Question 749493
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.
 
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+f(a)}}} and {{{x}}} with {{{x+a}}} and then solve for {{{y}}}
When we do that to
{{{y=f(x)=x^3 + alpha*x^2+bx+c}}} we get
{{{y+a^3+alpha*a^2+ba+cross(c)=(x+a)^3+alpha*(x+a)^2+b(x+a)+cross(c)}}}
{{{y+a^3+alpha*a^2+cross(ba)=(x^3+3ax^2+3a^2x+a^3)+alpha*(x^2+2ax+a^2)+bx+cross(ba)}}}
{{{y+cross(a^3)+cross(alpha*a^2)=x^3+3ax^2+3a^2x+cross(a^3)+alpha*x^2+alpha*2ax+cross(alpha*a^2)+bx}}}
{{{y=x^3+3ax^2+3a^2x+alpha*x^2+alpha*2ax+bx}}}
{{{y=x^3+(3a+alpha)x^2+(3a^2+2alpha*a+b)x}}}
The first and second derivatives of {{{f(x)=x^3 + alpha*x^2+bx+c}}} are
{{{"f'(x)"=3x^2+2alpha*x+b}}} and {{{"f''(x)"=6x+2alpha=2(3x+alpha)}}} so
{{{"f'(a)"=3a^2+2alpha*a+b}}} and {{{"f''(a)"=6a+2alpha=2(3a+alpha)}}}
Comparing to {{{y=x^3+(3a+alpha)x^2+(3a^2+2alpha*a+b)x}}} we see that the coefficient of {{{x}}} is indeed {{{"f'(a)"=3a^2+2alpha*a+b}}}
and {{{"f''(a)"/2=2(3a+alpha)/2=3a+alpha}}}is the coefficient of {{{x^2}}}
so {{{y=x^3+("f''(a)"/2)x^2+"f'(a)"*x}}} and {{{highlight(A=2)}}}
 
How would I use all of the above to find the coordinates of the point (B, C) in the graph of
{{{y=x^3-12x^2+48x-68}}} that when translated to the origin turns the function into {{{y=x^3}}}?
 
I wouldn't.
 
I would realize that {{{(x-4)^3=x^3-12x^2+48x-64}}} and that {{{y=(x-4)^3-4}}}
which is {{{y=x^3}}} translated 4 units to the left and 4 units down,
and that is the translation that would bring point (B, C) = (4, 4) to (0, 0).
That looks to me like the most efficient way to the solution.
 
Or maybe after being told that
{{{f(x)=x^3-12x^2+48x-68}}} translated turns into {{{y=x^3}}} and that
{{{"f''(4)"=0}}}
I would realize that {{{f(x)=x^3-12x^2+48x-68}}} must have just one inflection point, just like {{{y=x^2}}}.
Since I know that  {{{y=x^2}}} has its inflection point at (0, 0),
I would realize that the inflection point of {{{f(x)}}} at {{{x=4}}} must be the point translated to the origin.
Then I would know that {{{B=4}}} and would only need to calculate the y-coordinate of the inflection point, {{{C=f(4)}}}
{{{C=4^3-12*4^2+48*4-68}}} --> {{{C=64-12*16+192-68}}} --> {{{C=64-192+192-68}}} --> {{{C=64-68}}} --> {{{C=-4}}}
 
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
{{{f(x)=x^3-12x^2+48x-68}}} into {{{y=x^3+("f''(a)"/2)x^2+"f'(a)"*x}}}
and if {{{"f'(4)"=0}}} and {{{"f''(4)"=0}}} the equation
{{{y=x^3+("f''(a)"/2)x^2+"f'(a)"*x}}} transforms into {{{y=x^3}}}