Question

Describe the vertical asymptote and hole for the graph of x^2+x-6/x^2-9.

Answer 1

ok... you need to factor the numerator and denominator

are you able to do that..?

Answer 2

I can... should I do them individually and then but them together?

Answer 3

nope just factor them and leave it in fraction form...

can you show what you have... so I can check

Answer 4

(x - 2)(x + 3)/(x - 3)(x + 3)

Answer 5

ok... if you factorised you have

\[\frac{(x + 3)(x -2)}{(x+3)(x-3)}\]

as you can see (x+ 3) is a factor in the numerator and denominator

so this will result in a hole, or point of discontinuity

vertical asymptotes are the values of x that make the denominator zero...

so after removing the common factor of (x + 3)

your denominator is

x - 3

you need to solve

x = 3 = 0

to find the vertical asymptote.

Answer 6

so the vertical asymptote is x=-3?

Answer 7

so the point of discontinuity is when

x + 3 = 0 ..... or x = -3

substitute this into the simplified equation of

y = (x -2)/(x - 3)

and you'll find the hole, or point of discontinuity occurs at

(-3, 5/6)

hope this helps

Answer 8

oops the you need to solve

x - 3 = 0 for the vertical asymptote

Answer 9

sorry about the typo

Answer 10

can you help me with one more?

Answer 11

sorry I have to go....

but think about vertical asymptotes are the values that make the denominator zero

Answer 12

okay, thanks anyway