# How do you find the vertical, horizontal and slant asymptotes of: f(x)=(2x) /( x-5)?

##### 1 Answer
Jul 16, 2016

vertical asymptote x = 5
horizontal asymptote y = 2

#### Explanation:

The denominator of f(x) cannot be zero as this is 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 of x then it is a vertical asymptote.

solve: x - 5 = 0 → x = 5 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

$\frac{\frac{2 x}{x}}{\frac{x}{x} - \frac{5}{x}} = \frac{2}{1 - \frac{5}{x}}$

as $x \to \pm \infty , f \left(x\right) \to \frac{2}{1 - 0}$

$\Rightarrow y = 2 \text{ is the asymptote}$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (both of degree 1) Hence there are no slant asymptotes.
graph{(2x)/(x-5) [-20, 20, -10, 10]}