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High School
Calculus and Analysis
Calculus 1
Derivatives

ds522stlk

Answered question

2022-04-09

What is the equation of the tangent line of

f (x) = - 2 x^{3} + 6 x^{2} + 2 x - 3

at x=3?

Answer & Explanation

Mey9ci0

Beginner2022-04-10Added 14 answers

Explanation:
The equation of the line is y - b = m (x - a ) where m is gradient
and (a , b ) a point on the line.
To find m (gradient ) we differentiate f(x) as f'(x) gives the gradient of the tangent to the curve.

f^{'} (x) = - 6 x^{2} + 12 x + 2

and

m = f^{'} (3) = - 6 {(3)}^{2} + 12 (3) + 2 = - 54 + 36 + 2 = - 16

a = 3 : b = f (3) = - 2 {(3)}^{3} + 6 {(3)}^{2} + 2 (3) - 3

=-54+54+6-3=3
equation of tangent: y-b=m(x-a), m=-16, (a,b)=(3,3)
so y-3=-16(x-3) hence y-3=-16x+48

\Rightarrow y = - 16 x + 51

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