15 in materials. She sells the necklaces for $55 each. How many necklaces must she sell in a month to make twice as much money as she spends? How much will she spend and how much will she make?

Write and solve a system of equations. Let n be the number of necklaces that Bonnie makes and sells in a month. Let m be the amount of money Bonnie spends on the business that month. Write an equation to represent the amount Bonnie spends for the month if she makes n necklaces at her studio.

m=

+

Write another equation showing that the amount she makes by selling n necklaces is twice as much as she spends.

2m = SOLVE Solve the system by using substitution. The first equation has m isolated on the left side. So, substitute the expression on the right side for m in the second equation.

2m-

_

2(

Now, solve the resulting equation for n. n=

Now, substitute the value of n into either of the original equations to find the value of m. m— 2m =

Substitute the values of n and m into the original equations. Do the substitutions result in true equations? ^ If Bonnie makes

necklaces, she will spend $

and she will make $.

Lesson 15: Solving Systems of Linear Equations

113

Practice Determine if the given ordered pair is a solution to the given system.

1.

f 3 x + 7y= 12

2.

,6x - y ~ -4

• 2x - 7 - -y

3.

2X+3^

,-5x4- 13 (2,3)

(-3,3)

(4, -6)

Choose the best answer. 4.

A system of three equations is shown on the graph below.

5.

A baker rents space in a commercial kitchen for $210 per week. For each pie he bakes, he spends $4 on materials. He charges $7.50 per pie. The graph below shows the baker's costs and revenues for a week in which he sells p pies. m

o

400

' 300

3?c 20°

What is the solution to the system? A.

° 100

(2,-1) D

B. (-2, -3) C (0,3)

= 13

, 2 x - y = -3 elimination

114

Unit 2: Linear and Exponential Relationships

60

80

100

How many pies must he sell in a week in order to break even?

A. B. C.

20 40 60

D.

He will never break even.

Solve each system of equations by using the method suggested. -3x-5y

40

Pies Sold

D, The system has no solution.

6.

20

7.

Solve.

8

Sanjit has a collection of quarters and dimes worth $3.70. He has a total of 19 coins. How many quarters and how many dimes does Sanjit have?

9.

Sonya opened a savings account with $200 and deposits $10 each week. Brad opened a savings account with $140 and contributes $40 each week. After how many weeks will Brad's account balance be twice as much as Sonya's? What will the balance be in each account then?

Solve each system of equations by graphing on the coordinate grid.

11.

Solution:

Solution:

Answer the questions below. 12.

tj! How many solutions does the following system of equations have? How do you know?

Lesson 15: Solving Systems of Linear Equations

115

Using Functions to Solve Equations UNDERSTAND Solving a one-variable equation means finding the value of the variable that makes the equation true. So, solving 3x + 5 = -x - 3 means finding a value of x that makes the left side of the equation equal to the right side.

You can treat each side of the equation as a separate function and let the two functions form a system, like this: f f(x) - 3x + 5 1 g(x) - -x - 3 The graph of function f is the graph of y = f(x). This graph shows all the solutions for f. The graph of function g is the graph ofy = g(x). This graph shows all the solutions for g. The point where these two graphs intersect is the point at which one input, x, produces the same output for both functions. At this point f(x) = g(x], so the x-value for that point is the value of x that makes the equation 3x + 5 = — x — 3 true. You can find this value of x by graphing f(x) = 3x + 5andg{x) = -x - 3 on the same coordinate plane. The graph of f (x} = 3x + 5 has a y-intercept at (0, 5) and a slope of 3. The graph ofg(x) = -x - 3 has a y-intercept at (0, -3) and a slope of -1. Graph and label the two functions. Then find their point of intersection.

The graphs of/"and g intersect at ( — 2, -1). The x-value of that ordered pair is — 2, so the solution of 3x + 5 = — x — 3 isx = -2.

116

Unit 2: Linear and Exponential Relationships

Connect Solve the following equation for x by making a system of functions and graphing. 4x + l = 2x+ 3,

Treat the expression on each side of the equation as a function. Let f (x) = 4x + 1. Letg(x) - 2x + 3.

2 Graph each function in the system on the same coordinate plane. The graph of f is the graph ofy = f(x), or y = 4x + 1. This graph is a line with a y-interceptat(0,1) and aslope of 4. The graph of g is the graph ofy = g(x), ory = 2x + 3. This graph is a line with a y-intercept at (0, 3) and a slope of 2.

Find the x-coordinates of any points of intersection.

g(x) =

The graphs intersect at (1, 5). The x-coordinate of that ordered pair is 1. fc- The solution isx = 1.

Solve 4x + 1 = 2x + 3 algebraically and compare the solution to the one found above.

Lesson 16: Using Functions to Solve Equations

117

x-1

Use a graphing calculator to solve for x: 2

EXAMPL

=4.

Treat the expression on each side of the equation as a function and form a system. Graph the functions by using your graphing calculator and then find the point of intersection.

Letf(x) Let g(x)

4.

Press For Y1 enter 2 A {X - 1). For Y2 enter 4.

Your screen should show the following:

Look at tables of values on your calculator to verify the point of intersection. Press m'J~>r'm

si to view a table of values

for both graphs. X

Vj

&•• -1 0

1 2 3 4

.125 .25 .5 1 2 4 8

Y2 4 4 4 4 4 4 4

The point of intersection appears to be atx = 3.

X=-2

The tables show that when X is 3, Y] is equal to Y2 (both are equal to 4). The solution is x = 3.

Using pencil and paper (not a calculator), complete the tables of values below for these functions. Show all work. Use the tables to check that x = 3 is the solution for 2* ] = 4. X

f(x) = 2"

fif(x)

0 1

!

2

h

o

4

118

Unit 2: Linearand Exponential Relationships

\

=4

Problem Solving Cara and Cami are twins. They came up with a math puzzie. Cara says she is (—2x + 3) years old, and Cami says she is (-^x + 1) years old. What is the value of x? What are their ages?

Since Cara and Cami are twins, you can set their ages equal and solve for x. -2x + 3 - -|x + l Then evaluate one of the expressions (-2x + 3 or - x + 1)to determine their

SOLVE Use graphing to solve for x. Let f(x) Let g(x) Graph each function on the coordinate plane to the right. The point of intersection is ( So, x =

,

. The y-coordinate,

the twins'

). , represents

.

CHECK Substitute that value of x into the original problem to verify that the two ages are the same and that the ages are the ones you found. -2x

3 - -fx

Is this value of x the solution to the equation? t The value of x is

Each girl is

years old.

Lesson 16: Using Functions to Solve Equations 119

Practice Write a system of two functions, f and g, that could be graphed in order to solve the given equation. 1.

7x+ 11 = 8x- 1

2.

fx + 12-

2x-4

Assign each side to a function.

Solve each equation by using the given graph. *r«

X

'

£

o .A

x —-

REMEMBER Look for the point of intersection. 6.

x + 9 = |x + 1

g(x) = T

X -—

120

Unit 2: Linear and Exponential Relationships

7.

3.

3 f - 27

Solve each equation for x by using the given table. 8.

-x

2

9.

25

i.

Complete the tables to solve each equation for x. Show your work.

-2

11.

f(-2)=

-x+5-2x-l

x=

Lesson 16: Using Functions to Solve Equations 121

Define a system of two functions and graph them on the coordinate plane to solve for x. 12. x- 3 = -2x + 6 fix) -

13. g(x) =

fix)

x=

g(x) =

X —

15.

14. 4x+ 5 = 0.5x- 2

fix) -

-x+ 2 = - 3 x - 4

g(x) -

fix)

-I-"

• rr

_q: 4_.j_

rr

X =

X -—

16.

17. f(x) =

g(x) =

[£) -3 = 2 * - 3

fix) -

g(x) -

o-U-M-

122

Unit 2: Linear and Exponential Relationships

Choose the best answer. Use your graphing calculator to help you. 18.

Lucia correctly used a graphing calculator to solve an equation for x. Her screen is shown to the right. The solution was x = 2. Which could be the equation she solved? A.

-~x = 3x — 5

B. C.

19.

~x - 5 - 3x

Adler correctly used a graphing calculator to solve an equation for x. His screen is shown to the right. The solution was x = -1. Which could be the equation he solved? /1\ A. lil - 8 - -1 B. C.

D.

20.

|^| + 8 = 4 1

+8 = 8

Ling decided to sell cupcakes at the county fair. Her ingredients cost her about 25 cents per cupcake. Renting a booth costs $30 per day. She sells each cupcake for $1 . Ling's expenses can be modeled by the function c(x) = 0.25x + 30.00. Her income can be modeled by the function p(x) = 1 .OOx. How many cupcakes must she sell to break even?

50 -10 30 20

10

10

21.

20

30

40

50

plIWUlA' is there a value of x that makes 2* - - 2 true? Rewrite the equation as a system of two functions and graph the system. Use your graph to justify your answer.

