for the function
$y=2-4e^-(x-1)^2$
find the equation of the tangent line to the function y at =2

Question

anonymous · Answer

$$y=2-4e ^{-(x-1)^{2}}$$

campbell_st · Answer

can you find the 1st derivative..?

anonymous · Answer

$$8(x-1)e ^{-(x-1)^{2}}$$

campbell_st · Answer

ok... so this is the 1st derivative and also the equation of the slope of the tangent 

so substitute x = 2.... ( I think you left x = 2) out of the question

anonymous · Answer

$$8e ^{-1}$$

anonymous · Answer

does that look right?

campbell_st · Answer

great so that is the slope 

now you need a point on the curve 

substitute x = 2 into the original equation. to get a y value

anonymous · Answer

3.4715?

campbell_st · Answer

but you have a slight error in the equation for the slope 

$$y = 2 - 4e^{-(x -1)^2}$$

then 

$$\frac{dy}{dx} = -8(x -1)e^{-(x -1)^2}$$

can you leave the y value for the point as an exact value..?

campbell_st · Answer

so substituting x = 2

the slope is 

$$m = -8e^{-1}$$

anonymous · Answer

but isn't -2x-4=8?

campbell_st · Answer

now the point 

substituting x = 2  and getting an exact value 

its

$$y = 2 - 4e^{-1}$$

so now you have a point

$$(2, 2 - 4e^{-1})$$

and a slope... so you can find the equation of the tangent  using the point slope formula or slope intercept formula. 

hope it helps

campbell_st · Answer

oops you are correct... my mistake on the derivative

anonymous · Answer

what answer did you get for 2-4e^-1