College Algebra Extra Credit Worksheet

Fall 2011 Math W1003 (3)Corrin Clarkson

Due: Thursday, October 20th, 2011

Solutions

Contents

Contents 1

1 Instructions 2

2 Graphing composite functions 32.1 Order matters . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

2.1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 32.1.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 62.1.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.2 Completing the square . . . . . . . . . . . . . . . . . . . . . . 92.2.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 92.2.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 102.2.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.3 Piecewise functions . . . . . . . . . . . . . . . . . . . . . . . . 142.3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 142.3.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . 152.3.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . 15

1

List of Tables

1 Point distribution . . . . . . . . . . . . . . . . . . . . . . . . . 32 Transformations by linear functions . . . . . . . . . . . . . . . 43 Horizonal and vertical transformations . . . . . . . . . . . . . 4

List of Figures

1 Reordering transformations . . . . . . . . . . . . . . . . . . . 52 Graphs of quadratic functions . . . . . . . . . . . . . . . . . . 103 Section 2.2 exercise solutions . . . . . . . . . . . . . . . . . . . 124 Section 2.2 problem solutions . . . . . . . . . . . . . . . . . . 135 Section 2.3 exercise solutions . . . . . . . . . . . . . . . . . . . 166 Section 2.3 problem solutions . . . . . . . . . . . . . . . . . . 17

1 Instructions

Each section of this extra credit work sheet is broken into three parts. Thefirst part gives a brief explanation of a technique or concept as well as a fewexamples. The second part is a series of practice exercises involving thattechnique or concept and the final part is a series of problems that you mustsolve to receive credit.

When solving the Exercises, you may work with others and ask for help.However, you must solve the Problems on your own. Submit only yoursolutions to the Problems as the Exercises will not be graded.

Each problem in this worksheet is worth 3 points for a total of 21 points.If you choose to complete this worksheet, it will count as a fifth section of thefirst exam. Thus the total number of points on the exam will change from62 to 83 and your score will increase by the number of points that you earnon this worksheet. If your recalculated exam score is lower than the scoreyou originally earned, your exam score will remain the same.

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2 Graphing composite functions

Composing a function f : R → R with a linear function g(x) = mx + btransforms the graph of f in a predictable way. In order to understand thistransformation, we write g as a composition of simpler functions: g = g2 ◦ g1where g1(x) = mx and g2(x) = x+ b. Table 2 summarizes how the graph off is transformed when f is composed with one of these simpler functions.

2.1 Order matters

2.1.1 Introduction

The order in which functions are composed can have a significant effect onthe result of the composition; putting on your socks and then your shoes isnot the same thing as putting on your shoes and then your socks. Similarly,the order in which a sequence of transformations is applied to a graph cangreatly affect the result. (See Example 1.)

It is helpful to know which types of transformations can be reorderedwithout effecting the resulting graph. First we break the transformationsinto two groups: those that act vertically and those that act horizontally.(See Table 3.)This corresponds to distinguishing between the linear func-tions g and G in a composition G ◦ f ◦ g. In this composition the functionsG and g do not interact with one another, thus the horizontal and verticaltransformations act independently on the graph. In practice, this means thatgiven a list of transformations one can always apply all the vertical transfor-mations and then all of the horizontal transformations without changing theend result. (See Example 2.)

Example 1. Consider the following two sequences of transformations:

• First sequence

Table 1: Point distribution

Order matters 6 pointsCompleting the square 6 pointsPiecewise functions 9 points

Total 21 points

3

Table 2: Transformations by linear functions

(a) Adding a constant

g(x) = x+ b b < 0 b > 0f ◦ g shift right by |b| units shift left by |b| unitsg ◦ f shift down by |b| units shift up by |b| units

(b) Scaling by a constant

g(x) = mx m = −1 0 < m < 1 1 < m

f ◦ g reflect across expand horizontally compress horizontallythe y-axis by a factor of 1

mby a factor of m

g ◦ f reflect across compress vertically expand verticallythe x-axis by a factor of 1

mby a factor of m

Table 3: Horizonal and vertical transformations

Horizontal Verticalshift right or left shift up or downstretch horizontally stretch verticallycompress horizontally compress verticallyreflect across the y-axis reflect across the x-axis

4

1. reflect across the x-axis

2. stretch vertically by a factor of two

3. shift up by ten units

• Second sequence

1. shift up by ten units

2. reflect across the x-axis

3. stretch vertically by a factor of two

The only difference between these sequences is the order of the transforma-tions. None the less, they have completely different effects on the identityfunction idR(x) = x. The first sequence transforms the identity to the linearfunction G1(x) = −2x + 10 where as the second sequence transforms theidentity into the linear function G2(x) = −2(x+ 10) = −2x− 20.

