# How do you identify all asymptotes or holes for f(x)=(-x+1)/(x+4)?

Sep 17, 2016

vertical asymptote at x = - 4
horizontal asymptote at y = - 1

#### 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: $x + 4 = 0 \Rightarrow x = - 4 \text{ is the asymptote}$

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

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

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

$\Rightarrow y = - 1 \text{ is the asymptote}$

Holes occur when there are duplicate factors on the numerator/denominator. This is not the case here hence there are no holes.
graph{(-x+1)/(x+4) [-20, 20, -10, 10]}