supplementary section: transformations and graphing ... · 1.1 supplementary section:...

1.1 Supplementary Section: Transformations and Graphing Techniques 1

1.2aIf f (x) = x2 and g(x) = 2, we can think of each of the following as an algebraic combination of f and g.

y = 2x2 = g(x)f (x) y = x2 + 2 = f (x) + g(x) y = x2 __ 2 = f (x) ____ g(x)

So the graphing techniques we will learn next are closely related to algebra with functions. Our goal with the graphing techniques is to see how these combination of functions look on a graph.

Why Learn Graphing Techniques?

With all the graphing soft ware available to us, you may be wondering why it is nec-essary to learn any graphing techniques at all. Th e answer is that we want to obtain a basic, intuitive, sense of what our graphs will look like before we do any graphing. It is also important to know how diff erent numbers in the expression of a function aff ect the graph of that function. For example, you know intuitively what the graph of y = x 2 looks like. Do you also know how the number 2, in the functions below, aff ects the graph of y = x 2 ? y = 2x 2 y = x 2 + 2 y = (x + 2) 2 If not, you will aft er you study this section. Knowing how these numbers aff ect the graph of the underlying function means that you have a more complete under-standing of the relationship between a function and its graph.

We are not going to go into extreme detail with our techniques of graphing. We are going to do just enough to give you some good graphing tools that you can use in the next math class you take. To begin, let’s list the six basic graphs that you should memorize.

Six Basic Graphs

y = f (x) = x f (x) = x2 f (x) = x3

FIGURE 5a FIGURE 5b FIGURE 5c

Domain: All real numbers Range: All real numbers

Domain: All real numbers Range: {y | y ≥ 0}

Domain: All real numbers Range: All real numbers

x

y

−5−5−5 555

−−55

55

−5

5

x

y

−−55

55

−5

5

−5−5−5 555x

y

−−55 55

−−55

55

−5 5

−5

5

Supplementary Section:

Transformations and Graphing Techniques

dac_12a.indd 1dac_12a.indd 1 8/6/12 10:46 AM8/6/12 10:46 AM

2 Chapter 1

f (x) = 1 _ x

f (x) = |x| f (x) = √— x

FIGURE 5d FIGURE 5e FIGURE 5f

Domain: All reals except x = 0 Range: All reals except y = 0

Domain: All real numbers Range: {y | y ≥ 0}

Domain: {x | x ≥ 0} Range: {y | y ≥ 0}

Vertical Translations

Graph y = x2 − 4.SOLUTION We make a table as we did in the previous example. If the vertical trans-lation idea works for this type of equation, as it did with our straight lines, we expect this graph to be the graph of y = x2 translated down 4 units.

As you can see, the graph of y = x2 − 4 is the graph of y = x2 translated down 4 units.

Graph the equations y = − 1 __ 3 x, y = − 1 __ 3 x � 2, and

y = − 1 __ 3 x − 4.

SOLUTION We create a table, then we graph the information in the table. However, even before we start, we are expecting the graph of y = − 1 __ 3 x � 2 to be 2 units

x

y

−−55

55

−5

5

−5−5−5 555x

y

−5−5−5 555

−−55

55

−5

5

x

y

−5−5−5 555

−−55

55

−5

5

EXAMPLE 6

Input Calculate Using Output Form x the Equation y Ordered Pairs

−3 y = (−3)2 − 4 = 5 (−3, 5) −2 y = (−2)2 − 4 = 0 (−2, 0) −1 y = (−1)2 − 4 = −3 (−1, −3) 0 y = (0)2 − 4 = −4 (0, −4) 1 y = (1)2 − 4 = −3 (1, −3) 2 y = (2)2 − 4 = 0 (2, 0) 3 y = (3)2 − 4 = 5 (3, 5)

FIGURE 6

1–1–2–3–4–5 3 4 5

11

33322

4455

7766

8899

–1–1x

1 12–1–2–3–4–5 3 4 5

1

332

45

76

89

–1

–3–3–3–3–3–4

y

–2–2–2–2–2–2–2–2

y � x2

y � x2 � 3

(�1, �2)(�1, �3) (1, �2)(1, �3)

(1, �2)(0, �4)

(3, 9)(3, 6)(3, 6)

(2, 1)(2, 1)(�2, 1)2, 1)

(�3, 6)3, 6) (3, 9)(3, 5)

(2, 0)(�2, 0)

(�3, 5)

EXAMPLE 7 EXAMPLE 7



above the graph of y = − 1 __ 3 x and the graph of y = − 1 __ 3 x − 4 to be 4 units below the graph of y = − 1 __ 3 x .

