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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!

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Where did all the comments go? –  The Chaz 2.0 Aug 25 '11 at 18:34
    
Calculus sure does help a lot here. The calculus-free approach is a little bit involved... –  J. M. Aug 25 '11 at 20:18

3 Answers 3

up vote 6 down vote accepted

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

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there is no given (x,y) so how can i do? –  Sb Sangpi Aug 25 '11 at 14:20
    
Can you take the derivative of your cubic? That is the slope of the tangent line at each point. You want a slope of 3-that is the point of being parallel to y=3x. So set the derivative to 3 and solve the equation for x. –  Ross Millikan Aug 25 '11 at 14:39
    
aww..ok. so it is like this y=3x^3-1-12x^2-1+12x+2 ? thx –  Sb Sangpi Aug 25 '11 at 14:46
    
No-I'll add to the above. –  Ross Millikan Aug 25 '11 at 15:31
1  
@Ross: When presenting hints and/or solutions to problems like the present problem, I've found it helpful to notationally distinguish between the coordinates of the tangency point and the $x$, $y$ variables used in the functions, by letting $(a,b)$ be the coordinates of the tangency point. I don't have time now to write a detailed solution using this approach (and I probably shouldn't, but rather encourage Sb Sangpi to do this), but here's an example I wrote 7 months ago in another group: mathforum.org/kb/message.jspa?messageID=7363805 –  Dave L. Renfro Aug 25 '11 at 19:11

enter image description here

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

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

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step by step great help! looking forward more help! thx alot Tavares. God bless you! ;) –  Sb Sangpi Aug 26 '11 at 6:01
    
@Sb Sangpi: Glad I could help. –  Américo Tavares Aug 26 '11 at 8:39

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.

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Do you reckon that someone who asks this kind of question is up for Taylor series? –  Gerry Myerson Aug 26 '11 at 1:32
    
@gerry - yes I do. Derivatives of polynomials are trivial and the rest is algebra. Maybe this is an opportunity to learn something. –  ja72 Aug 26 '11 at 12:10
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I have 80 students this semester who think derivatives of polynomials are anything but trivial, and that algebra is not far removed from Sanskrit. –  Gerry Myerson Aug 26 '11 at 13:10
    
@gerry - rofl - On the other hand, somebody motivated enough to post on MathSE is already above the average student. BTW, do you grade on a curve? –  ja72 Aug 26 '11 at 14:47
    
Yes - but I'm not saying which curve. –  Gerry Myerson Aug 27 '11 at 1:34

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