MTHS MATHEMATICS DEPARTMENT

To: Precalculus Students

Re: Summer Project 2014

Date: June 2014

Welcome to Precalculus! Precalculus is a rigorous course that is the fourth course in the

regular college preparatory sequence. This course applies the skills obtained in

Algebra I, Geometry and Algebra II.

To help you begin your study of Precalculus, you will be completing a review project

this summer. This project consists of problems reviewing important concepts from

algebra, geometry or trigonometry. This assignment will help you keep your skills

sharp so that you will be successful next year.

Please follow these directions very carefully.

• Use pencil • Work in study groups and use your Algebra II notebook as a guide. These are the

problems that you may have forgotten how to do but are essential for success in Precalculus. Note: All students are to document their OWN work!!

• The project is due the second time your class meets. A quiz will be given on this material the third time your class meets.

• Please make a photocopy of the project before you hand it in. You will have a quiz on this material as well.

• Show all work neatly and clearly to receive full credit!! • Put answers on the line, if provided or circle the final answer.

Good luck and have a happy and safe summer. I’m looking forward to seeing you in September.

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42 Chapter I Functions and Their Graphs

Summary of Graphs of Parent Functions

One of the goals or this text is to enable you to build your imuition for the basic shapes of the graphs of diiTerenttypes of functtons. foor instance, from your study of lines in Section 1.1. you can determine the basic shape of the graph of the ltncar function/(.\) = Ill\ + h. Specifically. you know that the graph of thi~ function is a line whose slope is 111 and whose y-intercept is (0, b).

The six graph~ shown in Figure 1.40 represent the most commonly used functions in algebra. Familiarity with the ba~ic charaetcn~tics of these simple graphs will help you annlyte the shape!> of more complicated graphs.

·31~---1---- 3

- t

(a) Constant Function

(c) Absolute Value Function

(e) Quadratic Function

Figu re 1.40

(b) Identity Function

(d) Square Root Function

(f) Cubic Function

Throughout thts sec11on. you will dbcover how many compltcated graph~ arc derived hy shifting, stretching. shrinking, or relle~.:ting the parent graphs shown above. Shifh. stretches. shrinks. and reflections are called tmllsfomw­tionx. Many graphs of functions can be created from combination~ of these t.nmsfonnat ions.

What you should learn

• Re<ognize graphs of parent functions.

• Use vertical and horizontal shifts and reRections to graph functions.

• Use nonrigid transformations to graph functions.

Why you should learn it Re<ognizmg the graphs of parent functions and knowing how 10 shih, reRecl, and stretch graphs of functions can help you sketch a wide variety of simple funoions by hand.llus skill is useful in sketching graphs of functions that model real-life data. For example. In Exercise 57 on page 49, you are asked to sketch a function thai models I he amount of fuel used by vans, pkkups,and sport u1Hity vehicles from 1990 lhrough 2003.

lim lloylc/C"oeny htmgc>

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L lS.

Kl

iph

it tlons stretch ha nd.This nctlons ,in to oum or llinty

Section 1.4 Shifting, Reflecting, and Stretching Graphs 43

Vertical and Horizontal Shifts Many functions have graphs that are simple transformations of the graphs of parent functions summurized in Figure 1.40. For example, you can obtain the graph of

h(.r) = .r1 + 2

by shifling the graph of /(.r) = .l.J two units IIJIIIYIId. us shown in Figure 1.41. In

function notation, It and fa re related as follows.

h(.r) = .\.J + 2

= f(.r) + 2 I p\\.onl 'hill ul 1\\11 Ulllh

Similarly. you can obtain the graph of

g(.r) = (.r - 2)2

by shifting the graph off(.r) = x 1 two units to the right. as shown in Figure 1.42. In this case. the functions g <tndfhave the following relationship.

g(x) = (.r- 2p = f(.r - 2)

I

'•

.~ ~

•( I . I ) --1---f--f-~ .I

2 J

Figure 1.41 Vertical shift upward:

two units

Figure 1.42 Horizontal shift to the

right: two units

The following list summarizes vertical and horizontul shifts.

Vertical and Horizonlal Shifts

Let c be a positive real number. Vertical and horizontal shifts in the graph of y = f(x) are represented as follows.

