CALCULUS HELP NEEDED asap?

Question

Find the equations of the tangents to the curve y=x^3-6x^2+12x+2 which are parallel to the line y=3x.

I've tried differentiating and tried to make it equal to the gradient of the y=3x line, but I'm doing something very wrong. i just cant get the method of doing this question right. Could someone please show step by step how to solve this question, it has me stumped, and the examples given in my book don't help either. The answer at the back says :
y=3x+2
y=3x+6

Answer 1

The tangent at any point has a slope equal to the derivative.

All lines of the same slope are parallel.

We are looking for all lines passing through solutions to f(x) when f'(x)=3

f(x) = x^3 - 6x^2 +12x + 2
f'(x) = 3x^2 -12x + 12

f'(x) = 3

3x^2 - 12x +12 = 3
x^2 - 4x + 4 = 1
x^2 - 4x +3 = 0
(x-1)(x-3) = 0
x=1; x=3


f(x) = x^3 - 6x^2 +12x + 2
f(1) = 1^3 - 6(1^2) +12(1) + 2 = 1 - 6 + 12 +2 = 9
f(3) = 3^3 - 6(3^2) +12(3) + 2 = 27 - 54 + 36 + 2 = 11


Let y1 be the line tangent to f(1)
Let y3 be the line tangent to f(3)

y1 = 3x + b; (slope = 3)
when x=1, y1=9
9 = 3(1) + b; b=6
y1 = 3x + 6

y3 = 3x + b; (slope = 3)
when x=3, y3=11
11 = 3(3) + b; b=2
y3 = 3x + 2

Good luck.

Answer 2

We want to know places where the tangent to the curve has slope 3 so set the derivative equal to 3:
3x^2 - 12x + 12 = 3
3(x^2 -4x +4) = 3
x^2 -4x + 4 = 1
x^2 -4x + 3 = 0
(x-3)(x-1) = 0
x = 3, 1
so plug 3 and 1 into the original equation to get the points on the curve where thre slope of the tangent is 3, then use the point-slope method to get the equations of the tangents

Answer 3

4

Answer 4

first take the derivative
slope=y'=3x^2-12x+2
the slope of y=3x is 3 which is the coefficient of x

hence
3x^2-12x+2=3

or
3x^2-12x-1=0


a=3
b=-12
c=-1

delta=144+12=156

x1=(12+sqrt(156))/6
x2=(12-sqrt(156))/6

y-y1=m(x-x1)
y-y2=m(x-x2)

The numbers aren't suitable to find the answers you gave.

Answer 5

OH MY G0D THATS TOO HARD 4 me.