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- anonymous

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 .

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- anonymous

- katieb

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- anonymous

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?

- wolfe8

Well, dy/dx is the gradient at a specific point of a curve.

- anonymous

ok, so i need to find the equation of the tangent line

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- dumbcow

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)

- anonymous

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?

- anonymous

im meant y being negative...

- dumbcow

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

- anonymous

Got it. So i solved it and got: \[y(x) = x^2 +C\] What do you think?

- dumbcow

yep well close, i get constant in front
y = Cx^2

- anonymous

ahhhhh, yes. Because 2lnx + C becomes the exponent for e and e^C becomes C

- dumbcow

haha yeah was just going to say that :)

- anonymous

I always mess up whenever e or ln are involved

- anonymous

Thank you for your help, you got me out of a pickle :)

- dumbcow

no problem

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