When the first sequence of transformations is a applied to the graph off(x) = x3 the resulting graph is that of G1◦f . Similarly, applying the secondsequence of transformations to the graph of f gives the graph of G2 ◦ f . (SeeFigure 1.)

Figure 1: Reordering transformations

(a) No transformations

-3 -2 -1 1 2 3

-20

-10

10

20

fHxL � x3

(b) First sequence

-3 -2 -1 1 2 3

-40

-20

20

40

60

fHxL � 10 - 2 x3

(c) Second sequence

-3 -2 -1 1 2 3

-60

-40

-20

20

fHxL � -2 Ix3+ 10M

Example 2. The following two sequences of transformations have the sameeffect on a graph.

• First sequence

1. shift up by 1 unit

2. shift right by two units

5

3. stretch vertically by a factor of three

4. compress horizontally by a factor of four

5. reflect across the x-axis

• Second sequence

1. shift up by 1 unit

2. stretch vertically by a factor of three

3. reflect across the x-axis

4. shift right by two units

5. compress horizontally by a factor of four

When either of these sequences of transformations is applied to the graph ofa function f , the resulting graph is that of G◦f ◦g where G(x) = −3(x+1) =−3x− 3 and g(x) = 4x− 2.

2.1.2 Exercises

Exercise 1. Consider the linear function g(x) = −12(x + 2). Write g as a

composition of simple linear functions of the type described in Table 2.Solution: g = γ3 ◦ g2 ◦ g1 where g1(x) = x + 1, g2(x) = 3x and g3(x) =

−x. Alternatively, we can distribute the negative one half and write g(x) =−1

2x− 1. This leads to the decomposition g = h3 ◦h2 ◦h1 where h1(x) = 1

2x,

h2(x) = −x and h3(x) = x− 1.

Exercise 2. Let f : R → R be any function and let G and g be the linearfunctions G(x) = 2x + 1 and g(x) = −1

2x + 3. Find a sequence of transfor-

mations that transforms the graph of f to that of G ◦ f ◦ g.Solution: The following sequence is one of many solutions:

1. stretch vertically by a factor of two

2. shift up by one unit

3. reflect across the y-axis

4. stretch horizontally by a factor of two

5. shift left by three units

6

Exercise 3. When the following sequence of transformations is applied tothe graph of a function f , the resulting graph is that of the function G◦f ◦gwhere G and g are linear functions. Find G and g.

1. compress vertically by a factor of four

2. shift up by two units

3. reflect across the x-axis

4. stretch horizontally by a factor of three

5. shift right by one unit

Solution: G(x) = −(12x+ 2) and g(x) = 1

3x− 1

2.1.3 Problems

Problem 1. Let f : R → R be any function and let G and g be the linearfunctions G(x) = −1

3x − 1 and g(x) = 4x + 2. Find a sequence of transfor-

mations that transforms the graph of f to that of G ◦ f ◦ g.Solution: The following sequence is one of many solutions:

1. compress vertically by a factor of three

2. shift up by one unit

3. reflect across the x-axis

4. compress horizontally by a factor of four

5. shift left by two units

Problem 2. When the following sequence of transformations is applied tothe graph of a function f , the resulting graph is that of the function G◦f ◦gwhere G and g are linear functions. Find G and g.

1. compress horizontally by a factor of 5

2. shift up by four units

3. reflect across the x-axis

7

4. stretch vertically by a factor of three

5. shift right by two units

Solution: G(x) = −3(x+ 4) = −3x− 12 and g(x) = 5x− 2

8

2.2 Completing the square

2.2.1 Introduction

Every quadratic polynomial function f(x) = ax2+bx+c (where a, b and c arereal numbers and a 6= 0) can be written as a composition f = G◦F ◦g whereF (x) = x2 is the simplest quadratic function, and both G(x) = Mx+B andg(x) = mx + b are linear functions. Thus we can graph f by transformingthe graph of F (x) = x2. (See Figure 2a.)

The technique of completing the square can be used to decompose f intoG ◦ F ◦ g. Consider the following examples.

Example 3. Let f(x) = x2 − 2x. We complete the square by adding andsubtracting 1 from f(x):

f(x) = x2 − 2x= x2 − 2x+ 1− 1= (x− 1)2 − 1

From this computation, we conclude that f(x) = (G◦F ◦g)(x) where F (x) =x2, and G(x) = g(x) = x−1. Thus the graph of f is simply that of F shiftedup by one unit and to the right by one unit. (See Figure 2b.)

