Rational Functions and Applications
Now let’s consider the standard building blocks for rational functions.
To graph the reciprocal or inverse function y =
1
x
or y =
1
x2
, we create a table of values. The
only “trouble spot” for the domain in each case is the x value 0. Also, no matter how large or
small x may be, y could never quite equal 0. Thus, there are no x- and y-intercepts.
x
y
1
-4
-3
1
/ -3 or -0.333 . . .
1
-2
-1
-0.1
0
0.1
1
1
1
y
1
1
/ 16 or 0.0625
/ 9 or 0.111 . . .
-2
-1
-0.1
0
0.1
1
2
1
4
1
/ 4 or 0.25
1
3
y =
/ - 2 or -0.5
x
-3
1
x
/ 3 or 0.333 . . .
4
-4
y =
-1
-10
ERROR
10
1
1
/ 2 or 0.5
2
3
/ -4 or -0.25
1
/4 or 0.25
1
100
ERROR
100
1
x2
/4 or 0.25
/ 9 or 0.111 . . .
1
/ 16 or 0.0625
Notice intervals of increasing and decreasing y-values for each function, their domain and range,
the different symmetries, the vertical and horizontal asymptotes, and the odd and even functions.
As we consider variations of these two “building blocks”, you should see the transformation
ideas from previous lessons (e.g., translating the curve left or right, translating the curve up or
down, stretching the curve vertically, and so on).
Examples:
Graph the following rational functions. Include any vertical and horizontal asymptotes and
intercepts.
(1)
y =
1
x − 2
+ 5
The vertical asymptote is x = 2; the horizontal asymptote is y = 5; the y-intercept is 4.5
(shown in the table). The x-intercept is found by setting y equal to 0 and solving for x, as
shown: 0 =
+ 5 is equivalent to − 5 =
1
x − 2
1
x − 2
. Thinking of this as a proportion, we
set cross products equal and solve for x: − 5(x − 2) = 1 . This leads to − 5x + 10 = 1
and then to x = 1.8.
The curve is really just the rational curve y =
(2)
x
y
-8
0
1
1.9
2
2.1
3
4
12
4.9
4.5
4
-5
ERROR
15
6
5.5
5.1
y =
1
x − 2
2
1
x
shifted right 2 units and up 5 units.
y=5
x=2
+ 4
This curve has a horizontal asymptote y = 4 and range y: y > 4 . Its vertical asymptote
is x = 2, and its domain is x: x ≠ 2 . There are no x-intercepts, and the y-intercept is
4.25. The curve is identical to y = x12 shifted right 2 units and up 4 units.
x
y
-8
-3
-2
-1
0
1
4.01
4.04
4.0625
4.111 . . .
4.25
5
2
3
4
5
6
7
12
ERROR
5
4.25
4.111 . . .
4.0625
4.04
4.01
x=2
A rational function is any function that can be written as a ratio of polynomial functions. For
2x − 1
2x + 1
x 2 + 6x + 8
example, f(x) =
, g(x) =
,
and
h(
x
)
=
are all rational functions.
x + 2
x − 3
x2 − 1
All rational functions have asymptotes, and some have holes in the graph.
Graphing Tips:
(1)
First factor the numerator and denominator. Consider the function’s domain.
Canceled factors lead to holes in the graph; after the cancelling of factors, set the
denominator equal to zero to find vertical asymptotes.
x2 − 9
which has domain x ≠ -3 is really a line with a hole
x + 3
x2 − 9
x + 3 x − 3
in it. Since y =
=
= x − 3 , the apparent rational function is
x + 3
x + 3
really linear with a hole in the graph at (-3, -6).
For example, the function y =
x − 1
=
x2 − 1
x − 1
x + 1 x − 1
1
. The original
x + 1
1
function is rational with domain x ≠ 1, x ≠ -1. The graph would look just like y =
x
shifted left 1 unit. There is a hole at (1, 1) and a vertical asymptote at x = -1. The
horizontal asymptote is y = 0.
Another example: Consider y =
=
(2)
Find the x-intercepts (by setting y = 0 and solving for x) and the y-intercepts (by setting
x = 0 and solving for y). Setting the numerator equal to 0 is the essential step for finding
x-intercepts.
(3)
Find the vertical asymptote(s). [VA] Set the denominator equal to 0 and solve for x.
2x − 1
2x + 1
x 2 + 6x + 8
, g(x) =
,
and
h(
x
)
=
,
x + 2
x − 3
x2 − 1
the vertical asymptote(s) for f is x = -2, for g are x = 1 and x = -1, and for h is x = 3.
For example, given f(x) =
(4)
Find the horizontal asymptote. [HA] Consider the degree of the numerator (n) and the
degree of the denominator (d).
3 cases:
n<d
y = 0 is the only horizontal asymptote
leading coefficient of numerator
y =
is the HA.
leading coefficient of denominator
There is no HA.
n=d
n>d
2x − 1
2x + 1
x 2 + 6x + 8
, g(x) =
,
and
h(
x
)
=
,
x + 2
x − 3
x2 − 1
the horizontal asymptote for f is y = 2, for g is y = 0, and for h, there is none.
For example, given f(x) =
(5)
Use critical values (x-intercepts, holes, or vertical asymptotes) and a sign chart or an
actual table of values to complete the graph.
Putting it all together . . . examples:
(1)
y =
x − 2
x + 3
y=1
The vertical asymptote is x = -3; the horizontal asymptote is y = 1;
the y-intercept is − 23 and the x-intercept is 2 (shown in the table).
This curve is actually the y =
(x – 2) by (x + 3)]
1
x
x = -3
curve shifted left 3 units and up 1 unit. [Shown best by dividing
(2)
y =
x2 − 9
x + 3
As mentioned earlier, this has a surprising result. Since the numerator
factors nicely, the equation simplifies to be a linear function.
[y = x − x3 + x3 + 3 = x − 3 ] The slope of the line is 1, the x-intercept
(-3, -6)
is 3, and the y-intercept is -3. The only difference is that x cannot be -3,
so there is a “hole” in the graph at (-3, -6). A hole occurs anytime the
factors cancel.
(3)
y =
9
9 + x2
This also has an interesting result.
It is an even function with symmetry about the y-axis.
y=0
The y-intercept is 1. The curve approaches the horizontal
asymptote y = 0 as x gets very large or very small. There
is no x-intercept or vertical asymptote, and the curve is
roughly “bell-shaped”.
Exercises:
Give a complete graph of the following rational functions. Include any holes, asymptotes and
intercepts
x2 − 5 x + 4
2.
y =
x + 4
x − 2
3.
y =
5.
y =
x 2 − 16
x − 4
6.
y =
1.
y =
4.
y =
7.
The weight W, in pounds, of an object that is above the earth varies inversely as the
square of its distance d, in miles, from the center of the earth. Assume the radius of the
earth is 3960 miles. At sea level, an astronaut weighs W pounds. Find his weight when he
is x miles above the surface of the earth.
x2 − 1
1
x + 3
2
− 2
− 5
1
x + 2
x − 4
x 2 − 16
Use the given information to find the variation constant, then complete the chart, giving
weights to the nearest pound. Graph the relationship between distance above the earth (x)
and weight (W) over an appropriate domain. Place weight on the vertical axis.
x (mi)
0
d (mi)
3960
W (lb)
150
100
200
500
1000
2000
5000