From our previous two examples we can generalize as follows.

Match each relation with its graph, then indicate which relations are functionsa. y = � x � − 4 b. y = x2 − 4 c. y = 2x + 2

SOLUTION Using the basic graphs for a guide along with our knowledge of transla-tions, we have the following:

a. Figure 9 b. Figure 8 c. Figure 10

And, since all of these graphs pass the vertical line test, all are functions.

Input Output Output Output x f (x) g(x) h(x)

−3 1 3 −3

0 0 2 −4

3 −1 1 −5

x

y

1–1–2–3–4–5 3 4 5

1

–3–4–5

345

–2–1 1–1–2–3–4–5 3 4 5

1

–3–4–5

345

–2–1

y y = − = − x + x + 221133g(x) = − x + 21

3

y y = − = − x + x + 221133h(x) = − x − 41

3

y y = − = − xx1133f(x) = − x1

3(−3, −3)(−3, −3)

(0, −2)(0, −4)(3, −1)(3, −5)

FIGURE 7

If k > 0, then The Graph of Is the Graph of y = f (x) translated

y = f (x) + k k units up y = f (x) − k k units down

RULE Vertical Translations

EXAMPLE 8

x

y

1–2–3–4–5 3 4 5

–3–4–5

2345

–2–1 1–2–3–4–5 32 4 5

–3–4–5

2345

–2–1 2

FIGURE 10

x

y

1–2–3–4–5 3 4 5

1

–3–4–5

2345

–2–1 1–2 –2–1–3–4–5 32 4 5

1

–3–4–5

2345

–2–1 2

FIGURE 8

x

y

1–2–3–4–5 3 4 5

1

–3–4–5

2345

–2–1 1–2 –2–1–3–4–5 32 4 5

1

–3–4–5

2345

–2–1 2

FIGURE 9


4 Chapter 1

Horizontal Translations

Now consider the graphs of f (x) = (x + 2)2 and g(x) = (x − 2)2 shown in Figure 11. Compared with the graph of the basic function y = x2, the graph of f (x) = (x + 2)2 is shift ed two units to the left , as shown by the arrows. Similarly, the graph of g(x) = (x − 2)2 is shift ed two units to the right compared to the graph of y = x2.

We can generalize the information in Figure 11a and 11b as follows.

If h > 0, then The Graph of Is the Graph of y = f (x) translated

y = f (x + h) h units to the left y = f (x − h) h units to the right

RULE Horizontal Translations

x y = x2 f (x) = (x + 2)2

−3 9 1

−2 4 0

−1 1 1

0 0 4

1 1 9

2 4 16

3 9 25

x y = x2 g(x) = (x − 2)2

−3 9 25

−2 4 16

−1 1 9

0 0 4

1 1 1

2 4 0

3 9 1

x

y

–2–2 2

44

88

2

h(x)g(x)

FIGURE 11b

x

y

–2–2 2

44

88

2

hh((xx))f (x)

FIGURE 11a



State the domain and then graph the function.

h(x) = 1 _ (x − 3)2

SOLUTION Because we cannot divide by 0, our domain will be {x | x ≠ 3} because

3 is the only number that causes the denominator to be 0. Th e table shows that each y-value for h(x) occurs three units to the right of the same y-value for the graph of y = 1 _

x2 . Consequently, each point on the graph of y = h(x) is shift ed three

units to the right of y = 1 _ x2

, as shown in Figure 12.

Our next example involves both a horizontal translation and a vertical transla-tion of a graph.