1. Ve11ical shift c units upward:

2. Verticul shift c unirs dmt•nward:

3. Horizontal shift c units to lhe right:

4. Horizontal shift c units to the left:

h{.r) = .f(.r) + c

h(x) = f(x) - c

h(x) = f(x - c)

h(x) = f(x + c)

In items 3 and 4. be sure you see that h(.r) = f(x - c) corresponds to a rif(ht shift ~d h(.r) = f(x + c) corresponds to a le.fi shirt for c > 0.

-t:xploration

Use a graphing utility ro display (in the same viewing window) the graphs of' y = .r1 + 1'. where c = -2, 0. 2. and 4. Usc the result~ tO describe the errect that c has on the graph.

Use a graphing utility to display (in the same viewing window) the graphs of y = (.r + c)l. where c = - 2, 0. 2. and 4. Use the result-; to describe the effect that c has on the graph.

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44 Chapter I Functions and Their Graphs

Example 1 Shifts in the Graph of a Function

Compare I he g raph of each function with the g raph o l'f(x) = x3•

a . . t:(x) = \ ' I b. il(x) = (.\ - I )' c. k(x) = (.r + 2)3 + I

Solution

a. Graph f(x) = \'' and g(x) = \' I I see Figure 1.43(a) 1. Yo u can obtain the graph or M by shifti ng the graph off one unit downward.

b. Graphf(x) = r~ and h(x) = (x - I )3 I see Figure 1.43(h) l. You can obta in the graph of II by shifting the g raph of[ one unil to the right.

c. Graph /(.\) - '" ' and k(x) - (.\ I 2)' + I I sec Figure 1.43(c)]. You can obtain the grnph of k by shifting the graph o fj two units to the left and then o ne unit upward.

(a) Vertical shift: one unit downward

Figure 1.43

./nJEO POINT Now try Exerc ise 23.

(b) Horizontal shirt: one unit righ t

Example 2 Finding Equations from Graphs

(c) Two units left and one unit upward

The graph off(x) = x2 is shown in Figure 1.44. Each of' the graphs in Figure 1.45 is a transformation o f the graph of f Find an equatio n for each function .

6 1--~---+~---- 6

2 -2

(a) (b )

Figure 1.44 Figure 1.45

Solu tion a. The g raph o f g is a vertical shift of four units upward o r the g raph ofj(x) = x2•

So, the equation for g is g(x) = x2 + 4.

b. The graph o f II is a horizontal shift o f two units to the le ft. and a vertical shift of one unit downward, or the graph o f f(x) = .x2• So. the equation for h is h(x) = (A + 2)2 - I.

./ CHFCKPOINT Now try Exercise 17 .

Re Anc you imal

Ret

Reli as fc

Exam,

Thegr is a Lra

Figure 1

Solutic

a. T heJ uni ts

b. The ; reflec h(x)

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Section 1.4 Shifting, Renecting, and Stretching Graphs 45

Reflecting Graphs Another common type of transformation is called a reflection. For instance, if you consider the .r-axis to be a min·or, the graph of h(.r} = - .\.2 is the mirror image (or refl ection) of the graph of j(x) = A2 (see Figure 1.46).

\ ~ I ·

- 2

Figure 1.46

Rcncction~ in the Coordinate Axes

Reflections in the coordinate axes of the graph of y = f(x) are represented as follows.

1. Reflection in the x-axis: h(x) = - f(x)

2. Reflection in they-axis: h(x) = f( - x)

Example3 Finding Equalions rrom Graphs

The graph of f(x) = x 4 IS shown in Figure 1.47. Each of the graphs in Figure 1.48 1s a transformmion of the graph off Find an equation for each fum:tion.

3

- 1

-3 3 3

- 1 3

(a) (b )

Figure 1.47 Figure 1.48

Solution a. The graph of g is a rellcction in the x-ax i~.follol1'ed by an upward shift of two

uni ts of the graph olf(.r) = x 4 • So. the equation fo r g is g(x) = - .r4 + 2.

b. 'lllc gmph of ,, is a hori.w ntal shift or three units 10 the right followed hv a reflection in the x-axis of the graph of f(.r} = x 4

• So, the equation for II is h(x) = -(.\ - 3)4

•

/._, .,_uPOJNT Now try Exercise 19.

