Here's the question you clicked on:
EdG
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 .
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?
Well, dy/dx is the gradient at a specific point of a curve.
ok, so i need to find the equation of the tangent line
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)
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?
im meant y being negative...
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
Got it. So i solved it and got: \[y(x) = x^2 +C\] What do you think?
yep well close, i get constant in front y = Cx^2
ahhhhh, yes. Because 2lnx + C becomes the exponent for e and e^C becomes C
haha yeah was just going to say that :)
I always mess up whenever e or ln are involved
Thank you for your help, you got me out of a pickle :)