Question

How do you find an equation of the tangent and normal line to the curve `y=2xe^x` at the point (0,0)?

Answer 1

tangent line: `y = 2x`

normal line: `y = -1/2 x`

Explanation:

`y = 2xe^x, y(0) = 0`

For the tangent line, we will use the derivative:

Using the product rule:
`y' = 2e^x + 2x e^x = 2e^x (1+x), y'(0) = 2`

The tangent line is therefore `y = 2x`

The normal to line `y = mx + c` has slope `-1/m`. The normal also passes through the origin so it's equation is `y = -1/2 x`