- -Lxplor111 ion

Compare the graph of each function with the graph of f(x ) = .r2 by using a graphing utility to graph lhc function and fi n the same viewing window. Describe the transformation.

a. x(x) = - .\..2

b. h(x) = (- .rP

5

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46 Chapter 1 Funclions and TI1eir Graphs

Example 4 Rcncctions and Shirts

Compare the graph of each function with the graph of/h) - "~.

a. g(.\'} = - J;. h. 11(.1) F-; c. k(l) = - Jr + 2

Al~1cbraic <)olulion

a. Relative to the graph off(x) = h. the graph or g is a rcncction in the .\-aXIS hccau-.c

g(\) - - ,r; -_/'(.1).

h. The graph ol II i-. a reflection ol the graph ol f(x) -= J;. in the r-.txis becau:-.e

h(_\) = r--; = J( - x).

c. From the equation

.l.(x) =- J.1 + 2

= -J(x+2)

you can conclude that the graph or" i:-. a left :-.hilt of tWO UllliS. followed by a rcllcction in the \'-aXi\. of the graph of j'(x) = fi .

.fcl'tCM'OINT Now try Exercise 21

Graphh:al '-,olultlln

a. U:-.c a graphmg utility to graph I and g in the '\amc vtewing. window. From the graph in Figure 1.49. you can see that the graph or g t), a reflection or the graph of! in the 1'-axi:-..

b. Usc a graphing utility to graph .I and h in the same vtewlllg ""tndow. From the graph in Figure 1.50. you can :-.ee that the graph or h i' a rcllcction or the graph of fin the \'-axis.

c. u ... e a graphing utility to graph .I und k 111 the l>:tme vicwtng \\tndow. From the graph in f-igure 1.51. yuu can -;cc that the graph of/.; " a left shtlt of two units of the graph or f. followed by a rcllcction in the x-axi-;.

Figure 1.49 Figure 1.50

Figure 1.51

When graphtng luncuon~ mvolving :-.quare rooh, remember that the domain must be rc~trictcd 10 exclude negative numbers inside the radicaL ror in-.tance. here arc the domain:- of the fum:ttons 111 Example 4.

Domain nf ,11( 1) = JX: 1 ~ 0

Domain ol lt(x) = h: r s 0

Domnin of A(. I) - - ~: x ~ - 2

NOI Hnn. bee a only arc ll insta v= shri rcpn c: >

Co1

a. 1

b . .

Sol a .

b.

Ex,

Cc

is Pi

_ _____ L

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window. or g is a

window. or It is a

window. k i~ a left ln in the

Section 1.4 Sh ifting, Reflecting, and Stretching Graphs 47

Nonrigid Transformations Horizontal shifts. vertical shirts, and re llections are called rigid transformations because the basic shape of the graph i:-. unchanged. These trunsformations change only the position of the graph in the coordinate plane. Nonrigid transformations are those that cause a distortion- a change in the shape of the original graph. For mstance, a nonrigid tnu1sformation of the graph of y = f(.r) is represented by \' = cf(x), where the transformation is a vertical stretch if c > I and a vertical shrink if 0 < c < I. Another nonrigid transformation or the graph of y = f(x) is represented by h(x) = f(c:x) • where the transformation is a hol"izontal shr ink if r > I and a horizontal stretch if 0 < c < I .

fxample 5 Nonrigid Transformations

Compare the graph or each function with the graph of f(.r) = l.rl. a. h(x) = 3lxl

b. g(x) = ~lxl

Solution a. Relative tn the graph of f(x) = lxJ. 1 he graph of

h(x) = 3lxl = V(.r)

is a vc11icul stretch (each y-value is mu ltiplied by 3) of the graph off. (See Figure 1.52.)

b. Similnrly, the graph of

g(x) = ~lxl

= *f(x)

is a vertical shrink (each y-valuc is multiplied by *) of the graph or f. (See Figure 1.53.)

Now try Exercise 31.

