# How do you find vertical, horizontal and oblique asymptotes for f(x) = (3x + 5) /( x - 2)?

Aug 19, 2016

vertical asymptote at x = 2
horizontal asymptote at y = 3

#### 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 - 2 = 0 $\Rightarrow x = 2 \text{ is the asymptote}$

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(x)$= \frac{\frac{3 x}{x} + \frac{5}{x}}{\frac{x}{x} - \frac{2}{x}} = \frac{3 + \frac{5}{x}}{1 - \frac{2}{x}}$

as $x \to \pm \infty , f \left(x\right) \to \frac{3 + 0}{1 - 0}$

$\Rightarrow y = 3 \text{ is the asymptote}$

Oblique 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 oblique asymptotes.
graph{(3x+5)/(x-2) [-20, 20, -10, 10]}