# Graphs of Rational Functions

These are my notes and worked examples on graphs of rational functions.

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Steps for Analyzing a Graph
1. Factor the numerator and denominator and find the domain. If 0 is in the
domain, find the y-intercept, f(0), and plot it.

2. Write the function in lowest terms as $$\frac{p(x)}{q(x)}$$ and find the real
zeros of the numerator. That is finding the real solution of the equation
p(x)=0. These are the x-intercepts of the graph. Determine the behavior of the
graph near each x-intercept, using the same procedure as for polynomial
functions. Plot each x-intercept and indicate the behavior of the graph.

3. With the function written in lowest terms, find the real zeros of the
equation q(x)=0. These determine the vertical asymptotes of the graph. Graph
each vertical asymptote using a dashed line.

4. Locate any horizontal or oblique asymptotes using the procedure given in the
previous section. Graph the asymptotes using a dashed line. Determine the points
at which the graph intersects these asymptotes. Plot any such points.

5. Using the real zeros of the numerator and the denominator of the given
equation, divide the x-axis into intervals and determine where the graph is
above the x-axis and where it is below the x-axis by choosing a number in each
interval and evaluating the function there. Plot the points found.

6. Analyze the behavior of the graph near each asymptote and indicate this
behavior in the graph.

7. Put all the information together to obtain the graph of the function given.

Example 1
Analyze the graph of a rational function
$$r(x)=\frac{x-1}{x^2-4}$$

Solution:
1. We factor the numerator and denominator:
$$r(x)=\frac{x-1}{(x+2)(x-2)}$$
The domain is $${x|x \not=2, x \not =-2}$$
The y-intercept is $$r(0)=\frac{-1}{-4} = \frac{1}{4}$$
Plot the point $$0,\frac{1}{4}$$

2. Now that r is in lowest terms, the real zeros of the numerator satisfies the
equation x-1=0. The only x-intercept is 1.
Near 1: $$r(x)=\frac{x-1}{(x+2)(x-2)} = \frac{x-1}{(1+2)(1-2} = \frac-{1}{3}(x-1)$$
Plot the point (1,0) and indicate a line with slope $$-\frac{1}{3}$$ there.

3. R is in lowest terms. The real zeros of the denominator are the real
solutions of the equation (x+2)(x-2)=0, so -2 and 2. The graph of r has two
vertical asymptotes, the lines x=2 and x=-2. Graph each of these asymptotes
using dashed lines.

4. The degree of the numerator is less than the degree of the denominator, so r
is proper and the line y=0 (the x axis) is a horizontal asymptote of the graph.
Indicate this line by graphing y=0 using a dashed line. To determine if the
graph of r intersects the horizontal asymptote, we solve the equation r(x)=0.
$$\frac{x-1}{x^2-4}=0$$
$$x-1=0 >> x=1$$
The only solution is x=1. so the graph of r intersects the horizontal asymptote
at (1,0). We have already plotted this point.

5. The zero of the denominator, 1, and the zeros of the denominator, -2 and 2,
divide the x-axis into 4 intervals. Pick a point from each interval and solve
r(x). If the solution is negative then it is below the x-axis and if it is
positive then it is above the x-axis.

6. Next, we determine the behavior of the graph near the asymptotes.
A. Since the x-axis is a horizontal asymptote and the graph lies below the
x-axis for x<-2, we can sketch a portion of the graph by placing a small
arrow to the far left and under the x-axis.
B. Since the line x=-2 is a vertical asymptote and the graph lies below the
x-axis for x<-2, we continue by placing an arrow well below the x-axis and
approaching the line x=-2 on the left.
C. Since the graph is above the x-axis for -2<x<1 and x=-2 is a vertical
asymptote, the graph will continue on the right of x=-2 at the top.

Example 2
Analyze the graph of a rational function
$$r(x)=\frac{x^2-1}{x}$$

Solution:
1. The domain of r is $${x|x\not=0}$$. Because x cannot equal 0, there is not
y-intercept. Now factor r to obtain $$r(x)=\frac{(x+1)(x-1)}{x}$$

2. R is in lowest terms. Solving the equation r(x)=0, we find the graph has two
x-intercepts, -1 and 1.
Near -1: $$r(x)=\frac{(x+1)(x-1)}{x} = \frac{(x+1)(-1-1)}{-1} = 2(x+1)$$
Near 1: $$r(x)=\frac{(x+1)(x-1)}{x} = \frac{(1+1)(x-1)}{1} = 2(x-1)$$
Plot the point (-1,0) and indicate a line with slope 2 there. Plot the point
(1,0) and indicate a line with slope 2 there.

