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

Aug 21, 2016

vertical asymptotes at $x = - 2 , x = \frac{1}{2}$
horizontal asymptote at $y = \frac{1}{2}$

#### Explanation:

The denominator of f(x) cannot be zero as this would make f(x) 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: $2 {x}^{2} + 3 x - 2 = 0 \Rightarrow \left(2 x - 1\right) \left(x + 2\right) = 0$

$\Rightarrow x = - 2 \text{ and " x=1/2" are the asymptotes}$

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 ${x}^{2}$

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

as $x \to \pm \infty , f \left(x\right) \to \frac{1 - 0}{2 + 0 - 0}$

$\Rightarrow y = \frac{1}{2} \text{ is the asymptote}$

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