Need HELP . Find the equations of the tangents to the curve....?

Question

Need HELP . Find the equations of the tangents to the curve y=x^3-6x^2+12x+2 which are parallel to the line y=3x.
Could someone help me out step by step to work this out. Should show up with two linear equations with same gradient.

Answer 1

y' gives the slope of the tangent line.

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

We want the slope of the tangent to be 3, so:

3 = 3x^2 - 12x + 12

3x^2 - 12x + 9 = 0

x^2 - 4x + 3 = 0

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

x=1 or x=3

Corresponding to x=1, the y coordinate is (1)^3-6(1)^2+12(1)+2
y=9

Corresponding to x=3 the y coordinate is (3)^3-6(3)^2+12(3)+2
y=11

Now use the point slope formula to find the equations of the tangent lines.

Answer 2

First find the derivative of the curve:

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

This will define the slope of the tangent. You want the tangent that has a slope of 3, so set this equal to 3.

3x^2-12x+12 = 3

x^2-4x +3=0

x = 1, 3

Go back to the original equation and solve for y to get the two tangent points.

(1,9) and (3,11)

Use the point slope formula for the two equations:

y-9=3(x-1)

and

y-11=3(x-3)

Answer 3

y' = 3x^2-12x+12
all tangents parallel to y=3x will have slope 3:
3x^2-12x+12=3
x^2-4x+3=0
(x-1)(x-3)=0
x=1,3
y=9,11

a) y-9=3x-3,
y=3x+6
b) y-11=3x-9,
y=3x+2