Question

So here is the question:

Find and equation of the tangent line to the graph:

4)f(x)=e^(2x) passing through the point (0,0)

I find derivative being f'(x) =2e^(2x)

Then m = f'(0) = 2

put into point slope formula:

(y-y1)=m(x-x1)
y-0 = 2(x-0)
which comes out to
y=2x

0=2x ; so the new x coordinate is x= 0
new y is y= 2(0) = 0

new m = 2e^(2(0)) = 2e^(0) = 2(1) = 2

therefore final function is y = 2x

Is this right?

Answer 1

As you can see from the graphs at

http://www.wolframalpha.com/input/?i=plot+y%3De^%282x%29%2C+y%3D2x

you missed the mark. So, let's see what happened. At the point (x,y) on the curve, the slope is 2e^(2x).

A line with slope 2e^(2x) through the origin:

e^(2x)/x = 2e^(2x)
x = 1/2

So, the line is tangent to the curve at (1/2,e), with slope 2e. See

http://www.wolframalpha.com/input/?i=plot+y%3De^%282x%29%2C+y%3D2e*x

Answer 2

Thanks!