Graph y = � x − 3 � + 4.SOLUTION Our graph will be a transformation of the basic graph y = � x � . We pro-ceed according to the rule for order of operations which tells us to go within the fi rst grouping symbol fi rst and then work our way out.

EXAMPLE 9

x y = 1 __ x2 h(x) = 1 _______

(x − 3)2

−1 1 1/16

0 undefined 1/9

1 1 1/4

2 1/4 1

3 1/9 undefined

4 1/16 1

x

y

2

11

2

5–5 5–5

hh((xx))h(x)

FIGURE 12

EXAMPLE 10

Step 1: y = x

Step 2:

Translate 3 units right

y = x − 3 + 4

Step 3: Translate 4 units up

x

y

55

55

SteStepp 22

SteStepp 33 TTrraannssllaattee 44 ununiits ts SteStepp 11 iis s yy == xx

Step 2:

Step 3: Translate 4 units up is y = x

3

Step 1:

FIGURE 13


6 Chapter 1

Scale Factors

We have seen that adding a constant to the expression defi ning a function results in a translation of its graph. What happens if we multiply the expression by a con-stant? Consider the graphs of the functions f (x) = 2x2, g(x) = 1 _ 2 x 2 , and h(x) = − x 2 shown in Figure 14a, b, and c, and compare each to the graph of y = x 2 .

x y = x 2 f (x) = 2x2

−2 4 8

−1 1 2

0 0 0

1 1 2

2 4 8

x

y

5–5 5–5

––5

5

–5

5ff ((xx)) == 22xx22

yy == xx22

f (x) = 2x2

y = x2

FIGURE 14a

x y = x 2 g(x) = 1 __ 2

x 2

−2 4 2

−1 1 1 _ 2

0 0 0

1 1 1 _ 2

2 4 2

x

y

5–5 5–5

––5

5

–5

5 yy == xx22y = x2

gg((xx)) == 11 __ 22 xx22g(x) = 1 _ 2 x2

FIGURE 14b

x y = x 2 h(x) = − x 2

−2 4 −4

−1 1 −1

0 0 0

1 1 −1

2 4 −4

x

y

5–5 5–5

––5

5

–5

5yy == xx22y = x2

hh((xx)) == −−xx22h(x) = −x2

FIGURE 14c



Compared to the graph of y = x 2 , the graph of f (x) = 2x2 is expanded, or stretched,vertically by a factor of 2. Th e y-coordinate of each point on the graph has been doubled, as you can see in the table of values, so each point on the graph of fis twice as far from the x-axis as its counterpart on the basic graph y = x 2 . Th e graph of g(x) = 1 _ 2 x 2 is compressed vertically by a factor of 1 _ 2 ; each point is half as far from the x-axis as its counterpart on the graph of y = x 2 . Th e graph of h(x) = − x 2 is fl ipped, or refl ected, about the x-axis; the y-coordinate of each point on the graph of y = x 2 is replaced by its opposite.

In general, we have the following:

Compared with the graph of y = f (x), the graph of y = af (x), where a ≠ 0, is1. stretched vertically by a factor of � a � if � a � > 1,2. compressed vertically by a factor of � a � if 0 < � a � < 1, and3. refl ected about the x-axis if a < 0.

RULE Scale Factors and Refl ections

Th e constant a is called the scale factor for the graph.

Graph f (x) = 1 __ 4 (x + 4)2 + 2.

SOLUTION We identify the basic graph from the structure of the formula for f (x). In this case, the basic graph is y = x2. We will perform the translations separately, following the order of operations. Here is a diagram showing how we interpret the information in the equation, followed by Figure 15, which shows how we construct the graph.

EXAMPLE 11

Step 1: y = x2

Step 2:

Step 4: Translate 2 units up

x

y

–5–5–5

55

Step 1:Step 1: GGrraapph h yy == xx22

Step 2:Step 2:

Step 4:Step 4: ccuurrvve e uup 2 unp 2 uniitsts

Step 3:Step 3: FFllaatttten en ccuurrvve e bby y 11 __ 44 yy

Step 1: Graph y = x2

Step 2:

Step 4: curve up 2 units

Step 3: Flatten curve by 1 _ 4 y

FIGURE 15


8

Problem Set 1.2a

For each problem below, the equation of the red curve is given. Find the equation for the blue graph.

