Question

A function $f$ has a horizontal asymptote of $y = -4,$ a vertical asymptote of $x = 3,$ and an $x$-intercept at $(1,0).$ Part (a): Let $f$ be of the form $$f(x) = \frac{ax+b}{x+c}.$$Find an expression for $f(x).$ Part (b): Let $f$ be of the form $$f(x) = \frac{rx+s}{2x+t}.$$Find an expression for $f(x).$

Answer 1

1. $f$ has a horizontal asymptote at $y=-4$

This means that

$\displaystyle\lim_{x\to\pm\infty}f(x)-(-4)=0$

(for at least one of these limits)

2. $f$ has a vertical asymptote at $x=3$

This means that $f$ has a non-removable discontinuity at $x=3$. Since $f$ is some rational function, there must be a factor of $x-3$ in its denominator.

3. $f$ has an $x$-intercept at (1, 0)

This means $f(1)=0$.

(a) With

$f(x)=\dfrac{ax+b}{x+c}$

the second point above suggests $c=-3$. The first point tells us that

$\displaystyle\lim_{x\to\pm\infty}\frac{ax+b}{x-3}+4=0=\lim_{x\to\pm\infty}\frac{ax+b+4x-3}{x-3}=0$

In order for the limit to be 0, the denominator's degree should exceed the numerator's degree; the only way for this to happen is if $a=-4$ so that the linear terms vanish.

The third point tells us that

$f(1)=\dfrac{a+b}{1-3}=0\implies a=-b\implies b=4$

So

$f(x)=\dfrac{-4x+4}{x-3}$

(b) Since

$f(x)=\dfrac{rx+s}{2x+t}=\dfrac12\dfrac{rx+s}{x+\frac t2}$

we find that $\dfrac t2=-3\implies t=-6$, and $r=a=-4$ and $s=b=4$.