CLICK HERE!!!

- anonymous

CLICK HERE!!!

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- anonymous

@ShotGunGirl

- anonymous

DO you have any idea

- anonymous

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- anonymous

oh ok

- anonymous

@Rainbowpower

- anonymous

Do you know?

- anonymous

yep hold on im working the problem out :)

- anonymous

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)

- anonymous

putting it all together:

##### 1 Attachment

- anonymous

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 .

- anonymous

Hope this helps :) have a colorful day

- anonymous

@lala2

- anonymous

Sorry my computer wasnt working

- anonymous

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

- anonymous

@Rainbowpower ??

- radar

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.

- radar

Yes, that is what the problem is seeking.

- anonymous

so that is the answer

- anonymous

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

- anonymous

@radar ??

- anonymous

????

- radar

a division by zero would occur a x=3

- anonymous

so is it x=3 and x=-3

- radar

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

- anonymous

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

- anonymous

and you asked if there was a slope

- anonymous

here is the graph

- radar

|dw:1360084607949:dw|

- anonymous

##### 1 Attachment

- anonymous

what is that...

- anonymous

did you get the graph

- anonymous

@radar ?

- radar

I believe the 3rd choice may be the best, still not sure about the

- anonymous

about what?

- radar

about the x = -3

- anonymous

oh ok... and what about the question with the graph

- radar

I also don't know the answer to the link graph. Not too much help today.

- radar

Regarding the graph, you do have a y intercept at x=-3. It is a function as a vertical line will not intercept the y values twice. It is a non-linear function.

- radar

It is a direct variation y=3x

- anonymous

it want me to write an equation

- anonymous

its asking for an equation

- radar

and I provided one for you. y=3x

- radar

I gotta run for now (lunch)

- anonymous

oh ok thanks!

Looking for something else?

Not the answer you are looking for? Search for more explanations.