Afunction $f$ 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 $f(x).$

1. has a horizontal asymptote at

This means that

(for at least one of these limits)

2. has a vertical asymptote at

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

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

This means .

(a) With

the second point above suggests . The first point tells us that

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 so that the linear terms vanish.

The third point tells us that

So

(b) Since

we find that , and and .

why can't you?

step-by-step explanation:

sorry i just need to anwers this to get points

step-by-step explanation: