Finding the equation for the tangent line. Help Please!

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thomcart8

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Ok my equation is f(x) = (e^3x)((x^2)-5). I am supposed to find an equation for the tangent line at 1.

Ok so I was able to find f"(x) and that equals ((3x^2) + 2x - 15)(e^3x). So i know that is the formula I need to find my slope, but I kind of got lost after this point. Can anyone out there help me get farther on the problem? I greatly appreciate anything you can help me with.
 
\(\displaystyle f'(x)=e^{3x}(3x^{2}+2x-15)\), which you have. Now, sub in x=1 and find the slope there.

Then, sub x=1 into f(x) to find y at that point.

Sub these values into y=mx+b and solve for b. That's it.
 
Okay i did what you said and I found -4e^3 = -10e^3(1) + b. So b ended up being 6e^3. so the would the equation be f(1) =-10e^3(x)+6e^3?
 
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