# How do you find the vertical asymptotes and holes of f(x)=(x+2)/(x^2+3x-4)?

Nov 17, 2016

vertical asymptotes at x = - 4 and x = 1
horizontal asymptote at y = 0

#### 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 values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

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

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

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by the highest power of x, that is ${x}^{2}$

$f \left(x\right) = \frac{\frac{x}{x} ^ 2 + \frac{2}{x} ^ 2}{{x}^{2} / {x}^{2} + \frac{3 x}{x} ^ 2 - \frac{4}{x} ^ 2} = \frac{\frac{1}{x} + \frac{2}{x} ^ 2}{1 + \frac{3}{x} - \frac{4}{x} ^ 2}$

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

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

Holes occur when there are duplicate factors on the numerator/denominator. This is not the case here hence f(x) has no holes.
graph{(x+2)/(x^2+3x-4) [-10, 10, -5, 5]}