Chapter Sixteen — Graphing Functions

The Humongous Book of Algebra Problems

360

graph of r(x) is (–4,5), and any horizontal line below (and including) y = 5

intersects the graph. This is even true for y = 2. Although the graph excludes

point (1,2), the horizontal line y = 2 passes through the graph of r(x) at two

other points. Therefore, the range is r(x) ≤ 5.

Symmetry

Pieces of a graph are reections of each other

16.21 Complete the graph of f(x) in Figure 16-8 assuming it is y-symmetric.

Figure 16-8: A portion of the graph of f(x).

A y-symmetric function is a reﬂection of itself across the y-axis, so if the graph

passes through point (x,y), it passes through the corresponding point (–x,y) as

well. For instance, the graph of f(x) in Figure 16-8 passes through point (4,–3),

so it must pass through (–4,–3) as well. Figure 16-9 presents the completed

graph of f(x).

If you

graphed f(x) on

a sheet of paper

and folded the paper

along the y-axis, the

left and right sides

of f(x) would

overlap.

Chapter Sixteen — Graphing Functions

The Humongous Book of Algebra Problems

361

Figure 16-9: The graph of f(x) is y-symmetric, so the segments of the graph left and right

of the y-axis are reﬂections of one another.

16.22 Complete the graph of g(x) in Figure 16-10 assuming that it is symmetric

about the origin.

Figure 16-10: A portion of the graph of g(x).

Chapter Sixteen — Graphing Functions

The Humongous Book of Algebra Problems

362

Origin-symmetric functions have this property: If the graph passes through

point (x,y), then it passes through the corresponding point (–x,–y) as well. For

instance, the graph of g(x) passes through point (–3,5), so it must pass through

the point (3,–5) as well. The completed graph is presented in Figure 16-11.

Figure 16-11: The graph of g(x) is origin-symmetric.

Start

at the origin

and trace your

nger left along

the graph. Whatever

the function does

as you travel left, it

does the opposite as

you travel right of the

origin. In this case, as

you travel left from

the origin, the func-

tion goes up to (–3,5)

and then down to

(–6,0). The right side

does the opposite—

it goes DOWN to

(3,–5) and then

UP to (6,0).

Chapter Sixteen — Graphing Functions

The Humongous Book of Algebra Problems

363

16.23 Complete the graph of h(x) in Figure 16-12 assuming that it is symmetric

about the x-axis.

Figure 16-12: A portion of the graph of h(x).

If a relation is symmetric about the x-axis, then the segments of its graph above

and below the x-axis are reﬂections of one other. Therefore, an x-symmetric

function passing through point (x,y) will pass through the corresponding point

(x,–y) as well. For instance, h(x) passes through point (1,2), so it also must pass

through point (1,–2). The completed graph is presented in Figure 16-13.

h(x) isn’t a

function because

it fails the vertical

line test (as do most

x-symmetric graphs).

The vertical line test

states that vertical

lines drawn on graphs of

functions can intersect

those graphs only once

at most. Vertical lines

right of x = –5 will

intersect h(x) twice, so

h(x) is not a function.

Chapter Sixteen — Graphing Functions

The Humongous Book of Algebra Problems

364

Figure 16-13: The x-symmetric graph of h(x).

16.24 Verify that f(x) = 4x

3

– x is an odd function.

A function is odd if and only if its graph is origin-symmetric. According to

Problem 16.22, an origin-symmetric graph that passes through point (x,y) also

passes through point (–x,–y). In other words, substituting –x into f(x) should

result in –f(x).

Because f(–x) = –f(x), function f(x) is odd.

16.25 Verify that is an even function.

A function is even if and only if its graph is y-symmetric. According to Problem

16.21, a y-symmetric graph that passes through point (x,y) must pass through

the corresponding point (–x,y) as well. In other words, substituting x and –x into

g(x) produces the same result, so g(–x) = g(x).

Here’s

how to

gure out

if f(x) is odd:

replace all of

the x’s with –x. You

should end up with

the exact opposite of

f(x). In this case you

get –4x

3

+ x, which is

the opposite of 4x

3

– x

because the pairs of

corresponding terms

(–4x

3

and 4x

3

, +x

and –x) are

opposites.

If (x,y) and

(–x,y) are on the

graph of g(x), then

g(x) = y and g(–x) = y.

If you substitute –x

into g(x), you should

end up with the

same expression

(y).

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