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

Feb 3, 2017

Vertical asymptote $x = 3$ and oblique / slanting asymptote $y = x$

#### Explanation:

As $f \left(x\right) = \frac{{x}^{2} - 3 x + 2}{x - 3} = \frac{\left(x - 1\right) \left(x - 2\right)}{x - 3}$ and as $\left(x - 3\right)$ in denominator does not cancel out with numeraor, we do not ave a hole.

If $x = 3 + \delta$ as $\delta \to 0$, $y = \frac{\left(2 + \delta\right) \left(1 + \delta\right)}{\delta}$ and as $\delta \to 0$, $y \to \infty$. But if $x = 3 - \delta$ as $\delta \to 0$, $y = \frac{\left(2 - \delta\right) \left(1 - \delta\right)}{- \delta}$ and as $\delta \to 0$, $y \to - \infty$.

Hence $x = 3$ is a vertical asymptote.

Further $y = \frac{{x}^{2} - 3 x + 2}{x - 3} = \frac{{x}^{2} - 3 x}{x - 3} + \frac{2}{x - 3}$

= $x + \frac{2}{x - 3} = x + \frac{\frac{2}{x}}{1 - \frac{3}{x}}$

Hence as $x \to \infty$, $y \to x$ and we have an oblique or slant asymptote $y = x$
graph{(y-(x^2-3x+2)/(x-3))=0 [-17.34, 22.66, -8.4, 11.6]}