Question

Math help!

Answer 1

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

Answer 2

3x^2-2y+2x

Answer 3

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

Answer 4

So would it just be 2?

Answer 5

no. where did you get 2?

Answer 6

or how I should say?

Answer 7

I thought you would just derive 2y to get 2

Answer 8

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

Answer 9

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

Answer 10

yes

Answer 11

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

Answer 12

right. Now divide by -2y

Answer 13

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

Answer 14

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

Answer 15

Would I have to do the quotient rule for this?

Answer 16

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

Answer 17

Y would equal 0 then

Answer 18

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

Answer 19

So the point would be (0,0)?

Answer 20

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

Answer 21

(-1,0)?

Answer 22

yep :)

Answer 23

Okay thank you!

Answer 24

you're welcome!