3. R is in lowest terms, so the graph of r has the line x=0 (the y-axis) as a
vertical asymptote. Graph x=0 using a dashed line.

4. The rational function r is improper, since the degree of the numerator, 2, is
larger than the degree of the denominator, 1. To find any horizontal or oblique
asymptotes, use long division. There is no solution so the graph of r does not
intersect the line y=x.

5. The zeros of the numerator are -1 and 1. The zero of the denominator is 0. We
use these values to divide the x-axis into four intervals. Pick a value in each
interval and solve for r(x). If the value is negative then the point is below
the x-axis. If the point is positive then the point is above the x-axis.

6. Since the graph of r is below the x-axis for x<-1 and is above the x-axis for
x>1, the graph of r will approach the line y=x. Since the graph of r is above
the x-axis for -1<x<0, the graph of r will approach the vertical asymptote x=0
at the top to the left of x=0. Since the graph of r is below the x-axis for
0<x<1, the graph of r will approach the vertical asymptote x=0 at the bottom
right of x=0.

Example 3
Analyze the graph of a rational function
$$r(x)=\frac{x^4+1}{x^2}$$

Solution:
1. R is completely factored. The domain of r is $${x|x\not=0}$$. There is no
y-intercept.

2. R is in lowest terms. Since $$x^4+1=0$$ has no real solutions, there are no
x-intercepts.

3. R is in lowest terms, so x=0 (the y-axis) is a vertical asymptote of r. Graph
the line x=0 using dashes.

4. The rational function r is improper. To find any horizontal or oblique
asymptotes, use long division. We find the quotient to be $$x^2$$, so the graph
has no horizontal or oblique asymptotes. However, the graph of r will approach
the graph of $$y=x^2$$ as x approaches negative infinity and as x approaches
infinity. The graph of r does not intersect $$y=x^2$$. Graph $$y=x^2$$ using
dashes.

5. The numerator has no zeros, and the denominator has one zero at 0. We divide
the x-axis into the two intervals. Plot the points (-1,2) and (1,2).

6. Since the graph of r is above the x-axis and does not intersect $$y=x^2$$, we
place arrows above $$y=x^2$$. Also, since the graph of r is above the x-axis, it
will approach the vertical asymptote x=0 at the top to the left of x=0 and at
the top to the right of x=0.

Example 4
Analyze the graph of a rational function
$$r(x)=\frac{3x^2-3x}{x^2+x-12}$$

Solution:
1. We factor r to get $$r(x)=\frac{3x(x-1)}{(x+4)(x-3)}$$. The domain of r is
$${x|x\not=-4,x\not=3}$$. The y-intercept is r(0)=0. Plot the point (0,0).

2. R is in lowest terms. Since the real solutions of the equation 3x(x-1)=0 are
x=0 and x=1, the graph has two x-intercepts, 0 and 1. We determine the behavior
of the graph of r near each x-intercept.
Near0:$$r(x)=\frac{3x(x-1)}{(x+4)(x-3)}=\frac{3x(0-1)}{(0+4)(0-3)}=\frac{1}{4}x$$
Near1:$$r(x)=\frac{3x(x-1)}{(x+4)(x-3)}=\frac{3(1)(x-1)}{(1+4)(1-3)}=-\frac{3}{10}(x-1)$$
Plot the point (0,0) and show a line with slope $$\frac{1}{4}$$ there. Plot the
point (1,0) and show a line with slope $$-\frac{3}{10}$$ there.

3. R is in lowest terms. Since the real solutions of the equation (x+4)(x-3)=0
are x=-4 and x=3, the graph of r has two vertical asymptotes, the lines x=-4 and
x=3. Plot these lines using dashes.

4. Since the degree of the numerator equals the degree of the denominator, the
graph has a horizontal asymptote. To find it, we either use long division or
form the quotient of the leading coefficient of the numerator, 3, and the
leading coefficient of the denominator, 1. The graph of r has the horizontal
asymptote y=3. To find out whether the graph of r intersects the asymptote, we
solve the equation r(x)=3.
$$r(x)=\frac{3x^2-3x}{x^2+x-12}=3$$
$$3x^2-3x=3x^2+3x-36$$
$$-6x=-36$$
$$x=6$$
The graph intersects the line y=3 only at x=6, and (6,3) is a point on the graph
of r. Plot the point (6,3) and the line y=3 using dashes.

