What are the asymptote(s) and hole(s) of  f(x) = (2x-4)/(x^2-3x+2)?

May 1, 2017

$\text{hole at } x = 2$
$\text{vertical asymptote at } x = 1$
$\text{horizontal asymptote at } y = 0$

Explanation:

$\text{The first step is to factorise and simplify f(x)}$

$f \left(x\right) = \frac{2 \cancel{\left(x - 2\right)}}{\left(x - 1\right) \cancel{\left(x - 2\right)}} = \frac{2}{x - 1}$

The removal of the factor (x - 2) indicates a hole at x = 2

The graph of $\frac{2}{x - 1}$ is the same as $\frac{2 x - 4}{{x}^{2} - 3 x + 2}$

$\text{but without the hole}$

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.

$\text{solve "x-1=0rArrx=1" is the asymptote}$

Horizontal asymptotes occur as

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

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

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

as xto+-oo,f(x)to0/(1-0+0

$\Rightarrow y = 0 \text{ is the asymptote}$
graph{2/(x-1) [-10, 10, -5, 5]}