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The Humongous Book of Algebra Problems by W. Michael Kelley

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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 cant
have 0 in the
denominator, so the
domain must exclude
0. Theres 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 3stretches 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).

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