Question

I need to find a rational function that might have a certain given graph.
My zeros are -2 and 3, and the graph crosses at -2 and touches at 3. My vertical asymptotes are x=-3 and x=1, and my horizontal asymptote is y=2. The graph intersects y at 12. So far, I have
2(x-3)^2(x+2) / (x+3)(x-2)^2
This gives me the graph that I need, except that the graph intersects y at 3 instead of 12. I can't figure out how to get this graph to intersect at 12. Can anyone help, please?

Answer 1

ok so the function is asymptotic


well it looks like one of your vertical asymptotes is incorrect.

vertical asymptotes at x = -3 and x = 1

means the factors of the numerator are (x+3)(x-1)

for the oblique.. or in this case horizontal asymptote, the numerator and denominator are the same degree... its just that the coefficient of the leading term in the numerator is double that of the denominator.

so my view of the problem is

\[y = \frac{2(x +2)(x -3)^2}{(x +3)(x - 1)^2}\]

the power of 2 in the denominator could be associated with the factor ( x+3)

and I think it gives a y intercept of y = 12

as required,

hope this helps.