# How do you find vertical, horizontal and oblique asymptotes for (8x^2-5)/(2x^2+3)?

Mar 24, 2018

$\text{horizontal asymptote at } y = 4$

#### 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.

$\text{solve } 2 {x}^{2} + 3 = 0 \Rightarrow {x}^{2} = - \frac{3}{2}$

$\text{This has no real solutions hence there are no vertical}$
$\text{asymptotes}$

$\text{horizontal asymptotes occur as}$

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ ( a constant)}$

$\text{divide terms on numerator/denominator by the highest}$
$\text{power of x, that is } {x}^{2}$

$f \left(x\right) = \frac{\frac{8 {x}^{2}}{x} ^ 2 - \frac{5}{x} ^ 2}{\frac{2 {x}^{2}}{x} ^ 2 + \frac{3}{x} ^ 2} = \frac{8 - \frac{5}{x} ^ 2}{2 + \frac{3}{x} ^ 2}$

$\text{as } x \to \pm \infty , f \left(x\right) \to \frac{8 - 0}{2 + 0}$

$\Rightarrow y = 4 \text{ is the asymptote}$

$\text{Oblique asymptotes occur when the degree of the }$
$\text{numerator is greater than the degree of the denominator}$
$\text{this is not the case here hence there are no oblique}$
$\text{asymptotes}$
graph{(8x^2-5)/(2x^2+3 [-10, 10, -5, 5]}