Question

How do you graph `f(x)=(x^2+7x+12)/(-4x^2-8x+12)` using holes, vertical and horizontal asymptotes, x and y intercepts?

Answer 1

hole: `x=-3`
vertical asymptote: `x=1`
horizontal asymptote: `y=-1/4`
x and y intercepts: `(0,1) and (-4,0)`

Explanation:

`f(x)={x^2+7x+12}/{-4x^2-8x+12}={(x+3)(x+4)}/{-4[x^2+2x-3]}={cancel((x+3))(x+4)}/{-4[cancel((x+3))(x-1)]}=-{(x+4)}/{4(x-1)}`
`=>`
hole: `x=-3`

---

From now on we will work only with our new nicer `f(x)`:
`f(x)=-{(x+4)}/{4(x-1)}`
vertical asymptote: `4(x-1)=0 iff x=1`
horizontal asymptotes:
`lim_{x rarr +oo}f(x)=-1/4`
`lim_{x rarr -oo}f(x)=-1/4`
`=>`
horizontal asymptote: `y=-1/4`

---

x and y intercepts:
`f(0)=-{(0+4)}/{4(0-1)}=-4/-4=1`
`f(x)=0 iff -{(x+4)}/{4(x-1)}=0 => x=-4`
`=>`
x and y intercepts: `(0,1) and (-4,0)`

---

graph:
graph{(x^2+7x+12)/(-4x^2-8x+12) [-10, 10, -5, 5]}