Question

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

Answer 1

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