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anonymous

  • one year ago

Math help!

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  1. anonymous
    • one year ago
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    should be where dy/dx is undefined. Do you have a derivative?

  2. anonymous
    • one year ago
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    3x^2-2y+2x

  3. anonymous
    • one year ago
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    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

  4. anonymous
    • one year ago
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    So would it just be 2?

  5. anonymous
    • one year ago
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    no. where did you get 2?

  6. anonymous
    • one year ago
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    or how I should say?

  7. anonymous
    • one year ago
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    I thought you would just derive 2y to get 2

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

  9. anonymous
    • one year ago
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    So I would get -2y by itself by moving 3x^2 and 2x to the other side?

  10. anonymous
    • one year ago
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    yes

  11. anonymous
    • one year ago
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    So I would get -2ydy/dx=-3x^2-2x?

  12. anonymous
    • one year ago
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    right. Now divide by -2y

  13. anonymous
    • one year ago
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    So it would be 3x^2/2y+x/y

  14. anonymous
    • one year ago
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    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

  15. anonymous
    • one year ago
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    Would I have to do the quotient rule for this?

  16. anonymous
    • one year ago
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    no. just 2y = 0 to get the y-value

  17. anonymous
    • one year ago
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    Y would equal 0 then

  18. anonymous
    • one year ago
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    yes. then plug that into the original equation to find the x-coordinates

  19. anonymous
    • one year ago
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    So the point would be (0,0)?

  20. anonymous
    • one year ago
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    that's one of them. There's another x³ + x² = 0 x²(x + 1) = 0

  21. anonymous
    • one year ago
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    (-1,0)?

  22. anonymous
    • one year ago
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    yep :)

  23. anonymous
    • one year ago
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    Okay thank you!

  24. anonymous
    • one year ago
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    you're welcome!

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