Write a formula for each polynomial function graphed.
37.
38.
39.
40.
41.
42.
43.
44.
3.3 Graphs of Polynomial Functions 187
Write a formula for each polynomial function graphed.
45.
46.
47.
48.
49.
50.
51. A rectangle is inscribed with its base on the x axis and its upper corners on the
parabola
2
y = 5 − x . What are the dimensions of such a rectangle that has the greatest
possible area?
52. A rectangle is inscribed with its base on the x axis and its upper corners on the curve
4
y =16 − x . What are the dimensions of such a rectangle that has the greatest
possible area?
188 Chapter 3
Section 3.4 Rational Functions
In the last few sections, we have built polynomials based on the positive whole number
power functions. In this section we explore functions based on power functions with
negative integer powers, called rational functions.
Example 1
You plan to drive 100 miles. Find a formula for the time the trip will take as a function
of the speed you drive.
You may recall that multiplying speed by time will give you distance. If we let t
represent the drive time in hours, and v represent the velocity (speed or rate) at which
we drive, then vt = distance . Since our distance is fixed at 100 miles, vt = 100.
Solving this relationship for the time gives us the function we desired:
100
1
t( v) =
100 −
=
v
v
While this type of relationship can be written using the negative exponent, it is more
common to see it written as a fraction.
This particular example is one of an inversely proportional relationship – where one
quantity is a constant divided by the other quantity, like
5
y = .
x
Notice that this is a transformation of the reciprocal toolkit function,
1
f ( x) =
x
Several natural phenomena, such as gravitational force and volume of sound, behave in a
manner inversely proportional to the square of another quantity. For example, the
volume, V, of a sound heard at a distance d from the source would be related by
k
V =
2
d
for some constant value k.
These functions are transformations of the reciprocal squared toolkit function
1
f ( x) =
.
2
x
We have seen the graphs of the basic reciprocal function and the squared reciprocal
function from our study of toolkit functions. These graphs have several important
features.
3.4 Rational Functions 189
1
f ( x) =
1
f ( x) =
x
2
x
Let’s begin by looking at the reciprocal function,
1
f ( x) = . As you well know, dividing
x
by zero is not allowed and therefore zero is not in the domain, and so the function is
undefined at an input of zero.
Short run behavior:
As the input values approach zero from the left side (taking on very small, negative
values), the function values become very large in the negative direction (in other words,
they approach negative infinity).
We write: as
−
x → 0 , f ( x) → −∞ .
As we approach zero from the right side (small, positive input values), the function
values become very large in the positive direction (approaching infinity).
We write: as
+
x → 0 , f ( x) → ∞ .
This behavior creates a vertical asymptote. An asymptote is a line that the graph
approaches. In this case the graph is approaching the vertical line x = 0 as the input
becomes close to zero.
Long run behavior:
As the values of x approach infinity, the function values approach 0.
As the values of x approach negative infinity, the function values approach 0.
Symbolically: as x → ±∞ , f ( x) → 0
Based on this long run behavior and the graph we can see that the function approaches 0
but never actually reaches 0, it just “levels off” as the inputs become large. This behavior
creates a horizontal asymptote. In this case the graph is approaching the horizontal line
f ( x) = 0 as the input becomes very large in the negative and positive directions.
Vertical and Horizontal Asymptotes
A vertical asymptote of a graph is a vertical line x = a where the graph tends towards positive or negative infinity as the inputs approach a. As x → a , f ( x) → ±∞ .
A horizontal asymptote of a graph is a horizontal line y = b where the graph
approaches the line as the inputs get large. As x → ±∞ , f ( x) → b .
190 Chapter 3
Try it Now:
1. Use symbolic notation to describe the long run behavior
and short run behavior for the reciprocal squared function.
Example 2
Sketch a graph of the reciprocal function shifted two units to the left and up three units.
Identify the horizontal and vertical asymptotes of the graph, if any.
Transforming the graph left 2 and up 3 would result in the function
1
f ( x) =
+ 3 , or equivalently, by giving the terms a common denominator,
x + 2
3 x + 7
f ( x) =
x + 2
Shifting the toolkit function would give us
this graph. Notice that this equation is
undefined at x = -2, and the graph also is
showing a vertical asymptote at x = -2.
