# How do you identify all asymptotes or holes for f(x)=(x^3-4x^2+3x)/(3x^2-15x+18)?

Dec 29, 2017

See below.

#### Explanation:

First try to factor numerator and denominator:

(x^3-4x^2+3x)/(3x^2-15x+18)=(x(x-1)(x-3))/(3(x-2)(x-3))=(x(x-1))/(3(x-2)

Notice we can cancel the factor $\left(x - 3\right)$. This shows that $x = 3$ is a removable discontinuity. This is sometimes called a point discontinuity or a singularity.

So there is a hole at $x = 3$

Vertical asymptotes occur where the function is undefined. This occurs when $x = 2$ ( zero denominator )

So the line $x = 2$ is a vertical asymptote.

$\frac{x \left(x - 1\right)}{3 \left(x - 2\right)} = \frac{{x}^{2} - x}{3 x - 6}$

The degree of the numerator is greater than the degree of the denominator, so there will be an oblique asymptote. Will can find the equation of this by polynomial division.

$\frac{{x}^{2} - x}{3 x - 6} = \frac{1}{3} x + \frac{1}{3}$

We only need to divide until we have the equation of a line, so the remainder of 2 in this case is irrelevant to us.

So the line $y = \frac{1}{3} x + \frac{1}{3}$ is an oblique asymptote.

For end behaviour, we only need concentrate on:

${x}^{2} / \left(3 x\right) = \frac{x}{3}$

as $x \to \infty$ , $\frac{x}{3} \to \infty$

as $x \to - \infty$ , $\frac{x}{3} \to - \infty$

GRAPH: