# Thread: family of cubics question

1. ## family of cubics question

Q1. Find the equation of the form $\displaystyle y=a(x-h)^3+k$ using the coordinates (0,0),(24,9) and (60,10) where (24,9) is the point of inflection.

It seems obvious that using (24,9) as the point of inflection gives $\displaystyle y=a(x-24)^3+9$. Now substituting in (60,10) to find the dialation factor 'a'.

$\displaystyle 10=a(60-24)^3+9$

$\displaystyle 10=a(36)^3+9$

$\displaystyle 10=46656a+9$

$\displaystyle 1=46656a$

$\displaystyle a=\frac{1}{46656}$

therefore giving $\displaystyle y=\frac{1}{46656}(x-24)^3+9$

Q2. Generalise your result to find a family of cubic equations that satisfy these conditions

As we have 3 fixed points in a cubic I cannot start to think how a family of curves can be formed given these conditions. Anyone have an idea?

2. Originally Posted by pickslides
Q1. Find the equation of the form $\displaystyle y=a(x-h)^3+k$ using the coordinates (0,0),(24,9) and (60,10) where (24,9) is the point of inflection.

It seems obvious that using (24,9) as the point of inflection gives $\displaystyle y=a(x-24)^3+9$. Now substituting in (60,10) to find the dialation factor 'a'.

$\displaystyle 10=a(60-24)^3+9$

$\displaystyle 10=a(36)^3+9$

$\displaystyle 10=46656a+9$

$\displaystyle 1=46656a$

$\displaystyle a=\frac{1}{46656}$

therefore giving $\displaystyle y=\frac{1}{46656}(x-24)^3+9$

Q2. Generalise your result to find a family of cubic equations that satisfy these conditions

As we have 3 fixed points in a cubic I cannot start to think how a family of curves can be formed given these conditions. Anyone have an idea?

A general cubic has four degrees of freedom (three roots and a multiplier if you like, though two of the roots may be complex)

CB

3. Originally Posted by pickslides
Q1. Find the equation of the form $\displaystyle y=a(x-h)^3+k$ using the coordinates (0,0),(24,9) and (60,10) where (24,9) is the point of inflection.

It seems obvious that using (24,9) as the point of inflection gives $\displaystyle y=a(x-24)^3+9$. Now substituting in (60,10) to find the dialation factor 'a'.

$\displaystyle 10=a(60-24)^3+9$

$\displaystyle 10=a(36)^3+9$

$\displaystyle 10=46656a+9$

$\displaystyle 1=46656a$

$\displaystyle a=\frac{1}{46656}$

therefore giving $\displaystyle y=\frac{1}{46656}(x-24)^3+9$

[snip]
The graph of this equation does not pass through (0, 0). The given model cannot satisfy all of the given information. The question is flawed.

4. Originally Posted by mr fantastic
The graph of this equation does not pass through (0, 0). The given model cannot satisfy all of the given information. The question is flawed.
Its k that is the problem (the position of the point of inflection does not determine k.

CB