How do you find the horizontal and vertical asymptote of the following: f(x) = (2x-3)/(x^2+2)?

Feb 4, 2018

$\text{horizontal asymptote at } y = 0$

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 value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

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

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

$\text{horizontal asymptotes occur as}$

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

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

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

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

$\Rightarrow y = 0 \text{ is the asymptote}$
graph{(2x-3)/(x^2+2) [-10, 10, -5, 5]}