Lesson 16: Using Functionsto Solve Equations

123

Graphing Inequalities Graphing an Inequality A linear inequality is similar to a linear equation. The difference is that, instead of an equal sign, an inequality contains one of four inequality symbols: <, >, ^, or >. UNOERSTAN

It is important to note that a linear inequality is not a function. For example, for the linear inequality y > x, both 1 and 5 are possible values for y when x = 0. Since there are multiple outputs {y-values) for one input (x-value), the linear inequality y > xis not a function. You can, however, use the concept of a function to help you solve inequalities. If you replace the inequality symbol in a linear inequality with an equal sign, you get a related equation. y > 3x + 2 is a linear inequality. y = 3x + 2 is its related linear equation. Recall that you can think of y = 3x 4 2 as y = f(x) with f(x) = 3x + 2. Remember also that the solutions to a linear function can be graphed as a line on the coordinate plane. The solution to a linear inequality is a half-plane, the portion of the coordinate plane that lies on one side of a line called the boundary. The boundary is the graph of the related linear equation for the inequality. All of the points in the half-plane are solutions to the inequality. To graph a linear inequality in the coordinate plane, graph its related equation in order to find the boundary line. • If the symbol is < or >, draw a dashed line. Points on the boundary line are not solutions. • If the symbol is -^ or >, draw a solid line. Points on the boundary line are solutions. Then shade a region on one side of the boundary line. Put the inequality in slope-intercept form to determine where to shade. • If the inequality has the form y< mx + b ory < mx + b, shade below the line. • If the inequality has the form y > mx + fa ory > mx + b, shade above the line. You can also find the correct region to shade by choosing a test point and substituting its x- and y-values into the inequality. If the result is a true number sentence, such as 2 > 0, then shade the region that contains the test point. Otherwise, shade the other region. The graph shows the inequality y < ^x + 4. The dashed boundary line means that points on the line are not solutions of the inequality. Any point that lies below the line, in the shaded half-plane, is a solution of the inequality. The point (-3, -1) is a solution because it lies in the half-plane that shows all solutions to the inequality. The point (6, 8) is not a solution because it does not lie in the half-plane.

124

Unit 2: Linear and Exponential Relationships

--'

..

-•

.

.

Connect Graph the inequality y > 2x - 5.

Find the line for the related equation. To write the related equation, replace the inequality symbol ^ with an equal sign. The related equation is y — 2x — 5. The liney = 2x - 5 passes through the points (0,-5)and(l, -3).

Determine whether the line is solid or dashed. The inequality symbol is >. So, the line is solid. Points both on the line and in one half-plane are solutions of the inequality. Determine which half-plane to shade. The inequality is already in slope-intercept form. The inequality symbol is >. So, shade the half-plane above the line. Graph the inequality on a coordinate plane.

-6

The point (-1, 1) is in the half-plane. Substitute these values of x and y into the inequality to confirm that this coordinate pair is a solution.

Lesson 17: Graphing Inequalities

125

Graphing a System I UNDERSTAN The solution to a system of linear inequalities is also a portion of the coordinate plane. It consists of the points that are solutions for every inequality in the system. This is the part of the coordinate plane where all of the shaded regions overlap. In a system of two inequalities, the solution to the system is the intersection of the two half-planes that are solutions to the individual inequalities. All the points that lie in that intersection are solutions for both inequalities. The graph on the upper right shows the solutions to the following system of inequalities:

y> -fx + 4 The point (2, -4) is not a solution to either inequality. The point (-3, 0) is a solution to the first inequality, but not to the second inequality. The point (5, -1) is a solution to the second inequality, but not to the first inequality. The point (1, 4) is a solution to both inequalities. It is a solution to the system. In a system of more than two inequalities, the solution is the intersection of all the half-planes that are solutions to the individual inequalities. The graph on the lower right shows the solutions to the following system of inequalities:



<

• f

y> -4

i

!

-/

6

!

>

*

x< 5

4

, y
Unit 2: Linear and Exponential Relationships

.

/ ; i

f -i

">

6

4 \ 'd

•- 2} .1.



~

•~

i

-6

X

^'

126

-

(l

-e

_

-I

:

2

•+

r ,

'

..

i i

Connect Graph the solution for the following system of inequalities. <

y < 3x — 3

.y< -2* + 1 Graph the first inequality. The related equation, y = 3x - 3, is represented by the line through (0, -3} and (1,0).

Since the inequality symbol is <, use a dashed line and shade below the line. Graph the second inequality on the same coordinate plane. The related equation, y— —^x + 1, is represented by the line through (0,1) and (2,0). Since the inequality symbol is <, use a dashed line and shade below the line.

y

*""£_—_ *?•

*x

-s-t-r*.

1

Identify the solution. The darker region below both lines represents the solution set for the system of inequalities.

Is the point (4, —1) a solution to the system of inequalities y < 3x - 3 and

Lesson 17: Graphing Inequalities

127

EXAMPLE

Graph the solution to the following system of inequalities,

y > 2x + 3 -2y- 2 > -4x

Graph the first inequality. The related equation, y = 2x + 3, is represented by the line through (0, 3) and (-1,1). Since the inequality symbol is >, use a solid line and shade above the line.

Graph the second inequality on the same coordinate plane. Begin by solving the inequality for y. Remember to reverse the inequality sign when dividing both sides by a negative number. -2y- 2> -4x -2y> -4x+ 2 y < 2x - 1 The line for the related equation, y - 2x- 1, passes through (0, ~l)and (1,1). Since the inequality symbol is <, use a solid line and shade below.

Identify the solution. The two lines are parallel, which means that they will never intersect. ^ The two regions have no points in common. Thus, the system of inequalities has no solution.

Can a system of inequalities whose graph consists of parallel boundaries have a solution? If so, draw a graph to support your answer.

128

Unit 2: Linear and Exponential Relationships

Problem Solving A jewelry maker is creating a line of bracelets and necklaces with a new type of chain. The bracelets are 8 inches long, and the necklaces are 14 inches long. She has 280 inches of chain. It takes her 4 hours to make a bracelet and 3 hours to make a necklace. She can work no more than 120 hours this month. Write a system of inequalities to model the number of bracelets and the number of necklaces that the jewelry maker can create this month. Then determine how many necklaces and bracelets she can produce.

Write a system of inequalities to describe the situation. Let x be the number of bracelets and y be the number of necklaces she can make. Since these are numbers of real objects, they cannot be negative numbers. So, x ^

and y ^

It takes 8 inches of chain to make a bracelet and 14 inches to make a necklace. The total amount of chain used must be less than or equal to the total amount available, 280 inches. So,

280.

.x +

It takes 4 hours to make a bracelet and 3 hours to make a necklace. The total amount of time spent making the jewelry this month must be no more than 120 hours. So, 4x+ 3y

120.

SOLVE The boundary lines for this system of inequalities are graphed on the coordinate plane to the right. Shade the region that represents the solution. CHECK The point (15,

) lies within the solution region. Show that

it satisfies all 4 inequalities.

8(15) + 14( 4(15} + 3(

280

280 120

Lesson 17; Graphing Inequalities

129

Practice Determine whether each point is a solution to the inequality graphed below.

4

; : ;

1.

(-3,-!)

2.

. .. ^ ? : ;.

3.

(2, 0)

(6,4)

Points on a dashed boundary line are not included in a solution set, Points on a solid boundary line are included in a solution set

Determine whether each point is a solution to the system of inequalities graphed below.

m* 4.

5,0)

6.

5.

(1, -4)

7.

(0,4}

REMEMBER A solution to a system of inequalities must be a solution for each inequality in the system.

130

Unit 2: Linear and Exponential Relationships

Use the graph below to answer questions 8 and 9. Choose the best answer.

,

8.

Which point is not part of the solution set for this inequality?

A.

(0,3)

B.

(3, 3)

C.

(4,0)

D.

(-4,6)

9.

Which inequality is represented by the graph? A. y > 4 x + 3 B.

y<--U + 3

C.

y

D.

v

Use the graph below to answer questions 10 and 11. Choose the best answer.

10.

Which points are included in the solution set for this system of inequalities? A.

Monly

B.

MandN

C.

M, N, and P

D.

M and Q

11.

Which system of inequalities is represented by this graph? A.

fy>2x + 2

C.

[y<-2x + 2 B.

y>2x+ 2

fy>2x + 2 jy>-2x + 2

D.

y> -2x+ 2

Lesson 17: Graphing Inequalities

131

Use the graph below to answer questions 12-14.

-4

-X

U

•+-X

*

12.

Name a point that is part of the solution set.

13.

Name a point that is not part of the solution set.

14. Write the inequality represented by the graph.

Use the graph below to answer questions 15-17.

w

15.

Name a point that is part of the solution set.

16.

Name a point that is not part of the solution set.

17.

Write the system of inequalities represented by the graph.

132

Unit 2: Linear and Exponential Relationships

Graph each inequality. 19.

6x-2y<

18.

y<|x-l

20.

