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

1 Answer
Dec 26, 2017

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]}