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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
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
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) isnt 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).
Heres
how to
gure out
if f(x) is odd:
replace all of
the xs 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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