»|f['WF A farmer will plant corn and soy on his farm this year. He has a total of 25 acres available for planting. Each acreof corn costs $350 to plant, and each acre of soy costs $150 to plant. His costs must be no more than $5,250. Let x be the number of acres of corn to be planted and lety be the numberof acresof soy to be planted. Write a system of four inequalities to describe the situation. Then graph the system.

y

Crops Planted

411

35

_ . . _.

•5T 30 P » 0 25 CO

_c 20 0

I0

m

10 5 5

10 15 20 25 30 35 40 Corn (in acres)

Lesson 17: Graphing Inequalities

133

Translating Functions UNDERSTAN[p You can think of functions as being grouped into families. All functions in a family have similar characteristics. For example, the graphs of all functions in the family of linear functions are straight lines.

Each family of functions has a parent function, the most basic function in the family. The family of linear functions has the parent function f(x) = x. The function f(x) = ex is the general parent function for all exponential functions. However, it can often be easier to group the exponential functions into smaller subfamilies that have the same base, such as f(x) = 2* andf(x} = 23.5*. If you change the parent function by adding, subtracting, multiplying, or dividing by a constant, you transform the function and make a new function from the same family. For example, the function g(x) = x - 3 is different from the parent function f(x) = x, but it is still in the linear function family. Changing the equation of the function also changes the graph of the function. This change to the graph is called a transformation.

Adding to or subtracting from a function moves its graph up, down, left, or right on the coordinate plane. This kind of transformation is called a translation. Translation In a vertical translation, every point on the graph shifts up or down.

Algebraic Notation g(x) = f(x) + k

If k>0, shift the graph \k\s up.

A real number, k, is added to the output, f(x).

\fk< 0, shift thegraph \k\s down.

In a horizontal translation, g(x) = f(x + k) every point on the graph shifts left or right. | A real number, k, is added to the input, x.

134

Change to Graph

Unit 2: Linear and Exponential Relationships

If k < 0, shift the graph \k\s right. If k > 0, shift the graph \k\s left.

Connect The exponential function f(x) = 3X is graphed on the coordinate plane below. Make a table of values for the function g(x) = 3* + 2. Then graph function g on the same coordinate plane. Describe how function f could be translated to form function g and how translating a function affects its size and shape.

f(K) = 3"

Create a table of values for g(x) = 3* + 2. x

I

g(x) - 3" + 2

-2 g(-2) = 3-2 + 2 = ^ 4 -1

g(-1) = 3"1 + 2 = ^ +

19 9

19 9 7

0 1 g(0) = 3°+ 2 = 1 + 2 = 3

3

1 ;gO) = 3 ] + 2 = 3 + 2 = 5

5

2 ig(2}-3 2 + 2 = 9 + 2 = ll

11

Plot the ordered pairs for function g and connect them with a curve.

Compare the graphs. Each point on the graph of function g is 2 units above its corresponding point on function f. So, function g is the result of a vertical translation of function f 2 units up. Since all we are doing is sliding the graph in the coordinate plane, the size and shape of the graph have not changed.

Since you were given the graph of f(x) = 3X, could you have graphed g(x) = 3X + 2 without creating a table of values first? Explain.

Lesson 18: Translating Functions

135

Let f(x) = 2x and define a function g such that g(x) ~ f(x + 3). Graph both functions, f and g, on the same coordinate plane. Compare the two graphs and identify how function f could be translated to form function g.

Write function g in terms of x by using functional notation.

, Graph function f.

For the function g, use the expression for f(x) and replace x with (x 4- 3). g(x) = f(x + 3) g(x) = 2{x + 3) g(x) - 2x + 6

The graph of f is the graph of the equation y = f(x) or y = 2x. This equation has a slope of 2 and a y-intercept at {0, 0). Graph the y-intercept, use the slope to find another point, and draw a straight line through those points.

Graph function g. The graph of g is the graph of the equation y = g(x) or y = 2x + 6. This equation has a slope of 2 and a y-intercept at (0, 6). Graph the y-intercept, use the slope to find another point, and draw a straight line through those points.

Compare the graphs. ^ Each point on function g is 3 units to the left of the corresponding point on function f. This makes sense. When 3 is added to the input, as it was in g(x) - f(x + 3), the result is a translation of 3 units to the left. Notice that both lines have the same slope, 2. So, translating a line horizontally does not change its slope.

TRK

On the grid shown in Step 3 above, graph h(x) = f(x — 3). Describe how function f could be translated to form function h.

136

J M a » A linear function f is graphed below. On the same coordinate plane, graph the function g(x) = f(x) - 5. Identify the transformation.

Describe the translation. g(x) = f(x) - 5 is in the form g(x) = f(x) + k, where k = -5. When a numerical value, k, is added to an output, f{x), the result is a vertical shift. Since k is the negative number -5, shift the graph 5 units down.

Graph the function g. Shift two points on the graph of function fdown 5 units. Then draw a line through them. (0,1) is translated 5 units down to (0, —4). (2, 0) is translated 5 units down to (2, -5).

*

The transformation is a vertical translation 5 units down.

The equations for functions f and g are not given. Can you determine if their slopes are the same? Explain how.

Lesson 18: Translating Functions

137

EXAMPLE

The graph of f(x) = \j\s shown. Translate function f

to form the function n(x) = f(x - 4) + 2. Graph h and write its explicit equation.

Describe the translation by using words and symbols. Subtracting 4 from the input, x, indicates a horizontal translation. Translate the graph 4 units to the right. Adding 2 to the output, f(x ~ 4), indicates a vertical translation. Translate the graph 2 units up.

Graph h. Choose several points on the graph of f and translate each point 4 units to the right and 2 units up.

Connect the points with a smooth curve.

Write an equation for h. The translation involved subtracting 4 from the input and adding 2 to that output. So, Use a graphing calculator to check your work. Press For Y,, enter (1/2)AX. For Y2, enter (l/2) A (X-4) + 2. Press ESI SSB to bring up a table of values. Press 25322 to view the graph.

138

Unit 2: Linear and Exponential Relationships

subtract 4 from x, which is the exponent in 2 j , and then add 2 to the resulting expression. 2.

Functions f and g are graphed on the right. Using function notation, write an equation describing g(x) in terms of f(x). Then use the equation given for f(x) to write an equation for g(x) in terms of x.

Identify how f could be translated to form g. Choose a point on f, such as (0, 0). If this point is translated 2 units to the right, it would cover point (2, 0), which is on the graph of g.

Verify that any point on f, if translated 2 units to the right, has a corresponding point on g.

Write the function g(x) in terms of f(x). To represent a horizontal translation of 2 units to the right, add —2 to the input, x. Sog(x) = f(x-2). .

Write the function g(x) in terms of x. To find an explicit expression for g(x), substitute {x - 2) for x in the expression forf(x). g(x) = f(x - 2) g(x) = 3(x - 2) ^ g(x) = 3x - 6 Use a graphing calculator to check that g(x) = 3x — 6 is the correct equation for function g.

Lesson 18: Translating Functions

139

Practice Use words to describe how function f could be translated to form function g in one step. 1.

y

2.

y

How can (0,1) be translated to cover (0, 2)?

Use words to describe a horizontal translation that would transform function f into function g. Then describe a vertical translation that would transform function f into function g. 4.

3.

horizontal translation:

horizontal translation:

vertical translation: _

vertical translation: _

REMEMBER Horizontal means left and right. Vertical means up and down. Write true or foist? for each statement. If false, rewrite the statement to make it true. 5.

A translation is a slide of a graph to a new location on the coordinate plane.

6.

If g(x) = f(x) - k, then the graph off is translated k units down to form the graph of g.

7.

lfg(x) = f(x - k), then the graph off is translated k units left to form the graph of g.

140

Unit 2: Linear and Exponential Relationships

Translate the graph of f according to the verbal description to form g and draw the graph for g on the same coordinate plane. Then write an equation for g(x) in terms of x. 8.

Translate the graph off 5 units up to form g.

9.

Translate the graph of f3 units to the right to form g.

f(x) = 4

g(x) =

Choose the best answer. Use your graphing calculator to check your answer. 10.

The graphing calculator screen below shows the graph of f{x) - 1.5x and the graph of g.

11.

The graphing calculator screen below shows the graph of f(x) » 2 x and the graph of g.

Which equation could represent g(x) in terms of f(x)?

Which equation could represent g(x) in terms of f(x)?

A.

g(x) = f(x) + 6

A. fif(x) = f(x + 3) + 3

B.

g(x) = f ( x ) - 6

B.

g(x) = f(x-3) + 3

C. g(x) = f(x+6)

C.

g(x) = f(x + 3) - 3

D.

D.

g(x) = f(x - 3) - 3

g(x) = 6f(x)

Lesson 18: Translating Functions

141

Write an explicit expression in terms of xfor each function g{x) described below. For questions 12-17, f(x) = 5X. 12.

translation of f(x) 2 units up

13.

g(x) =

g(x)

14.

translation of f(x) 2 units left

15.

translation of f(x) 3 units right and 3 units up g(x) - _

translation of f(x) 2 units right

g(x) =

g(x)

16.

translation of f(x) 2 units down

_

17.

translation of f(x) 3 units left and 3 units down

g(x) =

__

Translate the graph of function fto form the translated function g described algebraically. Write an equation in terms of xto represent the translated image.

