Please advise...
Find the equation of tangent to curve g(x)=4x^3-2x^2+4 at the point (1;6)
Find the derivative of g(x)
g'(x) = 12x^2 -4x
Substitute 1,6 into g'(x)
12(1)^2 - 4 = rate of change at 1, 6 = 12 - 4 = 8
y = 8x + c
Substitute 1,6 into above
6 = 8(1) + c
c = -2
Tangent line
y = 8x - 2 = 2(4x-1)
IS THIS CORRECT???