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

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 `(x^2)/(x^3-4x)` or `(cancel(x) xx x)/(cancel(x)(x^2-4)` so there is a hole when `x = 0`.

In our case of `(2x^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 = (2x^2)/(x-3)`

`0 = 2x^2`

`0 = x^2`

`x = 0`

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

`y = (2(0)^2)/(0-3)`

`y = 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