# How do you find the vertical, horizontal or slant asymptotes for f(x)=( 2x - 3)/( x^2 - 1)?

Feb 5, 2017

The vertical asymptotes are $x = 1$ and $x = - 1$
The horizontal asymptote is $y = 0$
No slant asymptote

#### Explanation:

Let's factorise the denominator

${x}^{2} - 1 = \left(x - 1\right) \left(x + 1\right)$

Therefore,

$f \left(x\right) = \frac{2 x - 3}{{x}^{2} - 1} = \frac{2 x - 3}{\left(x - 1\right) \left(x + 1\right)}$

As you cannot divide by $0$, $x \ne 1$ and $x \ne - 1$

The vertical asymptotes are $x = 1$ and $x = - 1$

As the degree of the numerator is $<$ than the degree of the denominator, there is no slant asymptote

${\lim}_{x \to - \infty} f \left(x\right) = {\lim}_{x \to - \infty} \frac{2 x}{x} ^ 2 = i {m}_{x \to - \infty} \frac{2}{x} = {0}^{-}$

${\lim}_{x \to + \infty} f \left(x\right) = {\lim}_{x \to + \infty} \frac{2 x}{x} ^ 2 = i {m}_{x \to + \infty} \frac{2}{x} = {0}^{+}$

The horizontal asymptote is $y = 0$

graph{(2x-3)/(x^2-1) [-16.02, 16.01, -8.01, 8.01]}