# How do you find the vertical, horizontal or slant asymptotes for (x^2+x-12 )/( x^2-4)?

Feb 28, 2016

vertical asymptotes at x = ± 2
horizontal asymptote y = 1

#### Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation let the denominator equal zero.

solve:  x^2 - 4 = 0 → (x-2)(x+2) = 0 → x = ± 2

Horizontal asymptotes occur as lim_(x→±∞) f(x) → 0

If the degree of the numerator and denominator are equal , as they are in this case , both of degree 2. Then the equation can be found by taking the ratio of leading coefficients.

hence equation is $y = \frac{1}{1} = 1$ → y = 1
Here is the graph of the function.
graph{(x^2+x-12)/(x^2-4) [-10, 10, -5, 5]}