f(x)= 5x^3 + 20x^2 - 5x - 20 / x^2 + 6x + 8?

Question

I need the holes, vertical asynptote, and horizontal asynptote

Answer 1

f(x) = (5x^3 + 20x^2 - 5x - 20) / (x^2 + 6x + 8)

Factor 5 from the top, and factor the bottom normally.

f(x) = 5(x^3 + 4x^2 - x - 4) / [ (x + 4)(x + 2) ]

Factor the cubic by grouping.

f(x) = [5(x^2(x + 4) - (x + 4)) ] / [ (x + 4)(x + 2) ]

f(x) = [5(x^2 - 1)(x + 4)] / [(x + 4)(x + 2)]

Factoring as a difference of squares,

f(x) = [5(x - 1)(x + 1)(x + 4)] / [(x + 4)(x + 2)]

Notice (x + 4) cancels out (and this will be our hole).

f(x) = [5(x - 1)(x + 1)] / [(x + 2)

Vertical asymptote is what makes the denominator equal to 0; that is, x + 2 = 0 (which means x = -2 is our vertical asymptote).

Holes, are as we stated what canceled out; equate what cancelled out to 0; x + 4 = 0 (which means x = -4).

There are no horizontal asymptotes; if you perform long division, you'll find that you won't get a constant number as a result. That's the determining factor (because that constant after synthetic long division is what the horizontal asymptote is).

Answer 2

WOW!

GENIUS!