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 \(q(3)=-30\)

much help appreciated! :(((

Guest Mar 23, 2020

#1**+1 **

First, we observe that if the function \(q(x)\) has a hole at x = -2, that would indicate that both the values of the numerator AND denominator are 0. (0/0 produces the hole in the graph). The important takeway here is that the function \(q(x)\) then has a root of -2

If the equation \(q(x)\) has a vertical asymptote at x = 1, that would indicate that the solution present at 1 is undefined. We know from this that 1 is a root of \(q(x)\).

Since we have 2 roots of \(q(x)\) let's assume it's a quadratic with two roots**(anyone have a more rigorous proof than this?)**. We have the general form as:

ax^{2}+bx+c = 0

We know that -2 and 1 are roots of this quadratic, so by vietas formulas:

-2 + 1 = -b/a

-1 = -b/a

-a = -b

a = b

-2 * 1 = c /a

-2 = c/a

c = -2a

(we will use these later :P)

Next we look at the final thing the problem gives us, which is that \(f(3) = 30\)

That means that substituting into the expression, we get:

9a + 3b + c = 30

Here's where our previous expressions come in handy:

We can express everything in terms of b here. We then get:

9(b) + 3b + c = 30

c = -2a

a = b

c = -2b

9b + 3b -2b = 03

10b = 30

b = 3

Since a = b

a = 3

since c = -2b:

c = -6

we then have the quadratic:

**3x ^{2}+3x-6**

**on the bottom**

jfan17 Mar 24, 2020