Equation of tangent to the curve...

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???**

Re: Equation of tangent to the curve...

Quote:

Originally Posted by

**Scoplex** 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???**

Yes, it looks fine :)