document.write( "Question 750070: let a be a real number, let us translate the graph of the cubic function $\"y=f%28x%29=x%5E3+%2B+alpha\"$ $\"x%5E2%2Bbx%2Bc\"$ .....{1)
\n" ); document.write( "so that the point (a,f(a)) on the graph (1) is translated into the origin (0,0), and express the function of the translated graph in terms of f'(a) and f\"(a)
\n" ); document.write( "next we consder the translation which translates the point (a,f(a)) on the graph of (1) into the origin, we replace x with x+a and y with y+f(a) in (1), and obtain the expression
\n" ); document.write( " $\"y=x%5E3\"$ f\" $\"%28alpha%29%2FA%2Ax%5E2\"$ +f' $\"%28alpha%29\"$ $\"x\"$ \r
\n" ); document.write( "\n" ); document.write( "As an example, consider the function $\"f%28x%29=x%5E3-12x%5E2%2B48x-68\"$ ....(2)
\n" ); document.write( "f'(4)=0 and f\"(4)=0
\n" ); document.write( "we see that when we translate the graph of (2) so that the point (B,C) on the graph is moved to the origin, we get the graph of $\"y=x%5E3\"$
\n" ); document.write( "solve for [A] [B] and [C]
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Algebra.Com's Answer #456421 by KMST(5328) $\"\"$ $\"About$

You can put this solution on YOUR website!
This problem was posted as problem # 749493 (2013-05-16 12:20:44) and as problem # 750070 (2013-05-18 05:10:05). Each time something is was being lost in translation, but it helped to be able to listen to the message twice.
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\n" ); document.write( "One way to translate a graph so that point (a, f(a)) moves to the origin, point (0, 0), is to replace $\"y\"$ with $\"y%2Bf%28a%29\"$ and $\"x\"$ with $\"x%2Ba\"$ and then solve for $\"y\"$
\n" ); document.write( "When we do that to
\n" ); document.write( " $\"y=f%28x%29=x%5E3+%2B+alpha%2Ax%5E2%2Bbx%2Bc\"$ we get
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\n" ); document.write( " $\"y=x%5E3%2B3ax%5E2%2B3a%5E2x%2Balpha%2Ax%5E2%2Balpha%2A2ax%2Bbx\"$
\n" ); document.write( " $\"y=x%5E3%2B%283a%2Balpha%29x%5E2%2B%283a%5E2%2B2alpha%2Aa%2Bb%29x\"$
\n" ); document.write( "The first and second derivatives of $\"f%28x%29=x%5E3+%2B+alpha%2Ax%5E2%2Bbx%2Bc\"$ are
\n" ); document.write( " $\"%22f%27%28x%29%22=3x%5E2%2B2alpha%2Ax%2Bb\"$ and $\"%22f%27%27%28x%29%22=6x%2B2alpha=2%283x%2Balpha%29\"$ so
\n" ); document.write( " $\"%22f%27%28a%29%22=3a%5E2%2B2alpha%2Aa%2Bb\"$ and $\"%22f%27%27%28a%29%22=6a%2B2alpha=2%283a%2Balpha%29\"$
\n" ); document.write( "Comparing to $\"y=x%5E3%2B%283a%2Balpha%29x%5E2%2B%283a%5E2%2B2alpha%2Aa%2Bb%29x\"$ we see that the coefficient of $\"x\"$ is indeed $\"%22f%27%28a%29%22=3a%5E2%2B2alpha%2Aa%2Bb\"$
\n" ); document.write( "and $\"%22f%27%27%28a%29%22%2F2=2%283a%2Balpha%29%2F2=3a%2Balpha\"$ is the coefficient of $\"x%5E2\"$
\n" ); document.write( "so $\"y=x%5E3%2B%28%22f%27%27%28a%29%22%2F2%29x%5E2%2B%22f%27%28a%29%22%2Ax\"$ and $\"highlight%28A=2%29\"$
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\n" ); document.write( "How would I use all of the above to find the coordinates of the point (B, C) in the graph of
\n" ); document.write( " $\"y=x%5E3-12x%5E2%2B48x-68\"$ that when translated to the origin turns the function into $\"y=x%5E3\"$ ?
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\n" ); document.write( "I wouldn't.
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\n" ); document.write( "I would realize that $\"%28x-4%29%5E3=x%5E3-12x%5E2%2B48x-64\"$ and that $\"y=%28x-4%29%5E3-4\"$
\n" ); document.write( "which is $\"y=x%5E3\"$ translated 4 units to the left and 4 units down,
\n" ); document.write( "and that is the translation that would bring point (B, C) = (4, 4) to (0, 0).
\n" ); document.write( "That looks to me like the most efficient way to the solution.
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\n" ); document.write( "Or maybe after being told that
\n" ); document.write( " $\"f%28x%29=x%5E3-12x%5E2%2B48x-68\"$ translated turns into $\"y=x%5E3\"$ and that
\n" ); document.write( " $\"%22f%27%27%284%29%22=0\"$
\n" ); document.write( "I would realize that $\"f%28x%29=x%5E3-12x%5E2%2B48x-68\"$ must have just one inflection point, just like $\"y=x%5E2\"$ .
\n" ); document.write( "Since I know that $\"y=x%5E2\"$ has its inflection point at (0, 0),
\n" ); document.write( "I would realize that the inflection point of $\"f%28x%29\"$ at $\"x=4\"$ must be the point translated to the origin.
\n" ); document.write( "Then I would know that $\"B=4\"$ and would only need to calculate the y-coordinate of the inflection point, $\"C=f%284%29\"$
\n" ); document.write( " $\"C=4%5E3-12%2A4%5E2%2B48%2A4-68\"$ --> $\"C=64-12%2A16%2B192-68\"$ --> $\"C=64-192%2B192-68\"$ --> $\"C=64-68\"$ --> $\"C=-4\"$
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\n" ); document.write( "But maybe we are supposed to use the first part and realize that with $\"a=4\"$ it would man that translating (4, f(4)) into the origin would transform
\n" ); document.write( " $\"f%28x%29=x%5E3-12x%5E2%2B48x-68\"$ into $\"y=x%5E3%2B%28%22f%27%27%28a%29%22%2F2%29x%5E2%2B%22f%27%28a%29%22%2Ax\"$
\n" ); document.write( "and if $\"%22f%27%284%29%22=0\"$ and $\"%22f%27%27%284%29%22=0\"$ the equation
\n" ); document.write( " $\"y=x%5E3%2B%28%22f%27%27%28a%29%22%2F2%29x%5E2%2B%22f%27%28a%29%22%2Ax\"$ transforms into $\"y=x%5E3\"$ \n" ); document.write( "

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