# How do you find vertical, horizontal and oblique asymptotes for (x^2-4) /( x^2-2x-3)?

Mar 23, 2016

Vertical asymptotes x = -1 , x = 3
Horizontal asymptote y = 1

#### Explanation:

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

solve :  x^2-2x -3 = 0 → (x+1)(x-3) = 0

$\Rightarrow x = - 1 , x = 3 \text{ are the asymptotes }$

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

If the degree of the numerator and denominator are equal, as in this case. Then the equation is found by taking the ratio of leading coefficients.

$\Rightarrow y = \frac{1}{1} = 1 \text{ is the asymptote }$

Oblique asymptotes occur when the degree of the numerator is greater than the degree of the denominator. This is not the case here hence there are no oblique asymptotes.

Here is the graph of the function.
graph{(x^2-4)/(x^2-2x-3) [-10, 10, -5, 5]}