Chapter Seventeen — Calculating Roots of Functions
The Humongous Book of Algebra Problems
384
Reduce the fraction to lowest terms.
The roots of h(x) are x = 4, , and .
Leading Coefficient Test
The ends of a function describe the ends of its graph
17.14 Explain how the leading coefficient test classifies the end behavior of a
function.
The leading coefficient test classifies the end behavior of a function based on its
degree and the sign of its leading coefficient (that is, the coefficient of the term
containing the variable raised to the highest power).
If the degree of the function is even, the left and right ends of the function
behave the same way. Furthermore, a positive leading coefficient indicates
that both ends of its graph “go up” (that is, increase without bound), whereas
a negative leading coefficient indicates that both ends of the graph “go down”
(that is, decrease without bound).
The end behavior of a function with an odd degree differs—the ends behave
oppositely. Specifically, the graph of a function with a positive leading
coefficient has a left end that goes down and a right end that goes up. The
converse is true for functions with a negative leading coefficient.
Write each
complex root
as the sum or
difference of two
fractions with the
same denominator
(instead of writing them
as bigger fractions
that contain a sum or
difference in the
numerator).
End behavior
is what the
function graph
does at its far
left and right sides.
Polynomial functions
will either increase or
decrease without bound
(shoot upward or shoot
downward) at the
edges of the
graph.
Look for the
term containing x
raised to the highest
power. That power is
the degree, and the
coefcient of that
term is the leading
coefcient.
Here’s the leading coefcient test in a nutshell
(LC stands for leading coefcient):
* Even degree and + LC: Both ends go up.
* Even degree and – LC: Both ends go down.
* Odd degree and + LC: Left side goes down; right side goes up.
* Odd degree and – LC: Left side goes up; right side goes down.
Chapter Seventeen — Calculating Roots of Functions
The Humongous Book of Algebra Problems
385
17.15 Describe the end behavior of f(x) = x
3
– 2 using the leading coefficient test.
Verify your answer by plotting the graph of f(x).
Function f(x) consists of two terms: x
3
and –2. The term containing x raised to
the highest degree is x
3
, so the degree of f(x) is 3, and the leading coefficient
is 1. According to the leading coefficient test, the graph of a function with an
odd degree and a positive leading coefficient has a left end that goes down
and a right end that goes up. Plot the graph of f(x) to verify this conclusion, as
illustrated by Figure 17-1.
Figure 17-1: The graph of f(x) = x
3
– 2 is the graph of y = x
3
shifted down two units.
The negative side of the graph decreases without bound, and the positive
side of the graph increases without bound. Put simply, the left end “goes
down” and the right end “goes up.”
17.16 Describe the end behavior of g(x) = x
2
+ 2x + 1 using the leading coefficient
test. Verify your answer by plotting the graph of g(x).
The highest power of x in g(x) appears in the term x
2
. Thus the degree of g(x) is
2, and the leading coefficient is 1. According to the leading coefficient test, the
graph of a function with an even degree and a positive leading coefficient goes
up at both ends. This conclusion is verified by the graph of g(x) in Figure 17-2.
There’s
no coefcient
written in front
of x
3
, so its
coefcient
is 1.
You can
graph g(x) using
a table of values,
but if you factor you
get g(x) = (x + 1)
2
. You
can graph that using
a transformation—it’s
the graph of x
2
shifted left one
unit.
Chapter Seventeen — Calculating Roots of Functions
The Humongous Book of Algebra Problems
386
Figure 17-2: The graph of g(x) = x
2
+ 2x + 1 has right and left ends that “go up” (that
is, increase without bound).
17.17 Describe the end behavior of h(x) = –3x
4
+ 5x
2
– 2 using the leading coefficient
test.
Function h(x) has degree 4 and leading coefficient –3. According to the leading
coefficient test, the graph of a function with an even degree and a negative
leading coefficient goes down at both its right and left ends.
17.18 Describe the end behavior of j(x) = 9x
2
– x
3
– 6 using the leading coefficient
test.
Function j(x) has degree 3 and its leading coefficient is –1. According to the
leading coefficient test, the graph of a function with an odd degree and a
negative leading coefficient has a left end that goes up and a right end that goes
down.
17.19 Describe the end behavior of k(x) = x(2x
4
+ 11x
2
– 1) using the leading
coefficient test.
Write k(x) as a polynomial, rather than a product of polynomials, by
distributing x.
k(x) = 2x
5
+ 11x
3
– x
The function has degree 5 and leading coefficient 2. According to the leading
coefficient test, the graph of a function with an odd degree and a positive
leading coefficient has a left end that goes down and a right end that goes up.
The terms
of j(x) aren’t
written in order
of their exponents,
from highest to
lowest. Even though –x
3
is written in the middle
of the function instead
of in the front, it still has
the highest exponent.
It’s okay to rewrite
the function as
j(x) = –x
3
+ 9x
2
– 6
if you want.
Chapter Seventeen — Calculating Roots of Functions
The Humongous Book of Algebra Problems
387
17.20 Consider the function p(x) = a(x + b)
2
+ c, defined such that a, b, and c are
nonzero real numbers. Describe the end behavior of p(x) if the vertex of its
graph is (2,–3) and abc < 0.
To plot the graph of p(x), transform the graph of y = x
2
, which has
vertex (0,0). The graph of p(x) has points a units farther away from the
x-axis and is shifted –b units horizontally and c units vertically.
Stretching the graph of y = x
2
by a multiple of a does not affect its vertex:
(0,0 · a) = (0,0). However, the vertex is affected by the vertical and horizontal
shifts; the vertex of p(x) is (0 – b, 0 + c). The problem states that p(x) has vertex
(2,–3). Therefore, (0 – b, 0 + c) = (2,–3). Set the corresponding x- and y-values
equal to calculate b and c.
Substitute b = –2 and c = –3 into p(x) and expand the expression.
The degree of p(x) is 2, and the leading coefficient is a. The problem states that
abc < 0. Substitute b and c into that expression.
The leading coefficient of p(x) is negative. According to the leading coefficient
test, the graph of a function with an even degree and a negative leading
coefficient has left and right ends that go down.
Quadratics
like p(x) have
U-shaped graphs
called parabolas.
If the ends of the
parabola go up, the
vertex is the lowest
point of the U, the
bottom of the valley.
If the ends of the
parabola go down, the
vertex is the highest
point of the
upside-down U.
To review how
graph transformations
work, check out
Problems 16.31–16.40.
If b were equal
to 7, for example, you
would shift the graph
of y = x
2
seven units to
the left.
The leading
coefcient of
p(x) is a, and you
just proved that it’s
negative (a ‹ 0).
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