SOLUTION: find the equation of the normal to the curve with the equation y=e^3x-2 at the point (1,e)

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: find the equation of the normal to the curve with the equation y=e^3x-2 at the point (1,e)       Log On


   

Question 1136901: find the equation of the normal to the curve with the equation y=e^3x-2 at the point (1,e)
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
y=e^3x-2
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Is the exponent 3x ?
Or 3x - 2 ?
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Parentheses are free. Use some.
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f%28x%29+=+e%5E%283x-2%29
f'(x) = %283x-2%29%2Ae%5E%283x-2%29%2A3+=+%289x-6%29%2Ae%5E%283x-2%29 = slope
f'(1) = 3e
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slope of normal = -1/3e
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y-e = (-1/3e)*(x-1)
y+=+e+%2B+%281-x%29%2F3e