Question

Find the equation of the line tangent to the graph of f at (1,1) where f is given by f(x)=2x^3-2x^2+1

Answer 1

f(x)= 2x^3 - 2x^2 + 1
Find the derivative
f' = 6x^2 - 4x

The derivative is the slope of a tangent line to a curve, in this case, at (1,1).
slope m = 6x^2 - 4x at (1,1)
m = 6(1^2) - 4(1)
m = 6 - 4 = 2

Using the form of the equation as,
y = mx + b, m = 2
y = 2x + b

To find b,
P(1,1), m = 2
y = mx + b
1 = 2(1) + b
1 = 2 + b
b = -1

So the equation of the tangent line is,
y = 2x - 1