Rational functions

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Rational functions are functions defined by fractions of polynomials.

What are rational functions?

A rational function in the variable $x$ is a function the form $f(x) = \frac {p(x)}{q(x)}$ where $p$ and $q$ are polynomial functions. The domain of a rational function is all real numbers except for where the denominator is equal to zero.

Which of the following are rational functions?

$f(x) = 0$ $f(x) = \frac {3x+1}{x^2-4x+5}$ $f(x)=e^x$ $f(x)=\frac {\sin (x)}{\cos (x)}$ $f(x) = -4x^{-3}+5x^{-1}+7-18x^2$ $f(x) = x^{1/2}-x +8$ $f(x)=\frac {\sqrt {x}}{x^3-x}$

All polynomials can be thought of as rational functions.

What can the graphs look like?

There is a somewhat wide variation in the graphs of rational functions.

Here we see the the graphs of four rational functions.

Match the curves $A$ , $B$ , $C$ , and $D$ with the functions $\begin{align*} &\frac{x^2-3x+2}{x-2}, &&\frac{x^2-3x+2}{x+2}, \\ &\frac{x-2}{x^2-3x+2}, &&\frac{x+2}{x^2-3x+2}. \end{align*}$

Consider $\frac {x^2-3x+2}{x-2}$ . This function is undefined only at $x=2$ . Of the curves that we see above, $\answer [given]{D}$ is undefined exactly at $x=2$ .

Now consider $\frac {x^2-3x+2}{x+2}$ . This function is undefined only at $x=-2$ . The only function above that undefined exactly at $x=-2$ is curve $\answer [given]{A}$ .

Now consider $\frac {x-2}{x^2-3x+2}$ . This function is undefined at the roots of $x^2-3x+2 = (x-2)(x-1).$ Hence it is undefined at $x=2$ and $x=1$ . It looks like both curves $B$ and $C$ would work. Distinguishing between these two curves is easy enough if we evaluate at $x=-2$ . Check it out.

$\begin{align*} \eval{\frac{x-2}{x^2-3x+2}}_{x=-2} &= \frac{-2-2}{(-2)^2-3(-2)+2}\\ &= \frac{-4}{4+6+2}\\ &=\frac{-4}{12}. \end{align*}$

Since this is negative, we see that $\frac {x-2}{x^2-3x+2}$ corresponds to curve $\answer [given]{B}$ .

Finally, it must be the case that curve $\answer [given]{C}$ corresponds to $\frac {x+2}{x^2-3x+2}$ . We should note that if this function is evaluated at $x=-2$ , the output is zero, and this corroborates our work above.