Question

Can anyone explain to me how to find vertical asmptotes and holes for any graph the one i have is like (X+1)(X-2)*OVER*(X-2)(X+4)

Answer 1

Does this simplify at all?

Answer 2

uh i have no idea

Answer 3

do you see any common factors

Answer 4

Only X

Answer 5

how about x-2?

Answer 6

that appears twice

Answer 7

Oh So..this might be much to ask but can you explain how to solve it?...Sorry :/

Answer 8

\[\Large \frac{(x+1)(x-2)}{(x-2)(x+4)}\]

\[\Large \frac{(x+1)\cancel{(x-2)}}{\cancel{(x-2)}(x+4)}\]

\[\Large \frac{x+1}{x+4}\]

Answer 9

See how I'm cancelling those common factors?

Answer 10

I do

Answer 11

So

\[\Large \frac{(x+1)(x-2)}{(x-2)(x+4)}\]

simplifies to

\[\Large \frac{x+1}{x+4}\]

Answer 12

In other words, they are the same

Answer 13

But, in order for them to be the exact same, the domains must match

So we must specify that \(\Large x \neq 2\) and \(\Large x \neq -4\) to avoid dividing by zero

Answer 14

The vertical asymptote will be x = -4 because this causes a division by zero error in

\[\Large \frac{x+1}{x+4}\]

and there is a hole at x = 2 because there is no division by zero error in \[\Large \frac{x+1}{x+4}\] but we must make this exclusion to make sure the domains match

Answer 15

is +2 considered the hole and -4 the asmptote?

Answer 16

yes

Answer 17

Thank you Very much! Im understanding alot better now! i might be asking for a bit more help soon in the future! Please keep an Eye out! Thanks again!

Answer 18

alright, yw