
Joshua F.
asked 09/03/23Need help with finding a function
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.
2 Answers By Expert Tutors
Hediye G. answered 09/03/23
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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)

Josh F.
09/04/23

Hediye G.
09/04/23
James S.
09/06/23

Mark M. answered 09/03/23
Mathematics Teacher - NCLB Highly Qualified
In the numerator replace the x2 with (x+3)(x-1)
https://www.desmos.com/calculator/iynd4gqubo
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Doug C.
Hi Joshua, Check out this graph to see if it helps you understand. desmos.com/calculator/r095wqts57 Post a reply here if you still need some help.09/03/23