Mathematical differentiation?

The tangent line to y at x will have slope y'. Therefore, we should start by finding the derivative.

y' = 3x^2 -12x + 12

We want the tangent line to be parallel to y=3x. Parallel lines have the same slope, so you are being asked to find where the tangent line will have slope 3, or where y' =3.

3x^2 - 12x + 12 = 3

3x^2 -12x + 9 = 0

x^2 -4x + 3 = 0

(x-3)(x-1) = 0

so at x=3 and x=1, y'=3, which means the tangent line has slope 3.

To finish the question, you just need to find the equations of the tangent lines at x=3 and x=1.

You already know the slope is 3 for both.

Evaluate y at x=1: 1-6+12+2 = 9. So the tangent line has slope 3 and passes through (1, 9)

y-9 = 3(x-1)

y=3x + 6

Evaluate y at x=3: 27 - 54 + 36 + 2 = 11

y-11 = 3(x-3)

y=3x + 2

the slope of y = 3x is 3

y' = 3x^2 - 12x + 12

slope is also y' = 3

3 = 3x^2 - 12x + 12

1 = x^2 - 4x + 4

1 = (x-2)^2

+/- 1 = x - 2

x = 3 and x = 1

(3, 11) and (1, 9)

Slope forms:

y - 11 = 3(x - 3)

y - 9 = 3(x - 1)

y = 3x + 2

y = 3x + 6

dy/dx = 3x^2 - 12x + 12 = 3

x^2 - 4x + 4 = 1

(x - 2)^2 = 1

x - 2 = ± 1

x = 2 ± 1

x = 3 or 1

If x = 3, y = 11.

y - 11 = 3(x - 3)

If x = 1, y = 9.

y - 9 = 3(x - 1)