19.

18.

!i )

20. g(x) - f(x - 6) - 4

g(x) =

142

Unit 2: Linear and Exponential Relationships

21. g ( x ) - f ( x - 4 ) - 3

g(x)

Function f was translated to form function g according to the rule given. For each rule, identify the value of k. Include the sign. Briefly explain how you know.

22.

23.

vX

g(x) = f (x + k); k =

Examine the following situations and respond in complete sentences. 24. ^ * Macy graphed f(x) = -4x and parallel line g. She believes that since the rule for the translation is g(x) = f(x 4- 1), the equation for g must beg(x) = —4x + 1. Explain why Macy's reasoning is flawed. Then identify the correct equation for g.

25.

Zack used to charge only an hourly rate to mow lawns, as shown by the graph of function p. Because of rising costs, he now charges a set fee for each job in addition to his hourly rate, as shown by the graph of function n. Use algebraic notation to describe how p could be translated to form n. Use what you know about translations to explain how the new costs differ from the old costs. DESCRIBE.

Charges for Lawnmowing

2

4

6

;

Length of Job (in hours)

Lesson 18: Translating Functions

143

Reflecting Functions UNDERSTAN A reflection is a transformation that can flip the graph of a function over a line. That line is called the line of reflection. The new graph looks like a mirror image of the original graph. The image after a reflection is the same size and shape as the original graph. if gM = f(—x), the graph of g is the reflection of the graph of f across the y-axis. Changing the sign of the input, x, reflects the graph over the y-axis. This means that the y-axis is the line of reflection in this transformation.

*- X

Compare the two lines. The point (4, 6) is found on the graph off. The corresponding point (-4, 6) is found on the graph of g. The point (-4, 6) is the reflection of (4, 6) across the y-axis. If h(x) = -f(x), the graph of h is the reflection of the graph off across the x-axis. Changing the sign of the output, f(x), reflects the graph over the x-axis. This means that the x-axis is the line of reflection in this transformation.

Compare the two lines. The point (4, 6) is found on the graph of f. The corresponding point (4, -6) is found on the graph of h. The point (4, -6) is the reflection of (4, 6) across the x-axis.

144

Unit 2: Linear and Exponential Relationships

Connect The graph of the exponential function f(x) = 3X is shown. On the same coordinate plane, graph function g(x) = —f(x). Compare the graphs and describe the reflection.

1 Write the equation for g. We know that f(x) = 3* and that the reflected image is equal to g(x) = ~f(x). Substitute the expression for f(x) to find an expression for g(x).


Create a table of values for function g. g(x) = -(3")

X

9(x)

! g(-2) =

-(3 2) = - - 2 _ _l 9

Graph g(x) and compare it to f{x)

~1

g(-D

= -0 ') - "I1) ' 1 3

.

Consider (1, 3) and {1, - 3). Each has the same input, x, but the outputs have opposite signs. Also, notice that {1, 3} and {1, -3) are the same distance from the x-axis but lie on different sides of it.

1 9

~3 j

0 flf(O) = -(3°)= -(D = -1

-1

i

g(D - -(3)=

-3

2

g<2)=-(32)--(9)

-(3) = -3

-9

Use a graphing calculator to check your work. Press

. EnterY, = 3 A X.

Enter Y 2 = -3 A X.

This is true for every pair of corresponding points on the two functions.

Press EJ§ MsUdil to bring up a table of values, and compare them.

^ The graph of g is the result of a reflection of the graph of f across the x-axis.

Press ESE53- Compare the graphs to the ones shown on the left.

Lesson 19: Reflecting Functions

145

EXAMPLE A A linear function fis graphed. On the same coordinate plane, graph function g such that g(x) = f(-x).

Compare the graphs and describe the reflection.

Determine how to graph g. No explicit function was given for f, but the graph of g can be found by using the graph off. The algebraic notation g(x) = f(-x) means that the opposite value of each input, x, will be used. So, find points on the graph of f, change the sign of eachx-value, and graph the resulting points.

Find several points on the graph of g.

ICO'

(-2, 3) is a point on f. Find the opposite of the input (x-value) and keep the same output (y-value). (-2,3)-* (2, 3) So, (2, 3) is a point on g. Graph g and describe the reflection. Plot the points that you found and draw a straight line through them to graph g.

Each point on f and its corresponding point on g are the same distance from the y-axis, but they lie on different sides of it. ^ The graph of g is the result of a reflection of facross the y-axis.

146

Unit 2: Linear and Exponential Relationships

(2,1) is a point on f. So, (-2,1} is a point on g. (6, -1) is a point on f. So, (—6, —1) is a point on g.

Identify key features, such as the intercepts and slopes, of functions f and g. How are the key features of fand its reflected image similar? How are they different?

Exponential functions f and g are graphed. Use function notation to define g(x) in terms of f(x). If the equation for function f is f(x) = 2* - 2, write an equation for g. *- X

Describe the transformation. The graphs are mirror images of one another. So, this is a reflection.

Points (-2, 2) and (2, 2) lie on the same horizontal line, but they are on different sides of they-axis. Each is the same distance from the y-axis. Other pairs of points from f and g share this characteristic. So, f can be reflected across the y-axis to form g.

Use function notation to describe the reflection. For each pair of corresponding points, the inputs, x, are opposites and the outputs are the same. ^ Each point (x, y) on f has a corresponding point (-x, y) on g, so this reflection can be described as g{x) = f{—x).

Write an explicit equation for g(x). To write g(x), write the expression for f(x) and replace x with -x. If f(x) = 2"-2, then g(x) = 2

> g(x) - 2- * - 2

x

~ 2.

TRX

Use what you know about negative exponents to express g(x) = 2 - 2 in a different way. (Hint: The number raised to the exponent x will be less than 1.)

Lesson 19: Reflecting Functions

147

Practice Determine if each pair of functions f and g are reflections of one another across the x-axis, reflections of one another across the y-axis, or neither.

2.

1.

x

3.

Find pairs of corresponding points.

^J^ Which changed: the input or the output?

Choose the best answer. Use your graphing calculator to check your answer. 4.

148

The graphing calculator screen below shows f(x) - 2* and its reflection g. Which could represent g(x)?

5.

The graphing calculator screen below shows f(x) = - 3x - 4 and its reflection g. Which is not true of the functions?

A.

g(x) = 2*

A.

g(x) = f(-x)

B.

g(x) = 2- x

B.

C.

g(x) = -{2*}

Function f was reflected across the y-axis to form g.

D.

(x)--(2- x

C.

Both f and g have the same y-intercept.

D.

Both fand g have the same slope.

Unit 2: Linearand Exponential Relationships

Graph g. Then write an equation for g(x) in terms of x. 6.

g(x) = f(-x)

7.

g(x) = -f(x)

g(x) Answer the following questions. 8.

C ^ Maggie graphed functions f and g as shown. Maggie says that g is the result of a reflection of f across the x-axis. Jay says that g is actually the result of a reflection of facross the y-axis. Who is correct? Justify your answer. *- X

9.

WjMrf^ A graph can be reflected across a point, such as the origin. If j(x) = — f(—x), then j is the reflection of facross the origin. How could you use two different reflections to produce the same graph of7?

Lesson 19: Reflecting Functions

149

Stretching and Shrinking Functions Vertical Stretches and Shrinks Translations and reflections do not change the size of the graph being transformed, but stretches and shrinks do. After a stretch or a shrink, the new graph looks wider or narrower than the original graph. UNDERSTAN

Multiplying the output of a function, f(x), by a constant stretches or shrinks the graph in the vertical direction. Forg(x) = kf(x) where |k| > 1, the graph of g is a vertical stretch of the graph off. A vertical stretch pulls the points on the graph away from thex-axis.

The graphs off and g above illustrate the vertical stretch g(x) = 2f{x). Notice how each point on g is twice as far from the x-axis as its corresponding point on f. Forg(x) = kf(x) where 0 < |k| < 1, the graph of g is a vertical shrink of the graph off. A vertical shrink pushes the points of a graph toward thex-axis.

The graphs of f and g above illustrate the vertical shrink g(x) ~ ^f(x). Notice how each point on g is half as far from the x-axis as its corresponding point on f. Notice that, after a vertical stretch or shrink, the new function always has the same x-intercept as the original function. This point cannot be shrunk toward or stretched away from the x-axis because it is on the x-axis.

150

Unit 2: Linear and Exponential Relationships

Connect The exponential function f(x) - 2* is graphed on the right. Graph g(x) - 4{2X) on the same coordinate plane. Is g the result of a vertical stretch or a vertical shrink of f ? by what factor?

Create a table of values for g. X

-2

-1 g(- l) = 4(2~ 1 )-4(2) = :I

2

0

9(0} = 4(2°) = 4(1) = 4

4

1

9(1) - 4(2') = 4(2) = 8

8

2

9(2) = 4(22) = 4(4) = 16

16

Graph g. Plot those ordered pairs (x, g(x)) on the coordinate plane and connect them with a smooth curve.

