# How do you identify all vertical asymptotes for f(x)=(x^2-5x+4)/(x^2-4)?

Aug 9, 2017

Vertical asymptotes are at $x = 2 \mathmr{and} x = - 2$

#### Explanation:

$f \left(x\right) = \frac{{x}^{2} - 5 x + 4}{{x}^{2} - 4} = \frac{\left(x - 4\right) \left(x - 1\right)}{\left(x + 2\right) \left(x - 2\right)}$

For vertical asymptotes to form denominator is zero.

$x + 2 = 0 \therefore x = - 2 \mathmr{and} x - 2 = 0 \mathmr{and} x = 2$

So vertical asymptotes are at $x = 2 \mathmr{and} x = - 2$

graph{(x^2-5x+4)/(x^2-4) [-40, 40, -20, 20]} [Ans]

Aug 9, 2017

$\text{vertical asymptotes at } x = \pm 2$

#### 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.

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

$\Rightarrow x = \pm 2 \text{ are the asymptotes}$
graph{(x^2-5x+4)/(x^2-4) [-10, 10, -5, 5]}