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

Jan 18, 2018

Let's find the holes of this formula. A hole means that the same factor is in the numerator as denominator and they divide out. Such as $\frac{{x}^{2}}{{x}^{3} - 4 x}$ or (cancel(x) xx x)/(cancel(x)(x^2-4) so there is a hole when $x = 0$.

In our case of $\frac{2 {x}^{2}}{x - 3}$, there are no common factors, so there is no hole.

Vertical asymptotes occur when we try to divide a value by $0$. So let's see what value of $x$ makes the denominator equal to $0$:

$x - 3 = 0$

$x = 3$

So, there is a vertical asymptote at $x = 3$

Now let's see about the Horizontal asymptote.

I like to use this to help me remember:

BOBO - Bigger on bottom, y=0

BOTN - Bigger on top, none

EATS DC - Exponents are the same, divide coefficients

So in our case, the numerator (top) has a greater exponent (bigger). So there is no Horizontal asmptote (BOTN)

Now let's find the $x$- intercepts and $y$- intercepts:

$x$-intercept is the value of $x$ when $y$ equals $0$:

$0 = \frac{2 {x}^{2}}{x - 3}$

$0 = 2 {x}^{2}$

$0 = {x}^{2}$

$x = 0$

The $y$-intercept is the value of $y$ when $x$ equals $0$

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

$y = \frac{0}{-} 3$

$y = 0$

Now we have all the information we need

To check our answers, let's graph the equation
graph{y=(2x^2)/(x-3)}

We have an $x$ and $y$ intercept at $0$, that's right. There's no horizontal asymptote although there is an asymptote for $x = 3$

Our math is correct. Good job