# How do you graph f(x)=(3x^2-2x)/(x-1) using holes, vertical and horizontal asymptotes, x and y intercepts?

Apr 21, 2017

see explanation.

#### Explanation:

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.

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

Since the degree of the numerator > degree of the denominator there will be an oblique asymptote but no horizontal asymptote.

To obtain oblique asymptote, divide numerator by denominator.

$\text{ Consider the numerator}$

$\textcolor{red}{3 x} \left(x - 1\right) \textcolor{m a \ge n t a}{+ 3 x} - 2 x$

$= \textcolor{red}{3 x} \left(x - 1\right) \textcolor{red}{+ 1} \left(x - 1\right) \textcolor{m a \ge n t a}{+ 1}$

$\Rightarrow \text{ oblique asymptote is } y = 3 x + 1$

$\textcolor{b l u e}{\text{Intercepts}}$

$x = 0 \to y = 0 \to \left(0 , 0\right) \leftarrow \textcolor{red}{\text{ y-intercept}}$

$y = 0 \to 3 {x}^{2} - 2 x = 0 \to x \left(3 x - 2\right) = 0$

$\Rightarrow x = 0 , x = \frac{2}{3} \leftarrow \textcolor{red}{\text{ x-intercepts}}$
graph{(3x^2-2x)/(x-1) [-20, 20, -10, 10]}