Example 4. Let f(x) = 12x2 + 4x − 1. In order to complete the square

we must first factor out the coefficient on x2 and then add and subtract theappropriate constant:

f(x) = 12x2 + 4x− 1

= 12(x2 + 8x)− 1

= 12(x2 + 8x+ 4− 4)− 1

= 12(x2 + 8x+ 4)− 8− 1

= 12(x+ 4)2 − 9

It follows that f(x) = (G ◦ F ◦ g)(x) where F (x) = x2, G(x) = 12x − 9 and

g(x) = x+ 4. Thus the graph of f is that of F shifted to the left by 4 units,compressed vertically by a factor of 2, and then shifted down by 9 units. (SeeFigure 2c.)

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Figure 2: Graphs of quadratic functions

(a) Simplest quadratic

-3 -2 -1 1 2 3

2

4

6

8

FHxL � x2

(b) Example 3

-2 -1 1 2 3 4

2

4

6

8

fHxL � Hx - 1L2- 1

(c) Example 4

-10 -8 -6 -4 -2 2

-5

5

fHxL �

12

Hx + 4L2- 9

2.2.2 Exercises

Exercise 4. Decompose f(x) = x2+6x as a composition of the form G◦F ◦gand then graph f by transforming the graph of F .

Solution: Completing the square shows that f(x) = (x+ 3)2− 9. Thus fdecomposes as G ◦ F ◦ g where G(x) = x − 9 and g(x) = x + 3. The graphof f is that of F shifted left by three units and shifted down by nine units.(See Figure 3a.)

Exercise 5. Decompose f(x) = 13x2 − 2x as a composition of the form

G ◦ F ◦ g and then graph f by transforming the graph of F .Solution: f(x) decomposes as 1

3(x− 3)2 − 3. The functon f decomposes

as the composition G ◦ F ◦ g where G(x) = 13x− 3 and g(x) = x− 3. Thus

the graph of f is that of F shifted to the right by three units, compressedvertically by a factor of three and shifted down by three units. (See Figure3b.)

Exercise 6. Decompose f(x) = 4x2 − 4x + 6 as a composition of the formG ◦ F ◦ g and then graph f by transforming the graph of F .

Solution: f(x) can be rewritten as either (2x− 1)2 + 5 or 4(x− 12)2 + 5.

Thus f decomposes as either G1 ◦ F ◦ g1 or G2 ◦ F ◦ g2 where G1(x) =x + 5, g1(x) = 2x − 1, G2(x) = 4x + 5 and g2(x) = x − 1

2. From the first

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decomposition we concluded that the graph of f is that of F horizontallycompressed by a factor of 2, shifted right by one and then shifted up by fiveunits. Alternatively, the graph of f can be described as that F shifted rightby half of a unit, stretched vertically by a factor of four and then shifted upby four. (See Figure 3c.)

Exercise 7. Decompose f(x) = −x2 + 2x + 4 as a composition of the formG ◦ F ◦ g and then graph f by transforming the graph of F .

Solution: Completeing the square shows that f(x) = −(x− 1)2 + 5. Thefunction f decomposes as G ◦ F ◦ g where G(x) = −x+ 5 and g(x) = x− 1.Thus the graph of f is that F shifted to the right right by one unit, reflectedover the x-axis and shifted up by five units. (See Figure 3d.)

2.2.3 Problems

Problem 3. Decompose f(x) = 9x2 + 3x− 34

as a composition of the formG ◦ F ◦ g and then graph f by transforming the graph of F .

Solution:The following computation shows how to decompose f .

f(x) = 9x2 + 3x− 34

= 9x2 + 3x+ 14− 1

4− 3

4

= (3x+ 12)2 − 1

We conclude that f(x) = (G◦F ◦g)(x) where G(x) = x−1 and g(x) = 3x− 12.

Thus the graph of f is that of F compressed horizontally by a factor of three,shifted to the right by half of a unit and shifted down by one unit. (See Figure4a.)

Problem 4. Decompose f(x) = −2x2 + 12x − 14 as a composition of theform G ◦ F ◦ g and then graph f by transforming the graph of F .