Compare the graphs and identify the transformation. The point (0,1) on the graph of f is transformed to (0, 4) on the graph of g. The point (1, 2) is transformed to (1, 8). The point (2, 4) is transformed to (2,16). Notice that for each pair of points, the input is the same, but the output is 4 times as great. The equation of g also shows that the output, f(x), was multiplied by 4. ^ The graph of g is the result of a vertical stretch of f by a factor of 4.

On the coordinate plane above, graph h(x) = \(2X}. Is this a vertical stretch or a vertical shrink? by what factor?

Lesson 20: Stretching and Shrinking Functions

151

(Horizontal Stretches and Shrinks jUNDERSTAND Multiplying the input of a function, x, by a constant stretches or shrinks the graph in the horizontal direction. Forg(x) - f(kx) where \k\ 1, the graph of g is a horizontal shrink of the graph off. A horizontal shrink pushes the points of a graph toward the y-axis.

The graphs of f and g above illustrate the horizontal shrink g(x) = f (2x). Notice how each point on g is half as far from the y-axis as its corresponding point on f. For g(x) = f(kx) where 0 < |k| < 1, the graph of g is a horizontal stretch of the graph off. A horizontal stretch pulls the points of a graph away from the y-axis.

The graphs of fand g above illustrate the horizontal stretch g(x) ~ f k-x . Notice how each point on g is twice as far from the y-axis as its corresponding point on f. Notice that the factor of the stretch or shrink is the reciprocal of the constant. In the case of a horizontal stretch or shrink, k is not the factor by which a graph is stretched or shrunk; ^ is. Notice that, after a horizontal stretch or shrink, the new function always has the same y-intercept as the original function. This point cannot be shrunk toward or stretched away from the y-axis because it is on the y-axis.

152

Unit 2: Linear and Exponential Relationships

—*: Connect Graph the linear function f(x) = ^x + 2. Then graph its image, g, resulting from the transformation described below.

g(x) = f(3x) Is this an example of a horizontal stretch or shrink? by what factor?

Create a table of values for f. x+2

X

-6

f(~6) = 3j(-6) + 2

-3 + 2 ^

0 f(0) = \) 0) +=2\) = 0= +l 2=2 6

6) + 2 = 3 + 2 = 5

f(x) -1 2

Write an equation and create a table of values for g.

5

Since g(x) = f(3x), replace x with 3x to find the equation forg.

g(x) - f{3x) =5(3;

2=

2.

ofjrt-tr+a y\*/ 2

X

g(-2) = |(-2)

Use those points to graph the functions.

0M

+2

= -3 + 2 = -1 0

_,//~i^ — ^((J) ir\\ "> —A 0 g((J) -tZ — (ji n + 2i — 2. JL

2

!

2 \) = f (2) + 2 = 3 + 2 = 5

5

4 Identify the transformation and the factor. The points on the graph of g are closer to the y-axis than their corresponding points on the graph of f. A horizontal shrink pushes a graph toward the y-axis. Enter the equations for f and g into your graphing calculator, graph them, and call up tables of values for them. Do the graphs and tables match what appears on this page?

For g(x) = f(3x), k = 3. So, the factor . 1 1 Isk'or3-

^ Thegraph ofgisthe result of a horizontal shrink of fby a factor of •-•

Lesson 20: Stretching and Shrinking Functions

153

EXAMPLE

I Graph the linear functions f(x) - -4x - 3andg(x) = -2x- 3.

Use words to describe the transformation, including if it is a horizontal stretch or a horizontal shrink, and by what factor. Then write an equation for g(x) in terms of f(x).

Use what you know about slope-intercept form to graph each line. Inf(x) = -4x - 3, the slope is -4 and they-intercept is (0, -3). So, plot the point (0, -3). Then count 4 units down and 1 unit to the right and plot a second point, {1, - 7). Draw and label the line for f. lng(x) = -2x - 3, the slope is -2 and they-intercept is (0, -3). So, plot the point (0, —3). Then count 2 units down and 1 unit to the right and plot a second point, (1, -5). Draw and label the line for g.

Identify the transformation. The points on the graph of g are farther from the y-axis than their corresponding points on the graph of f. A horizontal stretch pulls a graph away from the y-axis.

«•

Describe the transformation in words and as a function.

Compare the point (-1,1) onfto its corresponding point (-2,1} on g. The point on g is twice as far from the y-axis as the point on f. The transformation was a horizontal stretch by a factor of 2. Since this transformation is horizontal, the stretch factor is equal to p and, in function notation, k is multiplied by the input, x.

k — ~ K-

g(x) = f Does a stretch or shrink of the graph of a line always change its slope? Explain.

154

Unit 2: Linear and Exponential Relationships

The graph of function f was transformed to create EXAMPLE the graph of function g, as shown. Was f stretched or shrunk, horizontally or vertically, and by what factor? If f(x) - 4X + 8, what is the equation of g(x)?

f

9

Examine the key features to determine how f was transformed. As you move left on the graph of f, the values of f(x) get very close to 8. So, f appears to have an asymptote at y = 8. Many points on the graph of g lie below that asymptote, so points from the graph of f must have been moved much closer to the x-axis. Thus, g appears to be a vertical shrink of f.

Identify the factor of the shrink. In a vertical shrink, the output is multiplied by a factor, k. So, compare points. A vertical shrink has the form g(x) = kf (x). So, k = Tr-rfor allx. /w The point (2, 24) on the graph of f was shrunk to (2, 3) on the graph of g.

9(2)

JL 24

Find the equation for g(x).

-:_j Suppose f(x) = 4" + 8 is shrunk horizontally by a factor of -~ to form function h. Will the equation of h be h(x) - 4^x + 8 or h(x) = 48* + 8? Explain.

Lesson 20: Stretching and Shrinking Functions

155

Practice Classify the graph of g as either a vertical stretch or a vertical shrink of the graph of f.

2.

1.

A vertical shrink draws points closer to the x-axis.

Classify the graph of g as either a horizontal stretch or a horizontal shrink of the graph of f. 4.

3.

*- x

o

; ^ *v Is g closer to or farther away from the y-axis?

Fill in each blank with an appropriate word, phrase, or expression. 5.

A horizontal

pushes the points of a graph toward the y-axis.

6.

A horizontal

pulls the points of a graph away from the y-axis.

7.

If g(x) = kf(x) and |k| > 1, then g is the result of a vertical stretch of g by a factor of _

8.

Ifg(x) = f{kx)and \k\ then g is the result of a horizontal shrink of g by a factor of

156

Unit 2: Linearand Exponential Relationships

Graph each function g on the coordinate plane below it. Classify each graph of g as either a vertical stretch or a vertical shrink of the graph off. Then identify the factor.

9.

g{x) = 2x - 6

10.

transformation: vertical

transformation: vertical

factor:

factor:

.

Graph each function g on the coordinate plane below it. Classify each graph of g as either a horizontal stretch or a horizontal shrink of the graph of f. Then identify the factor. 11.

g(x) = -5x

12.

transformation: horizontal

transformation: horizontal

factor:

factor:

Lesson 20: Stretching and Shrinking Functions

157

Match each verbal description of a translation of f (x) = 6X with the equation of its transformed image by writing the correct letter next to each description. 13. vertical stretch by a factor of 8

A.

14.

vertical shrink by a factor of Q

B.

15.

horizontal stretch by a factor of 8

C.

= 8(6J()

16. horizontal shrink by•* a factor of «•o Graph each function g on the coordinate plane below it. Classify each graph of g as a stretch or a shrink of the graph of f, horizontal or vertical, and by what factor. 17.

g(xHff(x)

18.

f(x = 4* +

•f-X

transformation:

transformation:

factor:

factor:

19.

158

20. g(x) = f(2x)

transformation:

transformation:

factor:

factor:

Unit 2: Linear and Exponential Relationships

Graph the functions described and provide the desired information about them. 21.

Graph f(x) - 2x - 2 and its image, g, after a horizontal stretch by a factor of 4. Write equations for g(x) in terms of f(x) and in terms of x.

g(x)

x -

-*-x

22.

DISTINGUISH The graph of f(x) = 3 x is shown. Graph g(x) = -^f(x). Then describe the two transformations needed to create the image, g, from the graph off.

•V

-

Lesson 20: Stretching and Shrinking Functions

159

Functions in Context People often use functions to model real-world relationships. These functions tell how one quantity changes in relation to another quantity. Representations of these functions, such as tables, graphs, or equations, can be used to answer questions about the relationships or to make predictions. EXAMPLE A The EZ Car Rental Company charges a set fee plus a daily rate to rent a car. It costs $90 to rent an economy car for 1 day and $170 to rent the same car for 3 days. Write a function to model the cost of renting an economy car for x days.

Decide if this relationship is linear or exponential. The relationship can be represented as: (total charge) = {set fee) + (cost per day) X (number of days, x) The variable, x, is multiplied by a constant rate, the cost per day. The set fee is then added to that product. Since the rate of change is constant, the function is linear.

