Question

how to idenify the horizontal and vertical asympotes:
(4x-8)/(x^2-x-6)

Answer 1

For vertical asymptotes
\[\lim_{x \rightarrow a} f(x)=\infty\]
then x=a is your vertical asymptote

for horizontal asymptotes
\[\lim_{x \rightarrow \infty} f(x)=b\]
then y=b is your horizontal asymptote

Answer 2

....huh? sorry im just confused

Answer 3

I know there's a formula for this but I don't remember it or can find it in my notes

Answer 4

\[\lim_{x \rightarrow a} \dfrac {4x-8} {x^2-x-6}=\lim_{x \rightarrow a} \dfrac {4x-8} {(x-3)(x+2)}\]
So, this limit goes to infinite when x=3 and x=-2, then you have two vertical asymptotes, x=3 and x=-2
\[\lim_{x \rightarrow \infty} \dfrac {4x-8} {x^2-x-6}=\lim_{x \rightarrow \infty} \dfrac {x(4-\dfrac {8} {x})} {x(1-\dfrac {3} {x}) (x+2)}=\lim_{x \rightarrow \infty} \dfrac {(4-\dfrac {8} {x})} {(1-\dfrac {3} {x}) (x+2)}\]
bla bla bla and it's equal to 0
Then yor horizontal asymptote is y=0

Answer 5

ohh i get it thanks

Answer 6

;)

Answer 7

(in the second limit you could have just used L'Hôpital rule if i remember well, it's an infinite/infinite limit)