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!
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!
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
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
$y=3x+2.\tag{7}$