# How do you find any asymptotes of g(x)=(2x-3)/(x^2-6x+9)?

Nov 1, 2016

The vertical asymptote is $x = 3$
There is no oblique asymptote

#### Explanation:

Let's factorise the denominator
${x}^{2} - 6 x + 9 = {\left(x - 3\right)}^{2}$
Therefore $g \left(x\right) = \frac{2 x - 3}{x - 3} ^ 2$
As you cannot divide by $0$, so $x \ne 3$
So $x = 3$ is a vertical asymptote
Limit $g \left(x\right) = \frac{2}{x} = {0}^{-}$
$x \to - \infty$
Limit $g \left(x\right) = \frac{2}{x} = {0}^{+}$
$x \to + \infty$

graph{(2x-3)/(x-3)^2 [-10, 10, -5, 5]}