Exponentials and tangents (1 Viewer)

[miss_b]

Find the equation of the tangent to the curve with equation y=1+2e^(x-1) at the point (1,3).

Thanks

 

[SoulSearcher]

Re: Not sure where this should go...

y = 1+2e^x-1
y' = 2e^x-1
at x = 1
y' = 2e^1-1
= 2e⁰
= 2
therefore equation of tangent is
y-3 = 2(x-1)
y-3 = 2x-2
y = 2x+1

 

[miss_b]

Re: Not sure where this should go...

Thanks heaps

 

[Slidey]

Re: Not sure where this should go...

This problem is at a 2unit level, for future reference.