Question

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

Answer 1

`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`