# How do you identify all asymptotes or holes for f(x)=(x^2-5x+6)/(x^2-4x+3)?

Dec 10, 2016

There is a hole at $x = 3$
The vertical asymptote is $x = 1$
The horizontal asymptote is $y = 1$

#### Explanation:

Let's factorise the numerator and denominator

${x}^{2} - 5 x + 6 = \left(x - 3\right) \left(x - 2\right)$

${x}^{2} - 4 x + 3 = \left(x - 1\right) \left(x - 3\right)$

So,

$f \left(x\right) = \frac{{x}^{2} - 5 x + 6}{{x}^{2} - 4 x + 3} = \frac{\cancel{x - 3} \left(x - 2\right)}{\left(x - 1\right) \cancel{x - 3}}$

$= \frac{x - 2}{x - 1}$

So, there is a hole at $x = 3$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{1\right\}$

As we cannot divide by $0$, $x \ne 1$

So, the vertical asymptote is $x = 1$

To calculate the limits as $x \to \pm \infty$, we take the terms of highest degree in the numerator and the denominator.

${\lim}_{x \to \pm \infty} f \left(x\right) = {\lim}_{x \to \pm \infty} \frac{x}{x} = 1$

So, the horizontal asymptote is $y = 1$

graph{(y-(x-2)/(x-1))(y-1)=0 [-11.25, 11.25, -5.625, 5.625]}