# How do you find the vertical, horizontal or slant asymptotes for f(x) = (x+1) / (x^2 +3x - 4)?

Jun 27, 2016

vertical asymptotes x = -4 , x = 1
horizontal asymptote y = 0

#### Explanation:

Vertical asymptotes occur as the denominator of a rational function tends to zero. To find the equation/s set the denominator equal to zero.

solve: ${x}^{2} + 3 x - 4 = 0 \Rightarrow \left(x + 4\right) \left(x - 1\right) = 0$

$\Rightarrow x = - 4 , x = 1 \text{ are the asymptotes}$

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by the highest power of x , in this case ${x}^{2}$

$\frac{\frac{x}{x} ^ 2 + \frac{1}{x} ^ 2}{{x}^{2} / {x}^{2} + \frac{3 x}{x} ^ 2 - \frac{4}{x} ^ 2} = \frac{\frac{1}{x} + \frac{1}{x} ^ 2}{1 + \frac{3}{x} - \frac{4}{x} ^ 2}$

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

$\Rightarrow y = 0 \text{ is the asymptote}$

Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 1 , denominator-degree 2 ) Hence there are no slant asymptotes.
graph{(x+1)/(x^2+3x-4) [-10, 10, -5, 5]}