How do you find the Vertical, Horizontal, and Oblique Asymptote given f(x)= (2x+1)/(x-1)?

Aug 22, 2016

vertical asymptote at x = 1
horizontal asymptote at y = 2

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: $x - 1 = 0 \Rightarrow x = 1 \text{ is the asymptote}$

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

$f \left(x\right) = \frac{\frac{2 x}{x} + \frac{1}{x}}{\frac{x}{x} - \frac{1}{x}} = \frac{2 + \frac{1}{x}}{1 - \frac{1}{x}}$

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

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

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{(2x+1)/(x-1) [-20, 20, -10, 10]}