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- anonymous

Math help!

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- anonymous

Math help!

- chestercat

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- anonymous

should be where dy/dx is undefined. Do you have a derivative?

- anonymous

3x^2-2y+2x

- anonymous

When you take the derivative of y² you have to use the chain rule/implicit differentiation, so multiply by dy/dx. Now solve for dy/dx
3x² - 2y(dy/dx) + 2x = 0

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- anonymous

So would it just be 2?

- anonymous

no. where did you get 2?

- anonymous

or how I should say?

- anonymous

I thought you would just derive 2y to get 2

- anonymous

oh ok. You actually have to solve the equation derivative for dy/dx by moving everything else to the other side.
\[3x^2-2y \frac{ dy }{ dx }+2x=0\]

- anonymous

So I would get -2y by itself by moving 3x^2 and 2x to the other side?

- anonymous

yes

- anonymous

So I would get -2ydy/dx=-3x^2-2x?

- anonymous

right. Now divide by -2y

- anonymous

So it would be 3x^2/2y+x/y

- anonymous

yes :) BUT it's easier to keep it as one fraction because you need to find the undefined locations and it's just easier with one fraction instead of multiples, especially on more complex problems.
\[\frac{ dy }{ dx }=\frac{ 3x^2+2x }{ 2y }\]
since we're looking for vertical tangent lines we need where dy/dx is undefined. That would be where the denominator is 0

- anonymous

Would I have to do the quotient rule for this?

- anonymous

no. just 2y = 0 to get the y-value

- anonymous

Y would equal 0 then

- anonymous

yes. then plug that into the original equation to find the x-coordinates

- anonymous

So the point would be (0,0)?

- anonymous

that's one of them. There's another
x³ + x² = 0
x²(x + 1) = 0

- anonymous

(-1,0)?

- anonymous

yep :)

- anonymous

Okay thank you!

- anonymous

you're welcome!

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