As x
2−
→ − , f ( x) → −∞ , and as
x
2+
→ − , f ( x) → ∞
As the inputs grow large, the graph appears
to be leveling off at output values of 3,
indicating a horizontal asymptote at y = 3.
As x → ±∞ , f ( x) → 3 .
Notice that horizontal and vertical asymptotes get shifted left 2 and up 3 along with the
function.
Try it Now
2. Sketch the graph and find the horizontal and vertical asymptotes of the reciprocal
squared function that has been shifted right 3 units and down 4 units.
In the previous example, we shifted a toolkit function in a way that resulted in a function
x +
of the form
3
7
f ( x) =
. This is an example of a more general rational function.
x + 2
3.4 Rational Functions 191
Rational Function
A rational function is a function that can be written as the ratio of two polynomials,
P(x) and Q(x).
2
P( x)
p
a + a x + a x ++ a x
0
1
2
f ( x)
p
=
=
2
Q( x)
q
b + b x + b x ++ b x
0
1
2
q
Example 3
A large mixing tank currently contains 100 gallons of water, into which 5 pounds of
sugar have been mixed. A tap will open pouring 10 gallons per minute of water into the
tank at the same time sugar is poured into the tank at a rate of 1 pound per minute. Find
the concentration (pounds per gallon) of sugar in the tank after t minutes.
Notice that the amount of water in the tank is changing linearly, as is the amount of
sugar in the tank. We can write an equation independently for each:
water = 100 + t
10
sugar = 5 + t
1
The concentration, C, will be the ratio of pounds of sugar to gallons of water
5 + t
C t() =
100 + t
10
Finding Asymptotes and Intercepts
Given a rational function, as part of investigating the short run behavior we are interested
in finding any vertical and horizontal asymptotes, as well as finding any vertical or
horizontal intercepts, as we have done in the past.
To find vertical asymptotes, we notice that the vertical asymptotes in our examples occur
when the denominator of the function is undefined. With one exception, a vertical
asymptote will occur whenever the denominator is undefined.
Example 4
2
+
Find the vertical asymptotes of the function
5 2
k( )
x
x =
2
2 − x − x
To find the vertical asymptotes, we determine where this function will be undefined by
setting the denominator equal to zero:
2
2
− x − x = 0
(2 + x 1
)( − x) = 0
x = − ,
2 1
192 Chapter 3
This indicates two vertical asymptotes, which a
look at a graph confirms.
The exception to this rule can occur when both the numerator and denominator of a
rational function are zero at the same input.
Example 5
−
Find the vertical asymptotes of the function
x 2
k( x) =
.
2
x − 4
To find the vertical asymptotes, we determine where this function will be undefined by
setting the denominator equal to zero:
2
x − 4 = 0
2
x = 4
x = 2,
− 2
However, the numerator of this function is also
equal to zero when x = 2. Because of this, the
function will still be undefined at 2, since 0 is
0
undefined, but the graph will not have a vertical
asymptote at x = 2.
The graph of this function will have the vertical
asymptote at x = -2, but at x = 2 the graph will
have a hole: a single point where the graph is not
defined, indicated by an open circle.
Vertical Asymptotes and Holes of Rational Functions
The vertical asymptotes of a rational function will occur where the denominator of the
function is equal to zero and the numerator is not zero.
A hole might occur in the graph of a rational function if an input causes both numerator
and denominator to be zero. In this case, factor the numerator and denominator and
simplify; if the simplified expression still has a zero in the denominator at the original
input the original function has a vertical asymptote at the input, otherwise it has a hole.
3.4 Rational Functions 193
To find horizontal asymptotes, we are interested in the behavior of the function as the
input grows large, so we consider long run behavior of the numerator and denominator
separately. Recall that a polynomial’s long run behavior will mirror that of the leading
term. Likewise, a rational function’s long run behavior will mirror that of the ratio of the
leading terms of the numerator and denominator functions.
There are three distinct outcomes when this analysis is done:
Case 1: The degree of the denominator > degree of the numerator
x +
Example:
3
2
f ( x) =
2
x + 4 x − 5
In this case, the long run behavior is
3
3
( )
x
f x ≈
= . This tells us that as the inputs grow
2
x
x
large, this function will behave similarly to the function
3
g( x) = . As the inputs grow
x
large, the outputs will approach zero, resulting in a horizontal asymptote at y = 0.
