Question

Medal &Fan
Describe the vertical asymptote and hole for the graph of x^2+x-6/x^2-9.
A.Asymptote:x=2;hole x=-3
B.Asymptote:x=3;hole x=2
C.Asymptote:x=-3;hole x=3
D.Asymptote:x=3;hole x=-3

Answer 1

x^2+x-6/x^2-9 =0
(x-3)(x+2) /(x-3)(x+3)=0
(x+2)/(x+3) =0 .............( x +3 )dnt=0 ok ?
x+2=0 , x=-2
got it untill nw ?

Answer 2

I think so ,yeahh

Answer 3

What do i do next?

Answer 4

( x +3 )dnt=0 ok
when
x+3=0 , x=-3 (it called vertical asymptote )

Answer 5

what about the hole part?

Answer 6

nw from this one
(x-3)(x+2) /(x-3)(x+3)=0 , u can see that we can cancelate (x-3) right ?

Answer 7

so the limite we cancelate
x-3=0
x=3
got it ??

Answer 8

Yess. Thank you so much!, i really appreciate it

Answer 9

good luck ^^

Answer 10

Thanks (:

Answer 11

np :)