1.

x

y

1–2–3–4–5 3 4 5

1

–3–4–5

2345

–2–1 1–2 –2–1–3–4–5 32 4 5

1

–3–4–5

2345

–2–1 2

y � xy � x

2.

x

y

1–2–3–4–5 3 4 5

1

–3–4–5

2345

–2–1 1–2 –2–1–3–4–5 32 4 5

1

–3–4–5

2345

–2–1 2

y � x

3.

x

y

1–2–3–4–5 3 4 5

1

–3–4–5

2345

–2–1 1–2 –2–1–3–4–5 32 4 5

1

–3–4–5

2345

–2–1 2

y � |x|y � |x|

4.

x

y

1–2–3–4–5 3 4 5

1

–3–4–5

2345

–2–1 1–2 –2–1–3–4–5 32 4 5

1

–3–4–5

2345

–2–1 2

y � � x2233y � x23

Graph each of the following. Use one coordinate system for each problem.

5. a. y = 2x

b. y = 2x + 3

c. y = 2x − 5

6. a. y = 1 __ 3 x

b. y = 1 __ 3 x + 1

c. y = 1 __ 3 x − 3

7. a. y = 1 __ 2 x2

b. y = 1 __ 2 x2 − 2

c. y = 1 __ 2 x2 + 2

8. a. y = 2x2

b. y = 2x2 − 8

c. y = 2x2 + 1

Identify each graph as a translation of a basic function, and write a formula for the graph.

9.

x

y

–2 2–2 2

22

–2–2

10.

x

y

–2 2–2 2

22

–2–2


1.1 Problem Set 9

11.

x

y

22 44

22

–2–2

12.

x

y

–2 2–2 2

22

–2–2

Sketch the basic graph and the graph of the given function on the same axes. Label the coordinates of two or three points on the graph of the given function.

13. f (x) = � x � − 2 14. g(x) = (x + 1)3 15. f (s) = s 2 + 3

16. G(t) = √— t − 2 17. G(r) = (r + 2)3 18. F(r) = 1 _ r − 4

19. H(d) = √— d − 3 20. h(v) = 1 _ v + 6 21. f (x) = 1 _ 3 � x �

22. H(x) = −3 � x � 23. g(z) = 2 _ z 24. G(v) = −3 √— v

25. f (s) = 1 _ 8 s 3 26. H(x) = 1 _ 3x

27. Match each graph with its equation.

a.

x

f(x)

–4 4–4 4

44

88

–4–4

b.

x

f(x)

–2 2–2 2

88

–8–8

c.

x

f(x)

–4 4–4 4

44

–4–4

d.

x

f(x)

44

44

i. f (x) = 3 √— x ii. f (x) = 3 _ x iii. f (x) = 2 x 3 iv. f (x) = 3 x 2

28. Match each graph with its equation.

a.

x

f(x)

–4 4–4 4

44

88

–4–4

b.

x

f(x)

–4 4–4 4

44

–4–4

c.

x

f(x)

–4 4–4 4

44

–4–4

d.

x

f(x)

–4 4–4 4

44

88

–4–4


10 Chapter 1

i. f (x) = x 3 − 2 ii. f (x) = � x � − 3 iii. f (x) = x 2 + 3 iv. f (x) = √— x � 3

Sketch the basic graph and the graph of the given function by hand on the same axes. Label the coordinates of two or three points on the graph of the given function.

29. f (x) = 2 + (x − 3)2 30. f (x) = (x + 4)2 + 1

31. g(z) = 1 _ z + 2 − 3 32. g(z) = 1 ____ z − 1 + 1

33. F(u) = −3 √—

u + 4 + 4 34. F(u) = 4 √—

u − 3 − 5

35. G(t) = 2 � t − 5 � − 1 36. G(t) = 2 − � t + 4 �


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