Solution: The following computation shows how to decompose f .

f(x) = −2x2 + 12x− 14= −2(x2 − 6x)− 14= −2(x2 − 6x+ 9− 9)− 14= −2(x2 − 6x+ 9) + 18− 14= −2(x− 3)2 + 4

We conclude that f(x) = (G ◦ F ◦ g)(x) where G(x) = −2x + 4 and g(x) =x − 3. Thus the graph of f is that of F shifted to the right by three units,

11

Figure 3: Section 2.2 exercise solutions

(a) Exercise 4

-6 -4 -2

-5

5

fHxL � Hx + 3L2- 9

(b) Exercise 5

-2 2 4 6 8

-2

2

4

6

8

fHxL �

13

Hx - 3L2- 3

(c) Exercise 6

-4 -2 2 4

20

40

60

80

fHxL � H2 x - 1L2+ 5

(d) Exercise 7

-2 -1 1 2 3 4

-4

-2

2

4

fHxL � 5 - Hx - 1L2

12

reflected across the x-axis, stretched vertically by a factor of two and shiftedup by four units. (See Figure 4b.)

Figure 4: Section 2.2 problem solutions

(a) Problem 3

-2 -1 1 2

10

20

30

40

fHxL � I3 x +12

M2- 1

(b) Problem 4

-4 -2 2 4 6 8 10

-80

-60

-40

-20

fHxL � 4 - 2 Hx - 3L2

13

2.3 Piecewise functions

2.3.1 Introduction

When graphing compositions of piecewise functions with linear functions, itis generally easier to think in terms of transforming the graph of the origi-nal piecewise function, rather than computing the result of the composition.No the less, it is important to be able to compute such a composition. Thefunction that results from composing a piecewise function with another func-tion is a piecewise function. However, this resulting piecewise function maybe defined on different intervals than the original piecewise function. (SeeExample 2.)

Example 5. Consider the piecewise function f defined below and the linearfunction g(x) = 2x+ 1

f(x) =

4 x < 012x− 1 0 ≤ x ≤ 2x2 x > 2

The composition f ◦ g is the piecewise function is defined by the formula

f(g(x)) =

4 g(x) < 012g(x)− 1 0 ≤ g(x) ≤ 2

(g(x))2 g(x) > 2

=

4 x < −1

2

x− 12

−12≤ x ≤ 0

4x2 + 4x+ 1 x > 0

By contrast the composition g ◦ f is the piecewise function defined by theformulas

g(f(x)) =

g(4) x < 0g(1

2x− 1) 0 ≤ x ≤ 2

g(x2) x > 2

=

9 x < 0x− 1 0 ≤ x ≤ 22x2 + 1 x > 2

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2.3.2 Exercises

In each of the following exercises, f : R→ R is the absolute value function:

f(x) = |x| ={−x x < 0x x ≥ 0

Exercise 8. Graph fSolution: See Figure 5a.

Exercise 9. Graph f ◦g where g(x) = 2x+3. Using the notation of piecewisefunctions write a formula for (f ◦ g)(x).

Solution: See Figure 5b for the graph of f ◦ g.

(f ◦ g)(x) =

{−2x− 3 x < −3

2

2x+ 3 x ≥ −32

Exercise 10. Graph G ◦ f where G(x) = −x + 4. Using the notation ofpiecewise functions write a formula for (G ◦ f)(x).

Solution: See Figure 5c for the graph of G ◦ f .

(G ◦ f)(x) =

{x+ 4 x < 0−x+ 4 x ≥ 0

2.3.3 Problems

In each of the following problems, f : R→ R is the piecewise function definedby the formula

f(x) =

−x+ 2 x < −2x2 −2 ≤ x ≤ 21 x > 2

Problem 5. Graph f .Solution: See Figure 6a.

Problem 6. Graph f◦g where g(x) = 13x−1. Using the notation of piecewise

functions write a formula for (f ◦ g)(x).Solution: See Figure 6b for the graph of f ◦ g.

(f ◦ g)(x) =

−1

3x+ 3 x < −3

19x2 − 2

3x+ 1 −3 ≤ x ≤ 9

1 x < 53

15

Figure 5: Section 2.3 exercise solutions

(a) Exercise 8

-10 -5 5 10

2

4

6

8

10

fHxL �  x¤

(b) Exercise 9

-10 -5 5 10

5

10

15

20

fHxL �  2 x + 3¤

(c) Exercise 10

-10 -5 5 10

-6

-4

-2

2

4

fHxL � 4 -  x¤

16

Problem 7. Graph G ◦ f where G(x) = −4x + 5. Using the notation ofpiecewise functions write a formula for (G ◦ f)(x).

Solution: See Figure 6c for the graph of G ◦ f .

(G ◦ f)(x) = |x| =

4x+−3 x < −2−4x2 + 5 −2 ≤ x ≤ 2

1 x > 2

Figure 6: Section 2.3 problem solutions

(a) Problem 5

-10 -5 5 10

2

4

6

8

10

12

f

(b) Exercise 6

-10 -5 5 10 15

123456

f ëg

(c) Exercise 7

-10 -5 5 10

-40

-30

-20

-10

G ëf

17