Find the cost per day.

Find the set fee and write the function.

The cost per day is the slope of the linear function. Find two points and use the slope formula to find the cost per day. A 1-day rental costs $90. So, (1, 90) is a solution for this function. A 3-day rental costs $170. So, (3,170) is also a solution. cost per day

170-90 3-1

f

= 40

Substitute 1 day and 3 days into your model function to check that it outputs $90 and $170, respectively.

160

Unit 2: Linear and Exponential Relationships

We can use an ordered pair, such as (1, 90), to determine the complete equation. f(x) = (set fee) + 40x = (setfee) 90 = (set fee)

40(1) 40

50 - (set fee) The cost, f(x), in dollars of renting an economy car for x days can be modeled by the equation f (x) = 50 + 40x.

EXAMPLE B Mr. Vega bought a new car for $20,000. He used a function to estimate how its value will depreciate, or decrease over time.

Age in Years, t

0

1

2

3

4

Value in Thousands of Dollars, v(t)

20

16

12.8

10.24

8.192

What type of function did Mr. Vega use to model this relationship? Describe the rate at which the car depreciates. Write an equation for the function.

Test to see if the function is linear. Examine the decrease in value from year to year. 1 year after purchase: v(0) - v(l) = 20 - 16 = 4, or $4,000 Between years 1 and 2: v(l) - v(2) = 16 - 12.8 = 3.2, or $3,200 Between years 2 and 3: v(2) - v(3) = 12.8 - 10.24 - 2.56, or $2,560 The rate of depreciation is not constant, so this model is not a linear function.

3 Test to see if the function is exponential. Compare the value of the car in successive years. v(\) 1 year after purchase: -rrr = 20 90 = Between years 1 and 2:

v(2) v(D v(3)

12.8 -0.8 16 10.24

Between years 2 and 3: v(2) 12.8 -0.8 Each year, the value of the car is 0.8, or 80%, of its value the previous year. This means the car's value is decreasing at a constant percent rate. Functions that decrease at a constant percent rate are exponential functions.

Determine an equation for the function. The constant percent rate is 80%, or 0.8. This is the factor by which the previous year is multiplied to get the next year. v(l) = v(0)-0.8 v(2) - v(l) - 0.8 - MO) - 0.8) • 0.8 = v(0)-(0.8)2 v(3) - v(2) • 0.8 - (v(0) • (0.8)2) - 0.8 You can see a pattern forming. For any year x after purchase, the car's value is given by v(x) = v(0)-(0.8) x . Substitute v(0), the initial value, 20.

Make a graph to illustrate the depreciation of the car over time.

Lesson 21: Functions in Context

161

Leah opened a checking account and bought a certificate of deposit (CD) on the same day. She deposits money into her checking account each month. The amount in her checking account, C, can be modeled by C(t) = 50(12)t, or C(t) - GOOr., where t is the time in years since the account was opened. EXAMPLE

Leah's CD has a set annual interest rate, and the interest is compounded monthly. The amount in her / 0 03 \' CD can be modeled by the function A(t) = 800(1 + -jy- ] , where t is the time in years since she deposited the money. Interpret the parameters of functions C and A in this situation.

2 Interpret the parameters of C(t).

Interpret the parameters of A(t).

The problem explains that t represents

The equation for A, an exponential growth

time in years, and there are 12 months in

function, shows an amount earning

T,

...

12 months ., ,

1 year. The quantity —yea?— x f years

compound interest. The formula for

gives a number of months.

calculating this amount, A, is

Thus, the 50 likely represents the amount deposited each month. $50 .., 12months x t years = the money in X year jDQOfttrr the checking account at time t

A = P(l + T))", where Pis the principal, r is the annual interest rate, and n is the number of times the interest is compounded per year. So, inA(t)

,

0.03 12

, 800 is P, the

principal; 0.03 is r, the interest rate; and 12 isn, the number of times the interest is compounded per year. This means that Leah deposited $800 in a CD at 3% annual interest, compounded monthly.

Combine functions Cand A above to build a function that shows the total amount of EXAMPLE money Leah has in both accounts at any time, t.

Combine both functions by adding them to form a new function, L C(t)+A(t) 0.03 Y2t

50(12)t+ 800 1 + 12 -600M- 800(1.0025)

I2t

The total amount in both accounts after f months is: L(t) =600t + 800(1.0025)'2'.

162

Unit 2: Linear and Exponential Relationships

Leah always keeps $200 hidden at home, which she calls her "emergency fund." Write a function for the total amount of money Leah has in her checking account, CD, and emergency fund at any time t.

Problem Solving Abdul buys a bus card with a value of $30. Each time he takes a bus ride, $1.50 is deducted from his card. Write a function that can be used to model this situation. Then use the function to determine the value of the bus card after Abdul takes 4 rides.

Since the value decreases at a constant rate per ride, mode! this with a(n] function. Write an equation and then use the equation to solve the problem.

SOLVE The initial value of his card is $30. So, when he has taken 0 rides, there is $30 left on his card. The point (0,

) is an ordered pair for this function. It is also the

-intercept.

Each time Abdul takes a bus ride, $1.50 is deducted from his card. So, the rate of change, or slope, is

Substituting

This rate should be negative because

form, the slope, and

for fa, the y-intercept, gives the equation:

c(x) = If Abdul takes 4 rides, the value of his card, in dollars, will be: c(4) - (

CHECK The initial value of the card is $30: c(0) = 30. The value decreases by $1.50 each time he rides the bus. After 1 ride, the value will be: c(l)-30.00-1.50- 28.50 After 2 rides, the value will be: c(2) -28.50 - 1.50 = After 3 rides, the value will be: c(3) = After 4 rides, the value will be: c(4) Is this the same value you found for c(4) when you used the equation? ^ This situation can be modeled by the equation c(x) — After taking 4 rides, the value of Abdul's card will be$-

Lesson 21; Functions in Context

163

Practice For each situation, identify the type of function (linear or exponential) that could model it. Then write a function to model the relationship. 1.

The highest possible grade for a report is 100 points. Each day the report is late, the teacher deducts 10 points.

Days Late, x

0

1

2

3

4

Starting Grade, g(x)

100

90

80

70

60

function type:

2.

Sixteen teams are participating in a tournament. Only the winning teams in each round advance to the next round.

Number of Rounds Completed, x

0

Teams Remaining, f (x)

16

g(x) =

function type:

REMEMBER If the rate of change is constant, the function is linear.

3.

As soon as a Web site went up, it received 1 hit After one minute, it had received 4 hits. The number of hits continued to quadruple each minute after that.

4.

A cake decorator charges a $30 set fee for each cake plus $20 for each color of icing required.

Colors of Icing Time Since Launch, i n minutes, x Number of Hits, h(x)

0

1

2

3

4

,

,,

*

\W

dollars, c(x) ,

.

,fi

64

n

->

/ W

h?v

o

c

Required, x CostofCakein 1 •V

1w V

256 function typp-

c(x) = ^

funrti<">n typ*3:

MY) =

Choose the best answer. 5.

164

A salesperson earns a weekly salary plus a commission on each appliance he sells. The function p(x) = 200 + O.OSx shows his weekly earnings if x represents his weekly sales, in dollars. Which is also true?

6.

The equation A(t) = 900(0.85)' represents the value of a motor scooter t years after it was purchased. Which statement is also true of this situation? A.

When new, the scooter cost $765.

A.

His weekly salary is $205.

B.

When new, the scooter cost $900.

B.

He earns $200 for each appliance he sells.

C.

The scooter's value is decreasing at a rate of 85% each year.

C.

He earns $0.05 for each appliance he sells.

D.

The scooter's value is decreasing at a rate of 0.15% each year.

D.

He earns a 5% commission for each appliance he sells.

Unit 2: Linear and Exponential Relationships

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Arithmetic Sequences UNDERSTAND A sequence is an arrangement of numbers or objects that follows a rule or a pattern. Take a look at the Fibonacci sequence: 1,1,2,3,5,8,13,21,34,... Each number in the sequence is called a term. The pattern of the Fibonacci sequence is that each term is equal to the sum of the two previous terms. This rule is an example of using a recursive process, because finding a term depends on knowing previous terms. When defining a recursive process, you must define at least one previous term. The variable a is often used to stand for the terms in a sequence. The first term is a,, the second is a2, and so on. The nth term of the sequence is written an, where n can be any positive integer. This notation is useful for writing the sequence's rule mathematically. The equations below represent the rule for the Fibonacci sequence. = 1

CL = 1

cr, = an - 1

an

- 2

UNOERSTAN In an arithmetic sequence, each term is found by adding a fixed number, called the common difference (d), to the previous term. The arithmetic sequence 3, 5, 7, 9,11,13, ... has a common difference of 2.

9, ...

