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anonymous
 2 years ago
Find the function y = y(x) such that the line tangent to the graph of y = y(x) at (x, y) intersects the xaxis at x/2 .
anonymous
 2 years ago
Find the function y = y(x) such that the line tangent to the graph of y = y(x) at (x, y) intersects the xaxis at x/2 .

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anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0i know i have to end up with dy/dx = something but i cant figure out how to get there with this clue... any suggestions?

wolfe8
 2 years ago
Best ResponseYou've already chosen the best response.0Well, dy/dx is the gradient at a specific point of a curve.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0ok, so i need to find the equation of the tangent line

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0tangent line \[yy_1 = m(xx_1)\] \[m = \frac{dy}{dx}\] 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)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Got you, I see where im going now. The only thing im not sure about is the sign on y. Are u positive is negative?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0im meant y being negative...

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0yes 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

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Got it. So i solved it and got: \[y(x) = x^2 +C\] What do you think?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0yep well close, i get constant in front y = Cx^2

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0ahhhhh, yes. Because 2lnx + C becomes the exponent for e and e^C becomes C

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0haha yeah was just going to say that :)

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0I always mess up whenever e or ln are involved

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Thank you for your help, you got me out of a pickle :)
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