Find the equation of the tangent line to the graph of the function f(x)= 2x^3−3x^2+4x+2 at x=2
I got y=16x-12 but I know it is wrong.
I get
f' = 6x^2 - 6x + 4
f'(2) = 16
f(2) = 14
so, you want the line with slope 16 passing through(2,14)
(y-14) = 16(x-2)
y = 16x - 18
where'd you mess up? It's pretty straightforward.