Chapter Sixteen — Graphing Functions

The Humongous Book of Algebra Problems

369

The domain and range of k(x) both consist of all real numbers except for 0.

The graph is symmetric about the origin, as the reciprocal of a number x and its

opposite –x are opposites: k(–x) = –k(x).

Graphing Functions Using Transformations

Move, stretch, squish, and ip graphs

16.31 Transform the graph of f(x) = x

2

to graph the function g(x) = x

2

– 5.

The only difference between functions f(x) and g(x) is a constant, –5. Adding

or subtracting a constant c from a function affects its graph—all points on the

graph are shifted c units up or down, respectfully, on the coordinate plane.

In this problem, the graph of g(x) is exactly 5 units below the graph of f(x), as

illustrated by Figure 16-19.

Figure 16-19: The graphs of f(x) = x

2

and g(x) = x

2

– 5 are equivalent, except that g(x)

is ﬁve units below f(x).

16.32 Transform the graph of f(x) = x

2

to graph the function h(x) = (x – 3)

2

.

Whereas only x is squared in function f(x), the expression x – 3 is squared

in h(x). When a constant c is added to, or subtracted from, the input x of a

function, the corresponding graph is shifted –c units to the right or left, as

illustrated by Figure 16-20.

You can’t

have 0 in the

denominator, so the

domain must exclude

0. There’s no number

you can divide into 1 to

get a quotient of 0, so

you have to exclude

0 from the range as

well.

If you subtract 3

from the input of a

function, it moves the

graph 3 units to the

RIGHT. Adding 3 to

the input would move

it 3 units to the LEFT.

That’s the opposite of

what you might think—

negative numbers move

things right and positive

numbers move things

left.

Chapter Sixteen — Graphing Functions

The Humongous Book of Algebra Problems

370

Figure 16-20: Move the graph of f(x) = x

2

three units to the right to create the graph of

h(x) = (x – 3)

2

.

16.33 Transform the graph of to graph the function .

The function g(x) is generated by adding 1 to f(x). According to Problem 16.31,

adding the constant 1 to a function moves its graph up one unit, as illustrated

by Figure 16-21.

Figure 16-21: Move the graph of up one unit to generate the graph of

.

Chapter Sixteen — Graphing Functions

The Humongous Book of Algebra Problems

371

16.34 Transform the graph of to graph the function .

The reciprocal function f(x) divides 1 by the number substituted into the

function, x; in h(x), a constant (1) is added to the input (x). Problem 16.32

states that adding a constant c to the input of a function shifts the graph –c

units horizontally, so the graph of h(x) is equivalent to the graph of f(x)

shifted one unit to the left, as illustrated by Figure 16-22.

Figure 16-22: The graphs of and .

16.35 Graph the function .

Transform the graph of , presented in Figure 16-17. Multiplying a

function by a constant—in this case multiplying by 3—stretches its graph

vertically. Each point on the graph of j(x) is three times as far from the x-axis

as the corresponding point on .

Let’s say

you know

the graph of

a function f(x).

That means you

automatically

know the graph of

f(x) + 2 (move the graph

of f(x) up 2 units),

f(x) – 3 (move it down 3

units), f(x + 6) (move

it left 6 units), and

f(x – 4) (move it

right 4 units).

The ˆ symbol

used in

is

meant to distinguish

it from the given

function j(x). The

symbol itself means

nothing. It’s describing

a different (but very

similar) function using

a different (but

very similar)

name.

Multiply the

y-values of the points

by 3. The graph of

contains the

points (0,0), (1,1) and (4,2), so the graph of

j(x) contains the points (0,0

˙

3), (1,1

˙

3),

and (4,2

˙

3).

Start Free Trial

No credit card required