Equation of the tangents 
What is the equation of the tangent to $y=x^3-6x^2+12x+2$ that is parallel to the line $y=3x$ ?

I have no idea, how to solve, no example is given in the book! 
Appreciate your help! 
 A: Hint:  what is the slope of $y=3x$?  Then what is the slope of $y=x^3-6x^2+12x+2$    (which depends upon $x$)?  You need to find an $x$ where the slopes match.  Then find a $y$ so that $(x,y)$ is a point on the cubic.  Now you have  a point and a slope, giving you the equation for the line.
Added:  the slope of the tangent is $y'=3x^2-12x+12\ \ $  ,  which we are told is $3\ \ $.  Solving $3=3x^2-12x+12\ \ $  gives $x=1 \text{ or } 3\ \ $  .  So the  points of tangency are $(1,9)$ and $(3,11)\ \ $.  The lines with slope $3$ that pass through these points are $y=3x+6\ $ and $y=3x+2\ \ $.   A figure is at Wolfram Alpha
A: 
The equation of the tangent to the graph of $f(x)$ at $(a,f(a))$ is given by
$$y=f(a)+f^{\prime }(a)(x-a)=f^{\prime }(a)x+f(a)-f^{\prime }(a)a.\tag{1}$$
Two lines with equations $y=mx+b$ and $y=m^{\prime }x+b^{\prime }$ are
parallel if and only if $m=m^{\prime }$. Hence, the family of lines parallel
to the line $y=3x$ is given by $y=3x+b$. So, we must have
$$f^{\prime }(a)x+f(a)-f^{\prime }(a)a=3x+b.\tag{2}$$
Equating coefficients we get $f^{\prime }(a)=3$ and $f(a)-f^{\prime }(a)a=b$. Since the derivative of $f(x)=x^{3}-6x^{2}+12x+2$ at $x=a$ is $f^{\prime
}(a)=3a^{2}-12a+12$, we obtain the system of two equations
$$3a^{2}-12a+12=3,\tag{3}$$
$$a^{3}-6a^{2}+12a+2-3a=b,$$
which is equivalent to
$$a=1,b=6\tag{4}$$
or
$$a=3,b=2.\tag{5}$$
Hence the equations of the two tangent lines are 
$$y=3x+6\tag{6}$$
and
$$y=3x+2.\tag{7}$$
A: Look at a taylor series (ref 1, 2) of your function about a point $x_c$ of order 1 (linear). 
Then find which $x_c$ produces a line parallel to $3x$, or has a slope of $3$. 
Hint! There are two solutions actually.