Example 6 Nonrigid Transrormations

Compare the graph of h(x) = .r{~.r) with the graph of f(.r) = 2 - x 3•

Solution

Figure 1.52

Figure 1.53

Relative to the graph of f(.r) = 2 - .r-'. the graph of -s t~-......... ....-....,f-'t-....,...~~

h(.l) = I(~x) = 2 - (~x)' = 2 - ~ -,-' is a horizontal stretch (each .r-value i ~ mu ltiplied by 2) of the graph of f. (See Figure 1.54.) Figure 1.54

I Hf.OiPOtNT Now try Exercise 39.

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What do a a nd n represent? a is the coefficient of the leading term. n is the exponent of the leading term.

ll1e degree of a polynomial function affects the shape of ils graph and determines the maximum number of turning points, or places where the graph changes direction. ll also affects the end behavior, or the directions of the graph to the far lefl and to the far right.

For polynomial functions of degree one or greater, there are four types of end behavior as you move to the left and move to the right, away from the origin: up and up, down and down, down and up, and up and riown.

Down and Down y

Down and Up Up and Down y y

X X

X X

y = x4 - 3x3 + 5x y = x2 + 6x y = - 0.3x3 + 4x + 2

You can determine the end behavior of a polynomial function of degree n from the leading term ax" of the standard form.

End Behavior of a Polynomial Function of Degree n With leading Term aX" (Moving Away From the Origin)

I• ' ... a Posit ive Up and Up Down and Up

a Negative Down and Down Up and Down

.Ailt'Ct§aJh Describing End Behavior of Polynomial Functions

Consider the leading term of each polynomial ftmclion. What is the end behavior of the graph? Check your answer with a graphing calculator.

fl y= 4x3 - 3x

The leading term is 4..\.~'i. Since n is odd and a is positive, the end behavior is down and up.

Check

The solution checks.

rn y = - 2x4 + 8x3 - Sx2 + 2

The leading term is -2x'1• Since n is even and a is negative, the end behavior is down and down.

Check

1he solution checks.

282 Chapter 5 Polynomials and Polynomial Functions

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Which function is substituted into the other? Use f(x) as the input for g.

Key Concept Composition of Functions

The composition of function gwith function/is written as g o f and is defined as (g o J)(x) = g(j(x)). The domain of g o f consists of the x-values in the domain off for which J(x) is in the domain of g.

(g o J)(x) = R( /{x)) 1. Evaluate f(x) first. I I

L___!__j 2. Then use f(x) as the input for g. -

Function composition is not commutative since f(g(x)) does not always equal g(j(x)).

~Rfi iDfDED RESPONSE

Let J(x) = x - 5 and g(x) = x2. What is (g o f)( -3)?

Method 1

(g o J)(x) = g(j(x))

= g(x - 5) = (x - 5)2

(g 0 /)( - 3) = ( - 3 - 5)2

= c -sr~

= 64

Method 2

(g 0 /)( - 3) = g(f( -3))

= g( - 3 - 5)

= g( -8)

= (-8)2

= 64

Got It? 3. What is (Jo g)( - 3) for the functions/and g defined in Problem 3?

