Find q(x) if the graph of (4x-x^3)/(q(x)) has a hole at x=-2, a vertical asymptote at x=1, no horizontal asymptote, and q(3) = -30.
+128475 (-x^3 + 4x) = x ( 4 -x^2) = x ( 2 - x) ( 2 + x)
Since we have "hole" at x = -2, then ( x + 2) must be a factor of q(x)
Also....since we have a vertical asymptote at x =1, then (x - 1) must also be a factor of q(x)
Since we have no horizontal asymptote, then we must have a slant asymptote, which means that we have a higher/ lower situation with respect to the degree of the polynomial in the numerator to the degree of the polynomial in the denominator
And since q(3) = -30
Then we must have that
-30 = a (3 - 1) ( 3 + 2)
-30 = a (2)(5)
-30 = 10a
a=-30/10 = -3
So
q(x) must be -3 ( x -1 ) ( x + 2) = (-3) (x^2 + x - 2) = -3x^2 -3x + 6
