# How do you find vertical, horizontal and oblique asymptotes for f(x) = (4x)/(x^2-25)?

Jun 28, 2016

vertical asymptotes x = ± 5
horizontal asymptote y = 0

#### Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s set the denominator equal to zero.

solve: x^2-25=0rArr(x-5)(x+5)=0rArrx=±5

$\Rightarrow x = - 5 , x = 5 \text{ are the asymptotes}$

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 ${x}^{2}$

$\frac{\frac{4 x}{x} ^ 2}{{x}^{2} / {x}^{2} - \frac{25}{x} ^ 2} = \frac{\frac{4}{x}}{1 - \frac{25}{x} ^ 2}$

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

$\Rightarrow y = 0 \text{ is the asymptote}$

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 1 and denominator-degree 2 ) Hence there are no oblique asymptotes.
graph{(4x)/(x^2-25) [-10, 10, -5, 5]}