Question

Find a point on the graph of y= e^(2x) at which the tangent line passes through the origin.

Answer 1

This was a tough one! First find the derivative.

y' = 2e^(2x)

The tangent line is y = mx+b
or y = 2e^(2x)*x+b, since y' = slope = m

The initial value of the tangent line is 0 since it passes through the origin, so b=0.
y = 2e^(2x)*x

The tangent line crosses with the original graph, so you can set them equal to each other.
2e^(2x)*x = e^(2x)
2x=1
x=1/2

There's your point :) This was tricky, feel free to ask if you don't understand