Tangent line that passes through origin

[crybllrd]

Homework Statement

Find a > 0 such that the tangent line to the graph of

f(x) = $x^{2}$e$^{-x}$ at x = a passes through the origin.

Homework Equations

The Attempt at a Solution

First I found the derivative to be:

$-e^{-x}(x-2)x$

, which is the slope of the function.

I know the tangent line must pass through the origin (0,0), but I'm a bit stuck.

I took a year off from math and am trying to get back into math mode. Any help to lead me to the next step would be great.

 

[Elliptic]

maybe this is the way
http://en.wikipedia.org/wiki/Tangent
find f '(x) and than substitute x=a
use point-slopre formula and substitude k=f '(a) inside, than epress from there y
than y=x=0 (through origin)
and ... tell me your result

 

[Joffan]

Formula for a line through the origin is y = mx

At a, y=f(a), m=f'(a), x=a

 

[Elliptic]

you are right (my k is yours m,)

 

[chapstic]

continuing from Joffan and Elliptic: find where f'(x)x = f(x) = a. find f'(a). at that point you are basically done.

 

[crybllrd]

Thanks guys for the responses.

Here's what I have so far:

$f(a)=a^{2}-e^{-a}$

$f '(a)=-e^{-a}(a-2)a$$y=f '(a)(x-a)+f(a)$

which gave me

$a=0,3$(zero is not in the domain, so three is the only solution)

There must be an error, just looking at it graphically I know that 'a' must be before the extrema (2).

 

[crybllrd]

I found my error.
I missed a negative sign :/
My final answer is a=1.
The problem doesn't require it, but I checked my answer by graphing the tangent line using y=f '(a)x, and it worked out.

thanks again for all your help.

 

[Elliptic]

Yes, a=1. Well done.