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\[y={a(x-a)^2}\div x^2-4a^2\]

I couldnot solve y'=0. I TRIED! :(

What did you get for y'

Oh and we also need to find where y' does not exist

Use quotient rule.

a(x-a)^2(2x-8a)-(x^2-4a^2)(2ax-2a^2)=0

hmmm.... You know a is a constant right?

Yep!

Derivative of a is 0.
Derivative of any constant *a is 0.

a^2 is also a constant since a is a constant
(a^2)'=0

Oops. Now I get it. O_o

Can you please show me what exactly y' would be? All these 'a' and brackets have confused me. :(

@experimentX : Can you please help me with this question?

I'm just asking you now to give me the derivative of
\[ (a(x-a))^2 \text{ and } (x^2-4a^2) \].

For (a(x-a))^2, it would be 2x-2a. For (x^2-4a^2), it would be 2x.

Would it be 2a(x-a)? :O

Now what factors in the numerator do the two terms on top have in common?

2a and x-a.

Yes. :) So factor that out.

\[2a(x-a)(x^2-4a^2)-2xa(x-a)^2\]
=
\[2a(x-a)[ ? - ?]\]

2a(x-a){(x^2-4a^2)-x(x-a)}

\[\frac{2a(x-a)(x^2-4a^2-x^2+xa)}{(x^2-4a^2)^2}\]
You have like terms. :)

Do you see the like terms to combine?

Yep. I'll end up with 2a(x-a)(xa-4a^2). Right?

Ok great now set equal factor equal to zero
You are almost there :)

So I'll have x=a and x=4a? What do I do now?

It said find the coordinates so find what y is if x=a
and find what y is if x=4a

If x=a, then y=0. If x=4a, y=9a^3. Is it correct?

No. Sorry. For x=4a, y is undefined.

Your first one is right.
Your second one is wrong.

So y is not undefined?

Oops. I found out my mistake. For x=4a, y=3a/4.
Thank you soooooo muchhhhh @myininaya ! :D

Great. :)