y' = 8x-6x^2
so, at x=a, y' = 8a-6a^2 = m
Use the above info to plug into the point-slope form of the lines. No sweat.
--- Find the slope m of the tangent to the curve y = 4 + 4x^2 − 2x^3 at the point where x = a.
---- Find equations of the tangent lines at the points (1,6) and (2,4).
(1,6) Y(x)=
(2,4) Y(x)=
1 answer