SOLUTION: Find the slope of the tangent line at the given value of x for questions below and then find the equation of the tangent line. f(x)=3x^2+x- 2 at x = 1 f(x)=(2x^2+1)(x

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Question 265306: Find the slope of the tangent line at the given value of x for questions below and then find the equation of the tangent line.
f(x)=3x^2+x- 2 at x = 1

f(x)=(2x^2+1)(x - 1) at x = 3

Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website!
Find the slope of the tangent line at the given value of x for questions below and then find the equation of the tangent line.

f(x)=3x^2+x- 2 at x = 1
f'(x) = 6x+1
f'(1) = 7 (slope of tangent)
f(1) = 2 --> tangent point (1,2)
y=mx+b
2 = 7*1+b
b = -5
y = 7x - 5 (equation of tangent line)
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f(x)=(2x^2+1)(x - 1) at x = 3
f'(x) = 4x(x-1) + 2x^2+1
f'(x) = 6x^2 - 4x + 1
At x=3, f' = 43 = slope of tangent line
f(3) = 38 --> tangent point (3,38)
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y = mx+b
38 = 43*3 + b
b = -91
--> y = 43x - 91 (eqn of tangent line)

SOLUTION: Find the slope of the tangent line at the given value of x for questions below and then find the equation of the tangent line. f(x)=3x^2+x- 2 at x = 1 f(x)=(2x^2+1)(x - 1)