If you know any term in an arithmetic sequence, you can add the common difference to it to find the next term. The following equation is a recursive definition of an arithmetic sequence. o

You can derive an explicit equation to find any term in an arithmetic sequence, as long as you know the first term.

o2 — a, + d

a3 - a2 + d = a} + d + d = a, + 2d a4= a3 + d = a}+ 2d + d = a,+ 3d Notice the pattern. To find a desired term, add the first term, a,, to a multiple of the common difference, d. The coefficient of d is 1 less than the number of the desired term. o. — a,

166

(n -

Unit 2; Linear and Exponential Relationships

Connect Below are the first four terms in an arithmetic sequence. 25,22,19,16 Use a recursive process to find the next 3 terms in the sequence. Plot the first seven terms on a coordinate plane.

Find the common difference. Solve the recursive equation for d.

j n = a n ,I + d Find the difference between successive terms. d = 2 2 - 2 5 = -3 d=19 - 22= -3 d = 1 6 - 1 9 = -3

So, the recursive formula for an is 3. a_ = a.

Find the next three terms. 5

= a 4 - 3 = 1 6 - 3 = 13

5

= a5- 3 - 1 3 - 3 - 1 0

7

= a 6 - 3 = 10-3 = 7 The next three terms are 13,10, and 7.

Plot a point to represent each term in the sequence. Plot each point (n, an) for the first seven terms. .

28

(1,25)

28

;M w

-V---1

.

^(2,22) *• s

20 18 16

S

lit

1 Kl

# (1, 10]

U

~* v (5 13) *- '

I?

10

* "+L$'

^)

: »

o

6 4

(7,7)

2

0

1

2

3

4

5

6

7

5

Based on the graph, what type of relationship exists between the n-values and the devalues? Use what you know about functions to help you explain the relationship.

Lesson 22: Arithmetic Sequences

167

I In a certain arithmetic sequence, each term is found by subtracting 4.5 from the previous term. If the first term in the sequence is 10, what is the 9th term in the sequence?

Identify the values of a, and d. You are told that the first term, a,, is 10. The next term is found by subtracting 4.5, so the common difference, d, is -4.5.

Write an explicit formula to find a . Substitute the values of a1 and d into the formula.

an = 10-4.5n + 4.5 a = 14.5 -4.5n Substitute 9 for n in the formula. Evaluate. on = 14.5-4.5n Qg

- 14.5 - 4.5(9) - 14.5 - 40.5 = -26

t The 9th term in the sequence is -26.

Use a recursive process to find the 9th term in the sequence. Which methodusing a recursive process or using an explicit formula—do you think is a better choice for solving this problem? Why?

168

Unit 2: Linear and Exponential Relationships

Problem Solving Ami is training for a long-distance race. She ran for 30 minutes per day on three days this week. Each week she will increase her daily running time by 5 minutes. By the 6th week, for how many minutes will she run each day that she trains?

She increases her running time by the same number of minutes each week. So, this is an arithmetic sequence. Write an explicit formula to represent the situation. Then find the sequence.

term in the

SOLVE Her daily running time during the 1st week is 30 minutes, She will increase her daily running time by 5 minutes each week, so d = _ Find a&, the number of minutes she will run each day during the 6th week.

30 + (

CHECK Use a recursive process to check the answer.

Do you get the same value for a ? By the sixth week, Ami will be running

minutes each day that she trains.

Lesson 22: Arithmetic Sequences

169

Practice Determine if each sequence is an arithmetic sequence. If it is, identify the common difference.

1.

5, 3, 1, -1, -3, ...

2.

" ^ it) :

2, 2, 4, 6,10, ...

3.

f, f, \ f, \,...

Is the same number added to each term to get the next term?

Write a recursive process for each arithmetic sequence. Then use it to find the specified term, 4.

5.

5,9,13,17,21, ...

6.

10, 3,-4, -11,-18,.

REMEMBER In a recursive process, an is defined by using previous terms.

Write an explicit formula for the nth term and use it to find the specified term. 7.

11,15,19, 23, ... Qn

8,

=-

Qrt — —

100, 88, 76, 64, ...

9.

on = ———

——

dirt

-—-

1.2,1.8, 2.4, 3,...

°n = .—.—-

tJi12

Use the given information to find the specified term in each arithmetic sequence. 10. a, = 14, d = 6

11. a 1 - 5 2 , d = - 5

12.

a.

a, - 0, d =

a15

Choose the best answer. 13. The formula an = 10 - 4n describes an arithmetic sequence. What are the first four terms in the sequence? A.

6, 2,-2,-6

For an arithmetic sequence, a, - 21. Its recursive formula is an - an - ,I +11. Which explicit formula can be used to find the nth term in the sequence?

B.

6,2,0,-2

A.

an= 10-lln

C.

10,6,2, -2

B.

on = 10 + lln

14, 18,22,26

C.

o ( = 21 -lln

D.

a = 21 + lln

D.

170

Unit 2: Linear and Exponential Relationships

14.

1 2'

Use the arithmetic sequence below for questions 15-17. 2,5,8,11,... 15.

Write an explicit formula in terms of n to show how to find the nth term in this sequence.

16.

Plot points (n, an) on the grid to represent the first six terms in the sequence.

0

1

2

3

4

5

6

What does the slope of the graph represent?

17.

Think of this sequence as a function. What type of function is it? What are its domain and its range? Explain your thinking.

'

Use sequences to describe the situation. 18.

Steve is buying a new tablet computer on layaway. He makes an initial payment of $50 and will increase the payment each month as shown by the table. Write a recursive formula and an explicit formula to describe this sequence. Explain how you determined the formulas.

Month (n)

Payment in $

K)

1

50

2

60

-^

70

4

80

5

90

Lesson 22: Arithmetic Sequences

171

Geometric Sequences in a geometric sequence, each pair of consecutive terms is related by a common ratio, r. To find a term in the sequence, multiply the previous term by the common ratio. The geometric sequence 1, 3, 9, 27, 81, ... has a common ratio of 3. Each term is multiplied by 3 in order to yield the next term.

81, .

If you know any term in a geometric sequence, you can multiply it by the common ratio to find the next term. The following is a recursive definition of a geometric sequence: o. Suppose you wanted to find the 100th term in this sequence. As with arithmetic sequences, using the recursive definition would require a lot of steps. Fortunately, you can derive an explicit equation to find any term in a geometric sequence, as long as you know the first term. Examine the formulas for the first few terms.

a, O-

a a

a-r-r

a-r

r = o1 • r • r = a,

a,

a - r • r = a.

Notice the pattern. To find a desired term, multiply the first term, a,, by a power of the common ratio, r. The exponent of r is 1 less than the number of the desired term. Written mathematically, this is:

a, • r

172

Unit 2: Linear and Exponential Relationships

Connect Below are the first four terms in a geometric sequence. 1,2,4,8 Use a recursive process to find the next 3 terms in the sequence. Plot the first seven terms on a coordinate plane.

Find the common ratio. Solve the recursive equation for r.

Find the quotient of successive terms. r =

2



~

Find the next three terms.

So, the recursive formula for an is

o7 = a6 • 2 - 32 • 2 - 64 ^ The next three terms are 16, 32, and 64. Plot a point to represent each term in the sequence. Plot each point (n, an) for the first seven terms.

68

-(7,64)^ i ('

60 56 52

48 44 40 36 32 28 24 20

16 12 8 4

- r i ( 1 1

_; __ .

. . __ ....

!. . • (6,32) •-

' /

---:-/

."-,-•- - - " - i

X(5, 16) r

(2, 2)

- * ^' ^

_••--*•"''

•s

Based on the graph, what type of relationship exists between the n-values and the c^-values? Use what you know about functions to help you explain the relationship.

Lesson 23: Geometric Sequences

173

EXAMPLE A geometric sequence has an initial value of 1,024, and each term in the sequence is half of the previous term. Write an explicit formula to find any term in the sequence. Then use that formula to find the 9th term in the sequence.

Identify the values of a, and r. You are told that the first term, a,, is 1,024. The next term is found by halving the previous term, or multiplying by ^, so the common ratio, r, is -

2 Write an explicit formula to find an. Substitute the values of a1 and r into the formula. a = a •r "

'

This can be simplified by using the laws of exponents.

a =1,024a =1,024

2

.

IV

a, = 2,048 • Find the 9th term in the sequence. f (9) = 2,048 -

= 2,048

=4

The 9th term in the sequence is 4.

Use a recursive process to find the 9th term. Compare this result to the answer above to check your work.

1/4

Unit 2: Linear and Exponential Relationships

Problem Solving Jenny started a chain letter by e-mail. She sent it to five of her friends and asked them to each send it to five of their friends. Assume that no one breaks the chain and that no person receives the e-mail twice. How many e-mails will be sent during the 6th generation? (Treat Jenny's e-mails as the 1st generation of the letter.)

Each person who receives the e-mail sends it to five friends. Each of those people also sends it to five friends. So, the total number of e-mails sent in each generation is

times the number of e-mails sent in the previous generation.

So, a recursive process can be used to find the answer.

