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Find the equation of the tangent line to the curve

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 at the given point. y=x^3-2x+1 at (3,22)?

asked Sep 25, 2014 in CALCULUS by anonymous

1 Answer

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The curve equation is y = x3 - 2x + 1

Differentiating on each side with respect of x .

y' = 3x2 - 2

Substitute the values (x , y) = (3, 22) in above equation.

y' = 3(3)2 - 2

y' = 27 - 2

y' = 25

This is the slope of tangent line to the parabola at (3, 22).

To find the tangent line equation, substitute the values of m = 25 and (x, y) = (3, 22) in the slope intercept form of an equation y = mx + b.

Where m is slope and b is y intercept.

22 = 25(3) + b

22 = 75 + b

b = - 75 + 22

b = - 53

Substitute m = 25 and b = - 53 in y = mx + b.

y = 25x - 53

Tangent line equation is y =  25x - 53.

 

answered Sep 25, 2014 by david Expert

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