how do I find the equation of the tangent line at y=(2x+2)e^x given the point p=2,6e^2

Question

tkhunny · Answer

What do you get when you evaluate the 1st derivative for a given value of x?

anonymous · Answer

the slope, and i got 2e^x(x+2) but i dont know what to do next

tkhunny · Answer

The slope of what?

anonymous · Answer

the equation of the tan line?

tkhunny · Answer

Very good.  If the derivative is correct, you are ready to proceed.

You have a point.
You have a slope.
You have the Point-Slope form of a line: $(y-y_{0}) = m(x-x_{0})$

Go!

anonymous · Answer

my answer has to be in y=mx+b form, and I cant seem to simplify it :( I have 4 options to choose from and my answer isnt matching up

tkhunny · Answer

x = 2
$y = (2(2)+2)\cdot e^{2} = 6\cdot e^{2}$

You have a point: $(2,6\cdot e^{2})$

As you correctly calculated:
$y' = 2(x+2)\cdot e^{x}$

Again, with x = 2
$y'(2) = 2(2+2)\cdot e^{2} = 8\cdot e^{2}$

You have a slope: $8\cdot e^{2}$

Now, we apply the Point-Slope form of the line!

tkhunny · Answer

Great!  Let's see what you get and how you get it.

anonymous · Answer

ohhhhhhhhh you plug the 2 back into the slope !! i was trying to figure out an answer with 2e^x(x+2) and it just wasn't working, so does 8e^2(x) - 10e^2 make sense?

tkhunny · Answer

$(y - 6e^{2}) = 8e^{2}(x-2)$

$y - 6e^{2} = 8e^{2}(x) - 16e^{2}$

$y = 8e^{2}(x) - 10e^{2}$

It's not pretty, but there it is!  Good work.

tkhunny · Answer

That's really awkward-looking.  I think I would prefer $8xe^{2}$ rather than $8e^{2}x$.

anonymous · Answer

no joke 8e^2x was how it was written in my text book as a possible answer, thanks so much though