1) The solution to the inequality \(\frac{x + c}{x^2 + ax + b} \le 0\) is \(x \in (-\infty,-1) \cup [1,2).\) Find a + b + c.
2) 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.
2) 4x - x^3 x ( 4 - x^2) x ( 2 - x) ( 2 + x)
_______ = __________ = _______________
q(x) q(x) q(x)
If we have a vertical asymptote at x = 1....then (x - 1) must be one of the factors of q(x)
And if we have a hole at x = - 2, then (x + 2) must also be a factor of q(x)
Since we have no horizontal asymptote, the q(x) must just be a second power polynomial....and we will have a "slant asymptote"
And if q(3) = 30 .....then q(x) = a ( 3 - 1) (3 + 2) = -30 .....so
a(2)(5) = -30
10a = - 30
a = -3
So q(x) = -3 ( x - 1) (x + 2) = -3 (x^2 + x - 2) = -3x^2 - 3x + 6
Here's a graph : https://www.desmos.com/calculator/vsc0kthufu
If you drag your cursor along the curve....you will find that at x = -2, y = undefined = "a hole" at this point