As x → ±∞ , f ( x) → 0
Case 2: The degree of the denominator < degree of the numerator
2
x +
Example:
3
2
f ( x) =
x − 5
2
In this case, the long run behavior is
3
( )
x
f x ≈
= 3 x . This tells us that as the inputs
x
grow large, this function will behave similarly to the function g( x) = 3 x . As the inputs grow large, the outputs will grow and not level off, so this graph has no horizontal
asymptote.
As x → ±∞ , f ( x) → ±∞ , respectively.
Ultimately, if the numerator is larger than the denominator, the long run behavior of the
graph will mimic the behavior of the reduced long run behavior fraction. As another
5
2
−
example if we had the function
3
( )
x
x
f x =
with long run behavior
x + 3
5
3 x
4
f ( x) ≈
= 3 x , the long run behavior of the graph would look similar to that of an
x
even polynomial, and as x → ±∞ , f ( x) → ∞ .
Case 3: The degree of the denominator = degree of the numerator
2
x +
Example:
3
2
f ( x) =
2
x + 4 x − 5
194 Chapter 3
2
In this case, the long run behavior is
3
( )
x
f x ≈
= 3 . This tells us that as the inputs grow
2
x
large, this function will behave like the function g( x) = 3, which is a horizontal line. As x → ±∞ , f ( x) → 3 , resulting in a horizontal asymptote at y = 3.
Horizontal Asymptote of Rational Functions
The horizontal asymptote of a rational function can be determined by looking at the
degrees of the numerator and denominator.
Degree of denominator > degree of numerator: Horizontal asymptote at y = 0
Degree of denominator < degree of numerator: No horizontal asymptote
Degree of denominator = degree of numerator: Horizontal asymptote at ratio of leading
coefficients.
Example 6
In the sugar concentration problem from earlier, we created the equation
5 + t
C t() =
.
100 + t
10
Find the horizontal asymptote and interpret it in context of the scenario.
Both the numerator and denominator are linear (degree 1), so since the degrees are
equal, there will be a horizontal asymptote at the ratio of the leading coefficients. In the
numerator, the leading term is t, with coefficient 1. In the denominator, the leading
term is 10 t, with coefficient 10. The horizontal asymptote will be at the ratio of these
values: As t → ∞ ,
1
C( t) →
. This function will have a horizontal asymptote at
10
1
y =
.
10
This tells us that as the input gets large, the output values will approach 1/10. In
context, this means that as more time goes by, the concentration of sugar in the tank will
approach one tenth of a pound of sugar per gallon of water or 1/10 pounds per gallon.
Example 7
Find the horizontal and vertical asymptotes of the function
( x − 2)( x + )
3
f ( x) =
( x − )(
1 x + 2)( x − )
5
First, note this function has no inputs that make both the numerator and denominator
zero, so there are no potential holes. The function will have vertical asymptotes when
the denominator is zero, causing the function to be undefined. The denominator will be
zero at x = 1, -2, and 5, indicating vertical asymptotes at these values.
3.4 Rational Functions 195
The numerator has degree 2, while the denominator has degree 3. Since the degree of
the denominator is greater than the degree of the numerator, the denominator will grow
faster than the numerator, causing the outputs to tend towards zero as the inputs get
large, and so as x → ±∞ , f ( x) → 0 . This function will have a horizontal asymptote at y = 0.
Try it Now
3. Find the vertical and horizontal asymptotes of the function
(2 x − )(
1 2 x + )
1
f ( x) =
( x − )(
2 x + )
3
Intercepts
As with all functions, a rational function will have a vertical intercept when the input is
zero, if the function is defined at zero. It is possible for a rational function to not have a
vertical intercept if the function is undefined at zero.
Likewise, a rational function will have horizontal intercepts at the inputs that cause the
output to be zero (unless that input corresponds to a hole). It is possible there are no
horizontal intercepts. Since a fraction is only equal to zero when the numerator is zero,
horizontal intercepts will occur when the numerator of the rational function is equal to
zero.
Example 8
x −
x +
Find the intercepts of
(
)(
2
)
3
f ( x) =
( x − )(
1 x + 2)( x − )
5