Chapter 4.5, Problem 59AYU

In Problems 51-54, find a rational function that might have the given graph. (More than one answer might be possible.)

To determine

To find: A rational function.

Answer to Problem 59AYU

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

Explanation of Solution

Given:

Since the graph is $x\text{-intercepts}$ to be 1 and 3, so the numerator must have the factors of $x-1$ and $x-3$.

Since the graph is vertical asymptotes to be $-1$ and $2$, so the denominator must have the factors of $x+1$ and $x-2$.

However, if $f\left(x\right)=\frac{\left(x-1\right)\left(x-3\right)}{\left(x+1\right)\left(x-2\right)}$, its not sufficient. In particular the $y\text{-intercept}$ will be $-\frac{3}{2}$ instead of 1 and the piece in the middle will be increasing instead of decreasing.

Therefore, we have to modify it. Since we don’t want any more vertical asymptotes, we can square each existing factors in numerator will not do since it won’t give us the $y\text{-intercept}$ 1.

Therefore, we have to use another strategy by introducing a new factor in the numerator but it won’t yield any more $x\text{-intercepts}$.

We can see that the degree of the denominator is 4 once we square each factor in the denominator, so we also need the degree of the numerator to be 4. Since the degree of the numerator so far is 2, we need to have the new factor to be degree 2 also. So this new factor must be a quadratic factor. So we can it to be of the form $x^2+a$ where a is a positive constant. In this case, $a=\frac{1}{3}$ will do since the $y\text{-intercept}$ is 1.

Therefore, $f\left(x\right)=\frac{\left(x-1\right)\left(x-3\right)\left(x^2+\frac{4}{3}\right)}{{\left(x+1\right)}^2{\left(x-2\right)}^2}$.