# What are the asymptotes for f(x) = (x^2 -1)/(2x^2 + 3x-2)?

Jul 13, 2015

Vertical asymptotes: $x = - 2$ and $x = \frac{1}{2}$;
Horizontal asymptote: $y = \frac{1}{2}$

#### Explanation:

We can have:
1] Vertical asymptotes; which are vertical lines at $x$ values that makes your denominator equal to zero. These are vertical lines that cannot be crossed by the graph of your function that instead get as near as possible to them.
To make the denominator equal to zero you can solve it as a 2nd degree equation:
$2 {x}^{2} + 3 x - 2 = 0$
using the Quadratic Formula you get:
$x = - 2$
$x = \frac{1}{2}$
these two are the equation of your two vertical asymptotes.

2] Horizontal asymptote; which are horizontal lines that again cannot be crossed by the graph of your function. Your function instead tends to get as near as possible to them.

To find them you look at the behaviour of your function for $x$ very large, using the Limit :
${\lim}_{x \to \infty} \frac{{x}^{2} - 1}{2 {x}^{2} + 3 x - 2} = {\lim}_{x \to \infty} \frac{{x}^{2} \left(1 - \frac{1}{x} ^ 2\right)}{{x}^{2} \left(2 + \frac{3}{x} - \frac{2}{x} ^ 2\right)} =$ where I collected ${x}^{2}$ in the numerator and denominator:
rearranging and taking the limit we get:
${\lim}_{x \to \infty} \frac{\cancel{{x}^{2}} \left(1 - \frac{1}{x} ^ 2\right)}{\cancel{{x}^{2}} \left(2 + \frac{3}{x} - \frac{2}{x} ^ 2\right)} = \frac{1}{2}$
(remember that as $x \to \infty$ the fraction $\frac{1}{x} ^ n \to 0$)
Your horizontal asymptote will then be:
$y = \frac{1}{2}$

Graphically you can see them as:
graph{(x^2-1)/(2x^2+3x-2) [-10, 10, -5, 5]}