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EdG Group Title

Find the function y = y(x) such that the line tangent to the graph of y = y(x) at (x, y) intersects the x-axis at x/2 .

  • 10 months ago
  • 10 months ago

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  1. EdG Group Title
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    i know i have to end up with dy/dx = something but i cant figure out how to get there with this clue... any suggestions?

    • 10 months ago
  2. wolfe8 Group Title
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    Well, dy/dx is the gradient at a specific point of a curve.

    • 10 months ago
  3. EdG Group Title
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    ok, so i need to find the equation of the tangent line

    • 10 months ago
  4. dumbcow Group Title
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    tangent line \[y-y_1 = m(x-x_1)\] \[m = \frac{dy}{dx}\] line crosses point (x/2 , 0) \[-y_1 = \frac{dy}{dx}(\frac{x_1}{2}-x_1)\] from here you can set up diff equ and separate variables to solve for y(x)

    • 10 months ago
  5. EdG Group Title
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    Got you, I see where im going now. The only thing im not sure about is the sign on y. Are u positive is negative?

    • 10 months ago
  6. EdG Group Title
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    im meant y being negative...

    • 10 months ago
  7. dumbcow Group Title
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    yes because the line is tangent to curve at (x_1, y_1) the "y" value of line is 0 at x intercept 0 - y_1 = -y_1

    • 10 months ago
  8. EdG Group Title
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    Got it. So i solved it and got: \[y(x) = x^2 +C\] What do you think?

    • 10 months ago
  9. dumbcow Group Title
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    yep well close, i get constant in front y = Cx^2

    • 10 months ago
  10. EdG Group Title
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    ahhhhh, yes. Because 2lnx + C becomes the exponent for e and e^C becomes C

    • 10 months ago
  11. dumbcow Group Title
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    haha yeah was just going to say that :)

    • 10 months ago
  12. EdG Group Title
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    I always mess up whenever e or ln are involved

    • 10 months ago
  13. EdG Group Title
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    Thank you for your help, you got me out of a pickle :)

    • 10 months ago
  14. dumbcow Group Title
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    no problem

    • 10 months ago
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