Given f(x)=-2x^3 + 3x^2 , find the equation of that tangent line of f at the point where x=2.

Step 1: Find f'(x):

f'(x) = -6x^2 + 6x

Step 2: Evaluate f'(2) to find the slope of the tangent line at x=2:

f'(2) = -6(2)^2 + 6(2) = -24 + 12 = -12

Step 3: Find f(2), so you have a point on y=f(x):

f(2) = -2·(2)^3 + 3·(2)^2 = -16 + 12 = -4

So, you have the point (2,-4) and the slope of -12.

Step 4: Find the equation of your tangent line:

Using point-slope form you'd have: y + 4 = -12 (x - 2)

That is the equation of the tangent line.

If your teacher is picky and wants slope-intercept, solve that for y to get:

y = -12 x + 20

i can't see the picture : /