Find the vertical asymptote of this function (if it exists)

f(x)= 4x^2+4x-24/(x^4-2x^3-9x^2+18x)

Please help? I'm thinking this has no vertical asymptote, but I dunno... Help?

Note that the numerator factors to:
4x^2 + 4x - 24 = 4(x^2 + x - 6)
= 4(x + 3)(x - 2),

while the denominator factors to:
x^4 - 2x^3 - 9x^2 + 18x
= x(x^3 - 2x^2 - 9x + 18)
= x[x^2(x - 2) - 9(x - 2)], by factoring by grouping
= x(x^2 - 9)(x - 2), by factoring out x - 2
= x(x + 3)(x - 3)(x - 2), via difference of squares.

So, we see that:
f(x) = (4x^2 + 4x - 24)/(x^4 - 2x^3 - 9x^2 + 18x)
= [4(x + 3)(x - 2)]/[x(x + 3)(x - 3)(x - 2)]
= 4/[x(x - 3)], by canceling x + 3 and x - 2.

Since we cannot simplify it anymore, set the denominator equal to zero to get:
x(x - 3) = 0 ==> x = 0 and x = 3.

Hence, your vertical asymptotes are x = 0 and x = 3.

I hope this helps!

f(x) = (4x^2 + 4x - 24) / (x^4 - 2x^3 - 9x^2 + 18x)
= 4(x^2 + x - 6) / [x(x^3 - 2x^2 - 9x + 18)]
= 4(x + 3)(x - 2) / [x(x^2(x - 2) - 9(x - 2))]
= 4(x + 3)(x - 2) / [x(x^2 - 9)(x - 2)]
= 4(x + 3)(x - 2) / [x(x - 3)(x + 3)(x - 2)]
= 4 / [x(x - 3)]

As x approaches 0 or 3, the denominator will approach 0, but the numerator will be constant. Therefore, depending on the direction of approach, f(x) will approach positive or negative infinity when x approaches 0 or 3. Thus, we have two asymptotes:

x = 0
x = 3

Factor and solve for x