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! :(((
+499 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:
ax2+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:
3x2+3x-6
on the bottom
