# How do you graph y =(2x(x-2)) / ((x-3) (x+1))?

Sep 19, 2015

#### Explanation:

First, find the vertical asymptote. The vertical asymptote is the value of x that will make the denominator equal to 0.

$\left(x - 3\right) \left(x + 1\right) = 0$
$x = 3$ or $x = - 1$

Graph x=3 and x=-1 using a dotted line.

Next, check whether the equation has a horizontal asymptote or a slant asymptote. The first thing you need to do is to find out the degree of the numerator and denomintor. Expand the equation first.

y=(2x(x−2)) / ((x−3)(x+1))
y=(2x^2−4x)/(x^2-2x-3)

Now you can see that both the numerator and denominator are both of degree 2. This means that the asymptote is horizontal. To solve for the horizontal asymptote, simply divide the coefficients of $2 {x}^{2}$ and ${x}^{2}$.

$y = \frac{2}{1}$
$y = 2$

Now that you're done with the asymptotes, you can start plotting some points. You can do this by simply inserting values. Try starting with simple ones such as by setting x to 0.

It will look like this:
graph{(2x^2 - 4x) / (x^2 - 2x - 3) [-40, 40, -20, 20]}