Need help with finding a function

Question

I'm trying to find a function with a Horizontal asymptote of -6, Vertical asymptotes of -1 and 3. It is supposed to have x intercepts of (-3,0) and (1,0). When I was using the asymptotes to find the equation I got (-6x^2)/(x^2-2x-3) but those don't appear to solve it.

Answer 1

The vertical asymptotes tell you that the denominator is (x+1)(x-3)=x^2-2x-3 and you are right.

x intercepts tell you that the numerator is c(x+3)(x-1)= cx^2+2cx-3c where c is a constant.

So far we have f(x)=(cx^2+2cx-3c)/ (x^2-2x-3)

Horizontal asymptote tells you that the leading terms of numerator divided by the denominator is 6. That is (cx^2)/(x^2)=c=6

Therefore: f(x)=(6x^2+12x-18)/ (x^2-2x-3)

Answer 2

-6(x+3)(x-1)/(x-3)(x+1)=

-(6x^2+12x -18)/(x^2-2x-3)

=(18-12x-6x^2)/(x^2-2x-3)

which has horizontal asymptote y = -6

vertical asymptotes x=3 and -1 which are also the x intercepts (3,0) and (-1,0)

Answer 3

In the numerator replace the x² with (x+3)(x-1)

https://www.desmos.com/calculator/iynd4gqubo