How do you graph f(x)=(x^2+3x+2)/(-3x-12) using holes, vertical and horizontal asymptotes, x and y intercepts?

1 Answer
Jul 9, 2018

Below

Explanation:

f(x)=(x^2+3x+2)/(-3x-12)

Using long division, we can rearrange the polynomial
f(x)=((-3x-12)(-1/3x-1/3)-2)/(-3x-12)

f(x)=-1/3x-1/3-2/(-3x-12)

This means that your oblique asymptote is y=-1/3x-1/3 which can be found by figuring out what happens when x approaches 0 and your horizontal asymptote is x=-4 after solving -3x-12=0 since your denominator cannot equal to 0.

To figure out your x and y intercepts, we let y=0 and x=0 respectively.

When x=0, y=-1/6 so your y-intercept is (0,-1/6)
When y=0, x=-2 and x=-1 so your x-intercepts are (-2,0) and (-1,0)

graph{(x^2+3x+2)/(-3x-12) [-10, 10, -5, 5]}
Above is the graph