# How many vertical asymptotes does the graph of f(x) have?

if f(x) is

____2x + 1______

x^3 + x^2 + 5x + 5

0, 1, 2, 3 or none of these

Relevance

ti has one, x = -1. because x^3+x^2+5x+5 = x^2(x+1)+5(x+1)= (x+1)(x^2+5)

x+1 = 0, for x=-1

x^2+5 >0 for any x. therefore, only 1

• You have to check when denominator = 0.

x^3 + x^2 + 5x + 5 = 0

x^2(x+1) + 5 (x+1) = 0

(x^2 + 5) (x+1) = 0

Denominator = 0 when x=-1.

lim _ ____2x + 1______ = + ∞

x>-1 x^3 + x^2 + 5x + 5

lim ____2x + 1______ = - ∞

x>-1+ x^3 + x^2 + 5x + 5

when y=f(x)--->∞ you have a vertical asymptote.

There is only one vertical asymptote: f(x) = -1.

• In this case, f(x) has a vertical asymptote when (x^3+x^2+5x+5)=0.

Or, if you simplify, (x+1)(x^2+5)=0.

This means that vertical asymptotes are at x+1=0 and x^2+5=0.

x+1=0 => x=-1

x^2+5=0 => x^2=-5 => (assuming you do not use complex numbers) => no solution from this.

Or, in other words: there is 1 vertical asymptote, at x=-1.

Good luck.

Source(s): High-school Math