## 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 . 11 months ago 11 months ago

1. EdG Group Title

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?

2. wolfe8 Group Title

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

3. EdG Group Title

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

4. dumbcow Group Title

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)

5. EdG Group Title

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?

6. EdG Group Title

im meant y being negative...

7. dumbcow Group Title

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

8. EdG Group Title

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

9. dumbcow Group Title

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

10. EdG Group Title

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

11. dumbcow Group Title

haha yeah was just going to say that :)

12. EdG Group Title

I always mess up whenever e or ln are involved

13. EdG Group Title

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

14. dumbcow Group Title

no problem