How to find the vertical asymptote for $F (x) = \frac{x^{2} + x - 12}{x^{2} - 4}$ $F(x)=(x^2+x-12)/ (x^2-4)$ ?

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Answer 1 · 2015-07-28T20:14:34

$F (x) = \frac{x^{2} + x - 12}{x^{2} - 4} = \frac{(x + 4) (x - 3)}{(x - 2) (x + 2)}$ $F(x) = (x^2+x-12)/(x^2-4) = ((x+4)(x-3))/((x-2)(x+2))$

So the vertical asymptotes are $x = 2$ $x=2$ and $x = - 2$ $x=-2$

The numerator $x^{2} + x - 12 = (x + 4) (x - 3)$ $x^2+x-12 = (x+4)(x-3)$ is zero when $x = - 4$ $x=-4$ or $x = - 3$ $x=-3$ .

The denominator $x^{2} - 4 = (x - 2) (x + 2)$ $x^2-4 = (x-2)(x+2)$ is zero when $x = \pm 2$ $x=+-2$ .

Since the denominator is zero at $x = \pm 2$ $x=+-2$ and neither of these is a zero of the numerator, both are vertical asymptotes.

graph{(x^2+x-12)/(x^2-4) [-10, 10, -5, 5]}

How to find the vertical asymptote for F(x)=(x^2+x-12)/ (x^2-4)F(x)=x2+x−12x2−4F(x)=(x^2+x-12)/ (x^2-4)?