# How do you find the vertical, horizontal and slant asymptotes of: f(x) = (2x^2 + 1) / (x^2 - 4)?

Sep 28, 2016

vertical asymptotes at x = ± 2
horizontal asymptote at y = 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: ${x}^{2} - 4 = 0 \Rightarrow \left(x - 2\right) \left(x + 2\right) = 0 \Rightarrow x = \pm 2$

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

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 ${x}^{2}$

$f \left(x\right) = \frac{\frac{2 {x}^{2}}{x} ^ 2 + \frac{1}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{4}{x} ^ 2} = \frac{2 + \frac{1}{x} ^ 2}{1 - \frac{4}{x} ^ 2}$

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

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

Slant 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 slant asymptotes.
graph{(2x^2+1)/(x^2-4) [-10, 10, -5, 5]}