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)=(ax+b)/(x+c)`

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

`\displaystyle\lim_(x\to\pm\infty)(ax+b)/(x-3)+4=0=\lim_(x\to\pm\infty)(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)=(a+b)/(1-3)=0\implies a=-b\implies b=4`

So

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

(b) Since

`f(x)=(rx+s)/(2x+t)=\frac12(rx+s)/(x+\frac t2)`

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