We are asked to graph the function `y=(x-4)/(x^2-3x) ` :

Factoring the numerator and denominator yields:

`y=(x-4)/(x(x-3)) `

There are vertical asymptotes at x=0 and x=3. The x-intercept is 4.

Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y=0.

The...

## Unlock

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

We are asked to graph the function `y=(x-4)/(x^2-3x) ` :

Factoring the numerator and denominator yields:

`y=(x-4)/(x(x-3)) `

There are vertical asymptotes at x=0 and x=3. The x-intercept is 4.

Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y=0.

The first derivative is `y'=(-(x-6)(x-2))/((x^2-3x)^2) ` ; y'=0 when x=2 or x=6. The function is decreasing on x<0 and 0<x<2, has a local minimum at x=2, inccreases on 2<x<3 and 3<x<6, has a local maximum at x=6, and decreases on x>6.

The graph: