# How do you find the vertical, horizontal and slant asymptotes of: f(x)=(x+7)/(x^2-49)?

Apr 23, 2016

vertical asymptote x = 7
horizontal asymptote y =0

#### Explanation:

The first step here is to factorise and simplify f(x)

$f \left(x\right) = \frac{x + 7}{\left(x + 7\right) \left(x - 7\right)} = \frac{\cancel{\left(x + 7\right)}}{\cancel{\left(x + 7\right)} \left(x - 7\right)} = \frac{1}{x - 7}$

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

solve : x - 7 = 0 → x = 7 is the asymptote.

Horizontal asymptotes occur as ${\lim}_{x \to \pm \infty} , f \left(x\right) \to 0$

divide terms on numerator/denominator by x

$\Rightarrow \frac{\frac{1}{x}}{\frac{x}{x} - \frac{7}{x}} = \frac{\frac{1}{x}}{1 - \frac{7}{x}}$

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

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

Slant asymptotes occur when the degree of the numerator > degree of denominator. This is not the case here hence there are no slant asymptotes.
graph{(x+7)/(x^2-49) [-10, 10, -5, 5]}