Graphing Rational Functions

If $f(x)$ is a rational function given by $f(x)=\frac{P(x)}{Q(x)}$ where $P(x)$ and $Q(x)$ are polynomials, we can use the following information to sketch the graph of $f$:
I) Asymptotes

A) Vertical Asymptotes

To find the vertical asymptotes, we can first cancel any common factors in $P(x)$ and $Q(x)$ and then take the vertical lines corresponding to the zeros of the denominator:

The line $x=a$ is a vertical asymptote for the graph of $f$ whenever $Q(a)=0$ and $P(a)\ne0$.

The y-coordinates of points on the graph of $f$ get arbitrarily large (in absolute value) as the graph approaches a vertical asymptote, and

the graph never crosses a vertical asymptote.

B) Horizontal Asymptotes

We can find the horizontal asymptotes by investigating the behavior of $f(x)$ as $x$ gets arbitrarily large (with either a plus sign or a minus sign):

1. If $\deg(P(x))<\deg(Q(x))$, then the line $y=0$ (the x-axis) is the horizontal asymptote for the graph of $f$.

2. If $\deg(P(x))=\deg(Q(x))$, and $a$ and $b$ are the coefficients of the highest powers of $x$ appearing in $P(x)$ and $Q(x)$, respectively, then the line $y=a/b$ is the horizontal asymptote for the graph of $f$.

3. If $\deg(P(x))>\deg(Q(x))$, then there is no horizontal asymptote for the graph of $f$.

The graph of $f$ will approach the horizontal asymptote (when there is one) as $x$ gets arbitrarily large (with either a plus sign or a minus sign).

To determine if the graph crosses a horizontal asymptote with equation $y=L$,

we need to solve the equation $f(x)=L$.

C) Slanted Asymptotes

If $\deg(P(x))=\deg(Q(x))+1$, then the graph of $f$ has a slanted asymptote; and we can find the slanted asymptote by dividing $P(x)$ by $Q(x)$:

If

$$\frac{P(x)}{Q(x)}=mx+b+\frac{R(x)}{Q(x)}$$

where $\deg(R(x))<\deg(Q(x))$, then the line $y=mx+b$ is the slanted asymptote.

To determine if the graph crosses a slanted asymptote, we need to solve

the equation $f(x)=mx+b$ or, equivalently, the equation $R(x)=0$.

Notice that if $f$ is a rational function, then its graph cannot have

a) two horizontal asymptotes or

b) both a horizontal asymptote and a slanted asymptote.

II) Intercepts

The intercepts correspond to the points where the graph intersects the two coordinate axes:

A) To find the y-intercept, set $x=0$ and solve for $y$; so the y-intercept is given by $f(0)$.

B) To find the x-intercepts, set $y=0$ and solve for $x$; so the x-intercepts are the values of $x$ for which $P(x)=0$ (and $Q(x)\ne0$).

III) Sign Chart for $f(x)$

The sign of $f(x)$ indicates where the graph is above or below the x-axis:

A) Where $f(x)>0$, the graph of $f$ is above the x-axis.

B) Where $f(x)<0$, the graph of $f$ is below the x-axis.

(In calculus, you will use sign charts for the first derivative $f^{\prime}(x)$ and the second derivative $f^{\prime\prime}(x)$ to get more detailed information about the graph of $f$.)

Ex 1 If $f(x)=(x+1)(x-1)^2(x-3)$,

find the asymptotes and intercepts for the graph of $f$, and then use this information and a sign chart for $f(x)$ to sketch the graph of $f$.

Sol 1) Since $f$ is a non-constant polynomial, there are no asymptotes for its graph. (Here $P(x)=(x+1)(x-1)^2(x-3)$ and $Q(x)=1$.)

2) a) $f(0)=-3$, so the y-intercept is -3.

b) $f(x)=0$ for $x=-1$, $x=1$, or $x=3$; so the x-intercepts are -1,1, and 3.

3) Using the facts that $f(0)=-3$ and that the sign of $f$ changes at -1 and 3 and does not change at 1, we get the following sign chart for $f(x)$:

The following is a sketch of the graph of $f$:

Ex 2 If

$$f(x)=\frac{2x^2-2}{x^2-9},$$

1) Find the asymptotes for the graph of $f$.

