can you show me how to this one?

Question

anonymous · Answer

@Yttrium

anonymous · Answer

What do you need help with? :)

anonymous · Answer

attachment

anonymous · Answer

Hmm. Not so sure how to do this one. I'll see if I can get someone else to help you though. :)

anonymous · Answer

ty

anonymous · Answer

no problem. @ganeshie8 @sapphire13 member still needs help! Thanks :)

anonymous · Answer

Hey

anonymous · Answer

ur in k12 too

anonymous · Answer

hey can you help and yes

anonymous · Answer

k

anonymous · Answer

I got D

anonymous · Answer

how tho i wanna be able to do t on my own

ganeshie8 · Answer

why not A

ganeshie8 · Answer

$$\dfrac{u+3}{u^2-9} = \dfrac{u+3}{u^2-3^2} = \dfrac{u+3}{(u+3)(u-3)}$$

anonymous · Answer

it could be A

ganeshie8 · Answer

^you need to knw below identity :
$a^2-b^2 = (a+b)(a-b) $

ganeshie8 · Answer

@sierraleone17

ganeshie8 · Answer

yes its either A or D, 
lets figure it out :)

anonymous · Answer

okay so what do i do. what did you just do?

ganeshie8 · Answer

The original function becomes meaningless for$ u = -3$
so we need to avoid both $u =3$ and $u = -3$

yttrium · Answer

^agreed sorry I forgot to see the original function

yttrium · Answer

@sierraleone17 do you know the fact that u^2 - 9 = u^2 - 3^2 = (u-3)(u+3) ???

anonymous · Answer

yes

yttrium · Answer

From that, we can simplify the original equation to 1/(u-3) agreed?

anonymous · Answer

i have that writted down my tutor says to cancel it BUT Idk it doesnt look right and agreed

yttrium · Answer

Well (u+3)/(u+3) = 1 so (u+3)/(u+3)(u-3) = 1/(u-3) you get it now?

anonymous · Answer

yes i do thank you the only thing im still fuzzy about is the restrictions

yttrium · Answer

okay. okay.whenever you are looking for restrictions, always focus on the original function.
the original function is $$\frac{ u+3 }{ u^2 - 9 }$$

anonymous · Answer

okay

anonymous · Answer

and the like what?

yttrium · Answer

As said, we are using restrictions to avoid denominator be equal to zero. Given that denominator is u^2 - 9, having $$u = \pm 3$$ will result to zero. Hence, $$u \neq \pm 3$$

anonymous · Answer

omg thank you!!!!!

yttrium · Answer

No problem. Another approach is that you can actually equate the denominator to zero and solve for its solutions, after doing that just put an inequality sign. :)