# How do you find vertical, horizontal and oblique asymptotes for (x^2+3x-4)/x?

Sep 15, 2016

vertical asymptote at x = 0
oblique asymptote y = x +3

#### Explanation:

The denominator of the function cannot be zero as this would make the function 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: $x = 0 \Rightarrow x = 0 \text{ is the asymptote}$

Horizontal asymptotes occur when the degree of the numerator$\le$ degree of the denominator. This is not the case here (numerator-degree 2 , denominator-degree 1 ) Hence there is no horizontal asymptote.

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. Hence there is an oblique asymptote.

divide numerator by x.

${x}^{2} / x + \frac{3 x}{x} - \frac{4}{x} = x + 3 - \frac{4}{x}$

as $x \to \pm \infty , x + 3 - \frac{4}{x} \to x + 3 - 0$

$\Rightarrow y = x + 3 \text{ is the asymptote}$
graph{(x^2+3x-4)/x [-20, 20, -10, 10]}