2) Find the intercepts for the graph.

3) Make a sign chart for $f(x)$.

4) Determine if the graph of $f$ crosses its horizontal asymptote, and if the graph has symmetry around the origin or the y-axis.

5) Use the above information to sketch the graph of $f$.

Sol

1) a) $f(x)=\frac{2x^2-2}{x^2-9}=\frac{2(x-1)(x+1)}{(x-3)(x+3)}$, so the vertical asymptotes are the lines $x=3$ and $x=-3$.

b) Since $P(x)$ and $Q(x)$ have the same degree, the horizontal asymptote is the line $y=2$.

2) a) $f(0)=2/9$, so the y-intercept is 2/9.

b) $f(x)=0$ for $x=1$ or $x=-1$, so the x-intercepts are -1 and 1.

3) Using the facts that $f(0)=2/9$ and all the exponents are odd, we get the following sign chart for $f(x)$:

4) a) Setting $f(x)=2$ and solving gives $\frac{2x^2-2}{x^2-9}=2$, so $2x^2-2=2(x^2-9)=2x^2-18$ or $-2=-18$. Therefore there is no solution, so the graph of $f$ does not cross the horizontal asymptote.

b) Since $f(-x)=f(x)$, $f$ is an even function and therefore its graph is symmetric around the y-axis.

5) Using this information, we get the following graph:

Ex 3 If

$$f(x)=\frac{x}{x^2-4},$$

1) Find the asymptotes for the graph of $f$.

2) Find the intercepts for the graph.

3) Make a sign chart for $f(x)$.

4) Determine if the graph of $f$ crosses its horizontal asymptote, and if the graph has symmetry around the origin or the y-axis.

5) Use the above information to sketch the graph of $f$.

Sol

1) a) Since $f(x)=\frac{x}{x^2-4}=\frac{x}{(x-2)(x+2)}$, the vertical asymptotes are the lines $x=2$ and $x=-2$.

b)Since $\deg(P(x))<\deg(Q(x))$, the horizontal asymptote is the line $y=0$ (the x-axis).

2) a) $f(0)=0$, so the y-intercept is 0.

b) $f(x)=0$ when $x=0$, so the x-intercept is 0.

3) Using the facts that $f(1)=-1/3<0$ and that the exponents are all odd, we get the following sign chart for $f(x)$:

4) a) Setting $f(x)=0$ and solving gives $x=0$, so the graph intersects the horizontal asymptote at $(0,0)$.

b) Since $f(-x)=-f(x)$, $f$ is an odd function and therefore its graph is symmetric about the origin.

5) Using the above information, we get the following graph:

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For each of the following functions,

1) Find the asymptotes for the graph of $f$.

2) Find the intercepts for the graph.

3) Make a sign chart for $f(x)$.

4) Determine if the graph of $f$ crosses its horizontal asymptote or slanted asymptote (if there is one), and if the graph has symmetry around the origin or the y-axis.

5) Use the above information to sketch the graph of $f$.

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Pr 1

$$f(x)=x^4-10x^2+9$$

Pr 2

$$f(x)=\frac{x+2}{x-2}$$

Pr 3

$$f(x)=\frac{x^2+3x+2}{x^2-x-2}$$

Pr 4

$$f(x)=\frac{x-4}{x^2+x-2}$$

Pr 5

$$f(x)=\frac{x^2}{16-x^2}$$

Pr 6

$$f(x)=x-4+3/x$$

Pr 7

$$f(x)=x+\frac{1}{x-2}$$

Pr 8

$$f(x)=\frac{x^2-x-6}{x-2}$$

Pr 9

$$f(x)=\frac{x^2-1}{x^2-4x+3}$$

Pr 10

$$f(x)=\frac{x^2-2x-8}{x^2-x-2}$$

Pr 11

$$f(x)=\frac{x^3-8}{x^2-3x-4}$$

Pr 12

$$f(x)=\frac{2x^2+4x-6}{x^2+3x-10}$$

Go to Solutions.

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