~jffl~t§.,(l Using Composite Functions

- . . . I~ ~ ~ ~~

3

~~·4 4 s)~ s

~~~ ~ ·~18 8 9)19 9

You have a coupon good for $5 off the price of any large pizza. You also get a 10% discount on any pizza if you show your student ID. How much more would you pay for a large pizza if the cashier applies the coupon first?

.. -. ..

0 0

li~ 1 l

' . 5 6

8 8 9 9

The coupon value and the discount rate

The difference between the results of applying the discount or coupon first

• Compose two functions in two ways. • Then find the difference in their

results.

St ep 1 Find functions C and D that model the cost of a large pizza.

Let x = the price of a large pizza.

Cost using the coupon: C(x) = x - 5

Cost using the 10% discount: D(x) = x - O.Ix = 0.9x

Step 2 Compose the functions to apply the discount and then the coupon.

( C o D)(x) = C(D(x)) Apply the discount, O(x), first.

= C(0.9x) = 0.9x - 5

400 Chapter 6 Radical Functions and Rational Exponents

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• inverse function • one-to-one function

(0, - 1) is ins. How do you find the corresponding pair in the inverse of s1 Switch the coordinates. ( - 1, 0} is in the inverse of s.

Inverse Relations and Functions

Objective To find the inverse of a relation or function

Mayor's Salary Restored At last night's meeting, the town

council approved a 20% increase in the mayor's salary. This follows last year's 20% decrease. The Mayor's comment was

If a relation pairs element a of its domain to element b of its range, the inverse relation pairs b with a. So, if (a, b) is an ordered pair of a relation, then (b, a) is an ordered pair of its inverse. If both a relation and its inverse happen to be functions, they are Inverse functions.

Essential Understanding The inverse of a function may or may not be a function.

Relation r Inverse of r This diagram shows a relation r(a function) and its inverse (not a function). The range of the relation is the domain of the inverse. The domain of the relation is the range of the inverse.

Domain Range Domain Range

1.2 ::=:::; 1 1.4

~:~- ~ 2

~jffiNf§,,jj Finding the Inverse of a Relation

f'.i.l What Is the inverse of relation s? Inverse of Relations Relations

0 - 1

2 0

3 2

4 3

Switch the x and y values to get the inverse. - 1 0

0 2

3

1 ... 1.2 ---. 1.4

2 ... 1.6 ---. 1.9

rG·i'Zf:Sfr'm?li§!';•luj I ~ 6-7 Inverse Relations and Functions 405 ~~=-----~~==~~~===-~

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Why do you solve for y7 If you solve the equation for y, you can use it to easily generate ordered pairs that are part of the inverse relation.

What does the graph of y = x2 - 1 look like? The graph of y = x2 - 1 is a translation of y = x2 down one unit

U) Wbat are the graphs of sand its inverse?

Reversing Relations the Ordered Pairs Inverse of s

ffll J: y d r 1

y ·-;-

·- - 4 - - - 1- - - -

2 -· -

X

- - 0 4

[ 1 2 U.Ll Got It? 1. a. What are the graphs oft and its inverse?

b. Reasoning Is t a function? Is the inverse oft a function? Explain.

- 4

L l

-2 0 2 4

- 2 l l

Relation t

0 2 3

- 5 - 4 -3 - 3

As shown in Problem 1, the graphs of a relation and its inverse are the reflections of each other in the line y = x. If you describe a relation or function by an equation in x andy, you can switch x and y to get an equation for the inverse.

~j#ftfit§,fj Finding an Equation for the Inve rse

What is the inverse ofthe rela tion described by y = x 2 - 1?

y = .\ 2- 1

\ = y2 - 1

X+ 1 = y2

±Y.X+l = y

Switch x andy.

Add 1 to each side.

Find the square root of each side to solve for y.

Got It? 2. What is the Inverse of y = 2x + 8?

What are the graphs of y = x2 - 1 and its inverse, y = ± Y.X+l?

The graph of y = x2 - 1 is a parabola that opens upward with vertex (0, - 1). The graph of the inverse is the reflection of the parabola in the line y = x.

Got It? 3. What are lhe graphs of y = 2x + 8 and its inverse?

X

406 Chapte r 6 Radical Functions and Rational Exponents ...... ______________________________________________ _. ______ ~--------~~----------~~~~~~-~~-----

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>uld a graph >u check your r? ld graph f- 1 and ~!her the graph :he vertical line t does, f- 1 is a I.

The inverse of a function/is denoted by / - 1. You read r 1 as "the inverse off" or as "/inverse:· The notation j(x) is used for functions, but the relation r 1 may not even be a function.