5. The zeros of the numerator, 0 and 1, and the zeros of the denominator, -4 and
3, divide the x-axis into 5 intervals. Choose a number in each interval, solve
r(x) for each number chosen. If solution is positive then the point is above the
x-axis but if the solution is negative then the point is below the x-axis.

6. Since the graph of r is above the x-axis for x<-4 and only crosses the line
y=3 at (6,3), as x approaches negative infinity the graph of r will approach the
horizontal asymptote y=3 from above. The graph of r will approach the vertical
asymptote x=-4 at the top to the left of x=-4 and at the bottom to the right of
x=-4. The graph of r will approach the vertical asymptote x=3 at the bottom to
the left of x=3 and at the top to the right of x=3. We do not know whether the
graph of r crosses or touches the line y=3 at (6,3). To see whether the graph
crosses or touches the line y=3, we plot an additional point to the right of
(6,3) at x=6. Because (6,3) is the only point where the graph of r intersects
the asymptote y=3, the graph must approach the line y=3 from below as x
approaches infinity.

Example 5
Analyze the graph of a rational function with a hole

Solution:
1. We factor r and obtain: $$r(x)=\frac{(2x-1)(x-2)}{x^2-4}$$
The domain of r is $${x|x\not=-2,x\not=2}$$. The y-intercept is r(0)=-1/2. Plot
the point (0,-1/2).

2. In lowest terms, $$r(x)=\frac{2x-1}{x+2}$$ $$x\not=-2$$
The graph has one x-intercept: 1/2.
Near 1/2: $$r(x)=\frac{2x-1}{x+2} = \frac{2x-1}{\frac{1}{2}+2} = 2/5(2x-1)$$
Plot the point (1/2,0) showing a line with slope 4/5.

3. Look at r in lowest terms. The graph has one vertical asymptote, x=-2, since
x+2 is the only factor of the denominator of r(x) in lowest terms. Remember,
though, the rational function is undefined at both x=2 and x=-2. Graph the line
x=-2 using dashes.

4. Since the degree of the numerator equals the degree of the denominator, the
graph has a horizontal asymptote. To find it, we use long division or form the
quotient of the leading coefficient of the numerator 2, and the leading
coefficient of the denominator, 1. The graph of r has the horizontal asymptote
y=2. Graph the line y=2 using dashes.
$$r(x)=\frac{2x-1}{x+2} = 2$$
$$2x-1=2(x+2)$$
$$2x-1=2x+4$$
$$-1=4$$ Not a solution. The graph doe snot intersect the line y=2.

5. Look at the given expression for r. The zeros of the numerator and
denominator, -2,1/2, and 2, divide the x-axis into four intervals. Plot these
points.

6. We know the graph of r is above the x-axis for x<-2. We know the graph of r
does not intersect the asymptote y=2. Therefore, the graph of r will approach
y=2 from above as x approaches negative infinity and will approach the vertical
asymptote x=-2 at the top from the left. Since the graph of r is below the
x-axis for -2<x<1/2, the graph of r will approach x=-2 at the bottom from the
right. Finally, since the graph of r is above the x-axis for x>1/2 and does not
intersect the horizontal asymptote y=2, the graph of r will approach y=2 from below as x approaches infinity.

The real zeros of the denominator of a rational function give rise to either
vertical asymptotes or holes on the graph.

Example 6
Constructing a rational function from its graph

Solution:
The numerator of a rational function in lowest terms determines the x-intercepts
of its graph. This graph has x-intercepts -2(even multiplicity, graph touches
the x-axis) and 5(odd multiplicity, graph crosses the x-axis). So one
possibility for the numerator is $$p(x)=(x+2)^2(x-5)$$.

The denominator of a rational function in lowest terms determines the vertical
asymptotes of its graph. The vertical asymptotes of the graph are x=-5 and x=2.
Since r(x) approaches infinity to the left of x=-5 and r(x) approaches negative
infinity to the right of x=-5, then x+5 is a factor of odd multiplicity in q(x).
Also, because r(x) approaches negative infinity on both sides of x=2, then x-2
is a factor of even multiplicity in q(x). A possibility for the denominator is
$$q(x)=(x+5)(x-2)^2$$. So far, we have:
$$r(x)=\frac{(x+2)^2(x-5)}{(x+5)(x-2)^2}$$.

The horizontal asymptote of the graph is y=2, so we know that the degree of the
numerator must equal the degree of the denominator and the quotient of leading
coefficients must 2/1. This leads to:
$$r(x)=\frac{2(x+2)^2(x-5)}{(x+5)(x-2)^2}$$