SOLUTION: let α be a real number, let us translate the graph of the cubic function {{{y=f(x)=x^3+ alpha}}}{{{x^2 + bx +c }}} .....{1) so that the point (α,f(α)) on the graph

Question 749493: let α be a real number, let us translate the graph of the cubic function
.....{1)
so that the point (α,f(α)) on the graph (1) is translated into the origin (0,0), and express the function of the translated graph in terms of f'(α) and f"(α)
next we consder the translation which translates the point (α,f(α)) on the graph of (1) into the origin, we replace x with x+α and y with y+f(α) in (1), and obtain the expression y=x^3+ f"(α).x^2/A + f'(α)x
As an example, consider the function
....(2)
f'(4)=0 and f"(4)=0
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
solve for [A] [B] and [C]
Answer by KMST(5328) (Show Source): 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.

One way to translate a graph so that point (a, f(a)) moves to the origin, point (0, 0), is to replace with and with and then solve for
When we do that to
we get

The first and second derivatives of are
and so
and
Comparing to we see that the coefficient of is indeed
and is the coefficient of
so and

How would I use all of the above to find the coordinates of the point (B, C) in the graph of
that when translated to the origin turns the function into ?

I wouldn't.

I would realize that and that
which is 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
translated turns into and that

I would realize that must have just one inflection point, just like .
Since I know that has its inflection point at (0, 0),
I would realize that the inflection point of at must be the point translated to the origin.
Then I would know that and would only need to calculate the y-coordinate of the inflection point,
--> --> --> -->

But maybe we are supposed to use the first part and realize that with it would man that translating (4, f(4)) into the origin would transform
into
and if and the equation
transforms into
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SOLUTION: let &#945; be a real number, let us translate the graph of the cubic function {{{y=f(x)=x^3+ alpha}}}{{{x^2 + bx +c }}} .....{1) so that the point (&#945;,f(&#945;)) on the graph

SOLUTION: let α be a real number, let us translate the graph of the cubic function {{{y=f(x)=x^3+ alpha}}}{{{x^2 + bx +c }}} .....{1) so that the point (α,f(α)) on the graph