Find \(q(x)\) if the graph of \(\frac{4x-x^3}{q(x)}\) has a hole at \(x=-2\), a vertical asymptote at \(x=1\), no horizontal asymptote, and \(q(3) = -30\).

Guest May 6, 2019

#1**+1 **

4x - x^3 x ( 4 - x^2) x ( 2 - x) ( 2 +x)

______ = ___________ = _____________

q(x) q(x) q(x)

If we have a "hole" at x = -2, then q(x) must have a factor of ( x + 2)

If we have a vertical asymptote at x = 1, then (x - 1) must also be a factor

If we have no horizontal asymptote, the q(x) must be a second degree polynomial - the rational fuction has a s;ant asymptote

And since q(3) = -30....then we have that

-30 = a(3 + 2) (3 - 1)

-30 =a (5)(2)

-30 = 10a

a = -3

So....q(x) = -3(x + 2)(x - 1) = -3(x^2 + x - 2) = -3x^2 - 3x + 6

Here's a graph : https://www.desmos.com/calculator/dz5hnv1yoa

CPhill May 6, 2019