Find an equation of the tangent line to the curve at the given point 3x^2-x^3, (1,2)

Question

Find an equation of the tangent line to the curve at the given point  3x^2-x^3, (1,2)

anonymous · Answer

take the derivative, 
$$f'(x)$$ then find 
$$f'(1)$$ for the slope and then use the point - slope formula

anonymous · Answer

Using the power rule to find the derivative. You should get 6x -3x^2.. You are given the point (1,2) so plug in the X value to find the slope of the tangent line which is 3. Use the point-slope formula to obtain the equation of the tangent line.

jim_thompson5910 · Answer

First derive y to find the derivative function. This will help you find the slope.


y = 3x^2-x^3


y' = 6x-3x^2


y' = -3x^2+6x


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Now find slope of tangent line at x = 1


y' = -3x^2+6x


y' = -3(1)^2+6(1)


y' = -3(1)+6(1)


y' = -3+6


y' = 3


So the slope of the tangent line at x = 1 is 3. So m = 3

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Now use the equation y = mx+b to find the equation of the line


y = mx+b


2 = 3(1)+b


2 = 3+b


2-3 = b


-1 = b


b = -1


So the equation of the tangent line is y = 3x-1