How do you identify all asymptotes or holes for f(x)=(-x^2+5x-4)/(x^2-7x+12)?

Sep 15, 2016

hole at $\left(4 , - 3\right)$
vertical asymptote at x = 3
horizontal asymptote at y = - 1

Explanation:

A hole or point discontinuity is indicated when there is a common factor on the numerator/denominator of f(x).

Factorising f(x)

$f \left(x\right) = - \frac{\cancel{\left(x - 4\right)} \left(x - 1\right)}{\cancel{\left(x - 4\right)} \left(x - 3\right)} = \frac{- \left(x - 1\right)}{x - 3}$

Since a factor (x-4) has been removed from the denominator this indicates a discontinuity and so x ≠4

substitute x = 4 into the 'simplified' f(x)

$\Rightarrow f \left(4\right) = \frac{- 3}{1} = - 3$

Hence (4 ,-3) is a hole, that is, does not exist in this graph.

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: $x - 3 = 0 \Rightarrow x = 3 \text{ is the asymptote}$

Horizontal asymptotes occur as

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

divide numerator/denominator by x

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

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

$\Rightarrow y = - 1 \text{ is the asymptote}$
graph{(-x+1)/(x-3) [-10, 10, -5, 5]}