# How do you graph #f(x)=(x-4)/(x^2-3x)# using holes, vertical and horizontal asymptotes, x and y intercepts?

##### 1 Answer

#### Answer:

Below

#### Explanation:

For vertical asymptotes, we look at the denominator and since the denominator cannot equal to 0, we let the denominator equal to 0 to see where the graph cannot be

Vertical Asmpytote

For horizontal asymptote, since the degree of the numerator is smaller than the denominator, then the line

In this case, since the degree of the numerator is greater than the denominator, then

For x-intercepts, let

For y-intercepts, let

But there is no solution since the denominator will equal to 0.

Your final graph should look something like this:

graph{(x-4)/(x^2-3x) [-14.24, 14.24, -7.12, 7.12]}