Question

CLICK HERE!!!

Answer 1

@ShotGunGirl

Answer 2

DO you have any idea

Answer 3

Not not really. I stink at this stuff. Sorry :(

Answer 4

oh ok

Answer 5

@Rainbowpower

Answer 6

Do you know?

Answer 7

yep hold on im working the problem out :)

Answer 8

Factoring polynomials never goes away. You have to factor numerator and denominator.

x^3+x^2-4x-4 can be factored "by grouping" as

X^3+x^2-4x-4=(x^3-4x)+(x^2-4)+(x^2-4)=(x+1)*(x^2-4)
It can be factored further because the difference of squares
x^2-4=x^2-2^2
factors as (x-2)*(x+2)

x^2+2x-3=(x+3)*(x-1)

Answer 9

putting it all together:

Answer 10

oh and the function is not defined (it does not exist) for x=1 and for x=-3 because the denominator is zero for those values of x .

Answer 11

Hope this helps :) have a colorful day

Answer 12

@lala2

Answer 13

Sorry my computer wasnt working

Answer 14

so would the answer be x=1 and x=-3

Answer 15

@Rainbowpower ??

Answer 16

x=1 and x=-3 would result in a division by zero, and as you know that is a "no-no" and would result in a discontinuity.

Answer 17

Yes, that is what the problem is seeking.

Answer 18

so that is the answer

Answer 19

And what about this one
Describe the vertical asymptote and hole for the graph of x^2+x-6/ x^2-9

Answer 20

@radar ??

Answer 21

????

Answer 22

a division by zero would occur a x=3

Answer 23

so is it x=3 and x=-3

Answer 24

I am not sure what occurs when x=-3 but x=3 and x=-3 is probably your best choice.

Answer 25

oh ok thanks... you know that question i sent you

Answer 26

and you asked if there was a slope

Answer 27

here is the graph