Question

Find equation of the normal line to the given curve at the specified point.

Answer 1

\[y=2xe^x \]
Point: ( 0, 0 )

Answer 2

Remember how to find slope of a `tangent line`? :)
`normal line` slope will be a similar process.

Answer 3

\[y' = 2e^x(x+1)\]

Answer 4

\[y'(0) = 2e^0(0+1) = 2 \times 1 \times 1\] = 2

Answer 5

The equation of the tangent line would be \[y - 0 = 2( x - 0)\]

Answer 6

That is what I have so far.

Answer 7

Mm k good.
Back up to the tangent slope that you found.
The normal line will be `perpendicular` to this tangent line.
So the slope will be the `negative reciprocal` of the tangent slope.

Answer 8

So if \(\large\rm tangent~slope=m\)
then \(\large\rm normal~slope=-\frac{1}{m}\)

Answer 9

ah okay, I got it now :)

Answer 10

yay c: