Find an equation for the line tangent.

Hi, I'm studying for a test and I am having troubles for finding the equations for the line tangents.

"Find an equation for the line tangent to y= -2x^3, (-1, 2)"

It would help me greatly if you explained it in a step by step process.

THANK YOU!

First find the derivative:
y = -2x^3
y' = -6x^2

Now evaluate the derivative at x = -1:
y'(-1) = -6(-1)^2
y'(-1) = -6

So, the slope of the tangent line at (-1,2) is -6. Now you know the tangent line passes through the point (-1,2) and has a slope of -6, so you can write the equation of it using point-slope form:
y - y1 = m(x - x1)
y - 2 = -6(x + 1)

If you want, you can rewrite that in slope-intercept form:
y - 2 = -6x - 6
y = -6x - 4

Hope that helps :)