How do you find the equation of a line tangent to the function y=x^3-2x^2+2 at (2,2)?

1 Answer
May 19, 2017

y = 4x - 6

Explanation:

First, find the derivative of y, which is

y = 3x^2 - 4x

Now, plug in the x-value (2) into this equation to find the slope of the tangent line:

y = 3(2)^2 - 4(2) = 4

Now plug the x-value (2) back into the original equation to find the y-coordinate of the tangential point:

y = (2)^3 - 2(2)^2 + 2 = 2

(Technically I didn't need to do this since the y-coordinate was given; I did it anyway so you could find the tangent line with just a given x-coordinate.)

Lastly, use the point-slope formula to find the equation of the line tangent to the function at x=2:

y-y_1 = m(x-x_1)

y - 2= 4(x-2)

y = 4x - 6