Omnm,e• Finding an Inverse Function

Consider the function f(x) = '\I".X=2.

a What are th e domain and range off?

The radicand cannot be negative, so the numbers x ;;::: 2 make up the domain. The principal square root Is nonnegative, so the numbers y ;;::: 0 make up the range.

g What is f- 1, the inverse off?

J(x) = '\I".X=2 y = '\I".X=2 X = v'_Y=2

il-=y - 2

Rewrite the equation using y.

Switch x and y. Since x equals a principal square root, x ;;::: 0.

Square both sides.

y = i2 + 2 Solve for y.

So, f - 1(x) = x2 + 2, for x ;;::: 0.

~What are the domain and range of J- 1?

Part (b) shows that the domain of r 1 is the range off-the numbers x ;;::: 0. Since :? ;;::: 0, x2 + 2 ;;::: 2. Therefore, the numbers y ;;::: 2 make up the range of f - 1• Note that the range of f - 1 is the same as the domain off

(!) Is f- 1 a function? Explain.

For each X in the domain (x ;;::: 0) of r 1, there is only one vaJue of yin the range. So

f - 1(x) = x2 + 2, x;;::: o, is a function.

Got It? 4. Let g(x) = 6 - 4x. a. What are the domain and range of g? b. What is the inverse of g? c. What are the domain and range of g - 1? d. Is g- 1 a function? Explain.

Functions that model real-world behavior are often expressed as formuJas with meaningful variables, like A = n?- for the area of a circle. Strictly speaking, the inverse formula wouJd be r = nA2, but this expresses a false relationship between A and r. It is better to leave the variables in place and solve for r as a function of A.

A = n?- Original formula .

r = g Same formula, but inversely expressed.

I Lesson 6-7 Inverse Relations and Functions

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Why shouldn't you interchange the variables? Interchanging the variables leads to a false relationship between distance and time.

. jfflNt§,,ij Finding the Inve rse of a Formula

The function d = 4.9t2 represents the distanced, In meters, that an object falls in t seconds due to Earth's gravity. Find the inverse of this function. How long, in seconds, does it take for the cliff diver shown to reach the water below?

d = 4.9r2

t2 = A.. 4.9

l = {d '/4.9 _ 124 - '/4.9

= 2.2

Solve fort. Do not switch the variables.

Time must be nonnegative.

Substitute 24 for d.

Use a calculator.

It will take about 2.2 seconds for the diver to reach the water.

24 meters

5. The function d = 1~6 relates the distanced, In meters, that an object has fallen to its velocity v, in meters per second. Find the inverse of this function. What is the velocity of the cliff diver in meters per second as he encers the water?

You know that for any functionf, each x-value in the domain corresponds to exactly one y-value in the range. For a o ne-to-one function, it is also true that each y-value in the range corresponds to exactly one x-value in the domain. A one-to-one function f has an inverse j-1 that is also a function. If/maps a to b, then j - 1 must map b to a.

Domain off Range of f - 1

Range off Domain of f-1

b

Key Concept Composition of Inverse Functions

Ifjand j - 1 are inverse functions, then

u - l 0 f)(x) = X and (j 0 j - l )(x) = X for X in the domains ofj and j - l, respectively.

1his says that the composition of a function and its inverse Is essentially the identity function, id(x) = x, or y = x.

408 Chapter 6 Radical Functions and Rational Exponents

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Is this a function? Yes. For each value of x, dlere is only one value lory.

O:m~t§,l.j Composing Inverse Functions

For f (x) = x !_ 1, what is each of the following?

O r 1(x)

J(x) = x _: 1

1 Y = x _ 1 Rewrite the equation using y.

1 x = Y _ 1 Switch x and y.

x(y - 1) = 1 Solve for y.

1 y - l =:x

y=i+I

(!} (J oJ - 1)(1) B U- 1of)(l )

(fo rt)(1) = f(J- l{l)) u-l 0 JJ(1) = r1uc l)J

= !(t + 1)

= !(2)

= 2 ~1= ]

= 1- ~c .: 1) = J-t(~) undefined

1 is not in the domain off Therefore

u-l 0 f)( 1) does not exist.

Got It? 6 . Let g(x) = x ! 2. What is each of the following? a. g- 1(x) b. (g o g- 1)(0) C. (g-l o g)(O)

' Lesson Check Do you know HOW? Do you UNDERSTAND?

Find the inverse of each function. Is the inverse a function?

1. j{x) = 4x + 3

2. j{x) = x2 - 1

3. j{x) = (x + 1)2

4. For h(x) = x ! 2 , find: a. h- 1(x) b. h-1(4) c. Value of x for which rhe equality (h o h - t )(x) = x

