# How do you find the asymptotes for f(x)=(5/(x-7)) +6?

Aug 16, 2016

vertical asymptote at x = 7
horizontal asymptote at y = 6

#### 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.

solve: $x - 7 = 0 \Rightarrow x = 7 \text{ is the asymptote}$

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by x

$f \left(x\right) = \frac{\frac{5}{x}}{\frac{x}{x} - \frac{7}{x}} + 6 = \frac{\frac{5}{x}}{1 - \frac{7}{x}} + 6$

as $x \to \pm \infty , f \left(x\right) \to \frac{0}{1 - 0} + 6$

$\Rightarrow y = 6 \text{ is the asymptote}$
graph{((5)/(x-7))+6 [-20, 20, -10, 10]}