SOLVE In the 1st generation, 5 e-mails were sent. So, a, - 5. In the 2nd generation, 5 times as many e-mails will be sent, and so on. 2nd generation: a? = 5 • 5 = 3rd generation: a3 =

IT

_-_

4th generation: a4 =

5=

5th generation: O5 =

5-

6th generation: a6 —

5-

CHECK The numbers of e-mails sent in each generation are the terms of a geometric sequence. So, check the answer by writing an explicit formula. and the common ratio, r, is 5.

We know that a, = on - a, • r a =

-

• (.

Use the formula to find o6. a.

Did you get the same value for a6? If the chain is not broken, then exactly sixth generation.

e-mails will be sent in the

Lesson 23: Geometric Sequences

175

Practice Determine if each sequence is a geometric sequence. If it is, identify the common ratio. 1.

5, 70, 15, 20, 25, .

2.

7, -14, 28, -56,112, .

3.

16, 24, 36, 54, 81,

REMEMBER A common ratio can be positive or negative. Write a recursive process for each geometric sequence. Then use it to find the specified term. 4.

1

1,6,36,216,...

6.

64' 256' '"

0,05,0.5,5,50,...

a, =

a, =

a

a

or What number is multiplied by each term to get the next term? Write an explicit formula for the nth term and use it to find the specified term.

7.

8.

3, 2, f , f ,...

3, -9, 27, -81, ...

9.

12, 24, 48, 96, ...

CL =

CL

Use the given information to find the specified term in each geometric sequence.

10. o, = 3, r - 20

11. o, - 5,000; r - 0.2

12. a, = 2, r - -4 o.

Choose the best answer. 13.

The formula on ~ —10 • (3)"' 1 describes a geometric sequence. Which recursive formula also describes this sequence? A.

a, =l; 0 f ] = o n _ 1 - - 30

R

/i

c.

Ol

=

); a n = a n - 1 - 3

D.

a, -

o,-=o.-,-30

14.

Which formula can be used to find the nth term in the sequence below? 128,96,72,54,... A.

B.

fl.-128.dr

c-

a fl.-i28.( 176

Unit 2: Linear and Exponential Relationships

Use the geometric sequence below for questions 15-17. 80,40,20,10, ... 15. Write an explicit formula in terms of n to show how to find the nth term in this sequence.

16.

Plot points {n, an) on the grid on the right to represent the first six terms in the sequence. Find the average rate of change between each adjacent pair of points.

17.

Think of this sequence as a function. What type of function is it? What are its domain and its range? Explain your thinking.

A petri dish contains 4 viruses. Each hour, the number of viruses increases, as shown in the table. The population change can be modeled by a geometric sequence. Write a recursive formula and an explicit formula that can model this sequence. Use the formulas to predict how many viruses will be in the dish by the 7th hour. Hour (n)

Population

K) .

...

4

2

12

3

36

4

108

5

324

;

Lesson 23: Geometric Sequences

177

fc

2

Review

Determine whether each relation represented is a function or not. Write yes or no. 1.

Input x « 0 -

2.

Output

Input -3 \ -2

7

3.

Output S



Input

Output

N.

x64

-100

k 125

o _

Use the arithmetic sequence below for questions 4-6. 18,14,10,6,...

.. f

4.

Write an explicit formula for the nth term.

5.

What is the tenth term in this sequence?

6.

Think of this sequence as a function. What type of function is it? What are its domain and its range?

Graph each function. Then identify the x- and y-intercepts of each graph.

7.

f(x)=~2x + 6

8.

f(x) =

y

2

4 . e -15 r -4 ; -2

178

x-intercept:

x-intercept:

y-intercept:

y-intercept:

Unit 2 Review

• 0

2

i 4

: fi

Fill in the blanks by writing an operation sign and a number to show how the y-values are changing over each interval. Then classify the function as linear or exponential.

9.

Function type: x

y

Write each expression in the requested form. 10.

Vx 2 in exponential form

11.

^j

(16a)^ in radical form

Use the graph and table below for questions 12 and 13. The graph represents an exponential function f. The table lists several ordered pairs for a linear function g. 9(x) = 5 x - 5 X

sKx)

-1

-10

0

_c

1

0

2

5

3

10

12.

Compare the intervals for which the functions are positive and negative.

13.

Compare and contrast the end behavior of the functions.

Solve each system of equations algebraically. 14. f x - 2y=ll 12x + 5y = 4

15.

|2x- 5y = 40

-4x + 3 = -10

Unit 2 Review

179

Graph g. Then write an explicit equation for g(x) in terms of x. 16.

g(x) - f(x + 2)

17. g(x) = -f(x)

f(x) = -3*+ 3: 2

4

G

a

g(x)

g(x) Choose the best answer. 18.

r.

Based on the graph on the right, which statement is not true? A.

Functions fand g have the same x-intercept.

B.

The ordered pair (2,14) is a solution for f(x).

C.

The ordered pair (2, 7} is a solution for g(x).

D. The value of f(x) begins to exceed g(x) during the interval between x = 1 and x = 2.

Graph each inequality or system of inequalities. Shade the portion of the graph that represents the solution set. 19.

2y-3x>4

180

Unit 2 Review

Rewrite each side of the equation as a function. Then graph the functions to solve for x. 21. 2 x + 6

g(x) = X =

-*x

Examine the following situations and respond in complete sentences.

22.

With each swing of a pendulum, the length of its swing becomes less, as shown in the table below. Write an explicit formula to model the length of the arc in terms of the number of the swing. Classify the list of arc lengths as either an arithmetic or geometric sequence and explain how you know. Number of Swing (n)

Arc Length

1

30

23 -

The cost of usinn thp at an Internet cafe equals a settee plus a certain rate per minute. Four minutes of Internet use cost $3, and eight minutes of Internet use cost $4. On the grid, make a graph to represent this situation and write an equation to model the situation. Find the slope of the graph and interpret its meaning in this situation. ex

Cost of Internet Use

27

24.3 21.87

19.683

o

T3 C

in o O 0

1 2 3 4 5 6 7 8 9 10

Time (in minutes)

Unit 2 Review

181

Performance Task

TAKM? CARE OF Business Working in parrs or individually, pretend you are starting your own business. 1.

The item I will self is

2.

Set a price per item that you will charge. Create a table, a graph, and an equation to show f, the amount you will collect if you sell up to 10 items. Represent the number of items as x. x

. The name of my new business is

f(x) =

1

,

P

ro due:t Sales

52 48 >

3 4 5 6 7

44

i

n

C 40 0-^-36 S « ?2 o _. 28 ^ -g 24

iIE:

§ c 20 O *=• 16

; !

E

12

<

8

8

4

9

0

±



!

I

j

x

10

1 2 3 4 5 6 7 8 910

Number Sold

3.

Explain how you determined the equation for f.

4.

Did you connect the points on the graph with a solid line? a dashed line? no line? Explain why.

The amount you collect from selling your product is not the same as the profit you earn from selling your product. 5.

Suppose that the amount of profit you earn is equal to-| of what you collect. So, if each product costs $5, then your profit for selling 1 product is: $5 • ~ = Using algebraic notation, represent this as a transformation: p(x) =

f(x).

What type of transformation off does it represent? ____—__—______^_^__

6,

182

Write an equation for p. Then graph it on the coordinate plane above, using a different colored pen or pencil than you used to draw the graph of f. Does this show the transformation you expected?

Unit 2 Performance Task

Profits are important to you, but your customers are only concerned with how much your product will cost them. They want to pay as little as possible. The graph below shows how much one of your competitors, Company G, charges for the same product, including shipping.

y

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1 2 3 4 5 6 7 8 9 10

Number Sold 7.

Based on the graph, how much does Company G charge per product? Explain how you know.

8.

Copy your original graph of function f from the previous page onto the graph above. Use a different color and label it f. After checking with your local shipper, you find that you will need to charge customers 15 dollars to ship any items. Transform the graph effusing this equation: MX) = f(x) + 15

Identify the transformation: Then write an explicit equation for h: h(x) =

Based on your graph, when would it be cheaper for Mr. Smith to order from Company G? When would it be cheaper to order from you? Would the price ever be the same from both companies?

Unit 2 Performance Task

183

Grade 8

Algebra I

Geometry & Algebra II

Functions Interpreting Functions

Expressions & Equations Understand connections between proportional relationships, lines, and linear equations.

Statistics and Probability Interpreting Categorical and Quantitative Data Functions Use functions to model relationships between quantities.

Statistics and Probability Investigate patterns of associations in bivariate data.

Summarize, represent, and intepret data on a single count or measurement variable. Summarize, represent, and intepret data on two categorical and quantitative variables. Interpret linear models.

Interpreting Categorical and Quantitative Data Summarize, represent, and intepret data on a single count or measurement variable. Making Inferences and Justifying Conclusions Make inferences and justify conclusions from sample surveys, experiments, and observational studies. Conditional Probability and the Rules of Probability Understand independence and conditional probability and use them to interpret data.

184

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Linear and Exponential Relationships

Unit 2 Linear and Exponential Relationships Lesson 10 Rational Exponents 70 Lesson 11 Functions 76 Lesson 12 Key Features of Functions 82 Lesson...

30MB Sizes 2 Downloads 25 Views

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