does not hold.

5. Vocabulary Does every function have an inverse which is a function? Does every relation have an inverse which is a relation?

6. Reasoning A function consists of the pairs (2, 3), (x, 4), and (5, 6). What values, if any, may x not assume?

7. Error Analysis A classmate says that (f o g) -l(x) = (f- 1 o g- 1 )(x). Show that this is incorrect by finding examples of J(x) and g(x) for which the equation does not hold.

Q •l\'ZM6WlfS.l£13•1u. I Lesson 6-7 Inverse Relations and Functions 4 09

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Concept Byte For Use With Lesson 6-1

Properties of Exponents

Exponents indicate powers. The table below lists the properties of exponents. Assume thal no denominator is equaJ to zero and that m and n are integers.

~ Properties Properties of Exponents

• a0 = 1, a::/= 0 111

• f!_ = 0 m-n an

Example

• a- " = _!_ an • (ab)11 = a11b11

• (am)" = am"

Simplify and rewrite each expression using only positive exponents.

a. (5a3)( -3a-4) b. ( -4x-3y5)2

• am . a" =am+" • (f!)n = 0 n

b b11

c 4ab6dl · a5bdl

(Sa3)( -3a-4 ) = 5( - 3)a(3+(- <1)) ( -4x- 3y5)2 = ( -4)2(x-3)2(y5)2 4ab6dl = 4a(l - s)b(6 - I)c(3- 3)

a5bdl = - lsa- 1 = 16x-6yl 0 = 4a-4b5c0

- - 15 0 __!§_ 16yl0

-a,ra =~ =4~

a"

Exercises Simplify each expression. Use only positive exponents.

1. (2a3)(5a4) 2. ( - 3x2)( - 4x- 2) 3. (3x2y3)2

4. (3x-4r? 4a0 6 ]~~ 5. - 4 2a · 4x- 1

7 (6.x3)0 . 3xy2 a.(~r 9. ( -4m2n3)(2mn)

10. (2x3y1)- 2 (3r-2s-lt0)-3

11 · 3rs 12. (h7JC3)0

T2s4t6 14. 7. l~x 15. (s4 t)2(st) 13. ,J4 - 6 s I

( 1 tl 16. h - 2 • h3 17 _ l_ (a2b-3)-1 • a2b- 3

(r- ls2c3)-J 18. - 2 0 I r s l

19. Reasoning Your friend tells you that (k!)-5 = -k10. Did sh e apply tbe properties of exponents correctly? Explain why or why not.

360 Concept Byte Properties of Exponents

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1. Match each equation with its appropriate graph. Hint: Use end-behaviors and x-intercepts. A B C D _______ 3 42y x x= − _______ 3( 2)( 1)y x x= + − _______ 3 23y x x= − +

_______ 31 42

y x x= −

2. The graph of ( )y f x= is shown. Match each graph with its equation.

( )y f x= I II

_______ _______

A) ( )y f x= − B) ( )y f x= − C) ( ) 2y f x= − D) ( 2)y f x= +

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3. Find the minimum value and the zeros of the function algebraically, then sketch the function.

f (x) = x2 − 6x +5

Minimum value is __________.

The minimum occurs when x =________.

Zeros _____________________

4. Let f (x) = 2x2 , g(x) = 2x + 4 , and ( ) 2xh x = .

Find each of the following:

a) h( f (x)) : _______________ b) f (g(x)) :______________

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5. Let f (x) = 1

2x −1.

a) Find the inverse function f

−1(x) : _______________________

b) Sketch the graphs of f and f −1 .

6. Simplify using powers of the same base.

92n i 27n

3−n

_____________________

7. Simplify (2−2 + 4−2 )−1 _____________________

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8. Solve: 4x = 25 ⋅84 9. Solve: (8+ x)3 = 64 ___________ ___________

10. Solve for x : log x2 = 2

____________

11. Solve for x : ex−1 = 2

____________

12. Condense the expression using the properties of logarithms.

log2 6+ 2log2 4 _________________________

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13. Solve: 28−x = 41+x

_____________________

14. Suppose that $1200 is invested at an interest rate of 3.5%. How much

is the investment worth after 18 months if interest is compounded quarterly?

________________________

15. Give the domain, range and zeros for the function:

f (x) = x + 4

Domain: ________________

Range: _________________

Zeros: _________________

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16. Graph the function f (x) = 2− x . Identify at least three points that lie on the curve. Include these in your graph.

( , )

( , )

( , )

Factor each of the following. Be careful, there will be no partial credit.

17. a2 −18a +81

________________________

18. 16x2 − 9y2

________________________ 19. x3 − x2 + 2x − 2

_______________________ Simplify each of the following. All radicals must be in simplest form. There will be no partial credit.

20. 36x8 21.

1575

___________ ___________

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