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

Aug 4, 2016

vertical asymptotes x = -1 , x = 3
horizontal asymptote y = 1

#### Explanation:

The denominator of f(x) cannot be zero as this is undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve:${x}^{2} - 2 x - 3 = 0 \Rightarrow \left(x - 3\right) \left(x + 1\right) = 0$

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

Horizontal asymptotes occur as

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ (a constant)}$

divide terms on numerator/denominator by the highest power of x.that is ${x}^{2}$

$\frac{{x}^{2} / {x}^{2} - \frac{4}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{2 x}{x} ^ 2 - \frac{3}{x} ^ 2} = \frac{1 - \frac{4}{x} ^ 2}{1 - \frac{2}{x} - \frac{3}{x} ^ 2}$

as $x \to \pm \infty , f \left(x\right) \to \frac{1 - 0}{1 - 0 - 0}$

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

Slant asymptotes occur when the degree of the numerator > degree of the denominator. Tis is not the case here ( both of degree 2) Hence there are no slant asymptotes.
graph{(x^2-4)/(x^2-2x-3) [-10, 10, -5, 5]}