# How do you find the Vertical, Horizontal, and Oblique Asymptote given f(x) = (2x - 3)/( x^2 - 1)?

Aug 2, 2016

vertical asymptotes x = ± 1
horizontal asymptote y = 0

#### 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^2-1=0rArr(x-1)(x+1)=0rArrx=±1

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

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 ${x}^{2}$

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

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

$\Rightarrow y = 0 \text{ is the asymptote}$

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( numerator-degree 1 , denominator-degree 2) Hence there are no oblique asymptotes.
graph{(2x-3)/(x^2-1) [-10, 10, -5, 5]}