Curve Fitter 2

Stage: 5 Challenge Level: Challenge Level:2 Challenge Level:2

Part 1:
$y=f(x)-a$
This transformation shifts the function along the y axis by -a, it will therefore leave the distance between the turning points unchanged.

$y=af(x)$
This transformation will stretch all the y values of the function by a scale factor of a. The x values of the turning points are hence unchanged but the y values are not, this will therefore change the distance between the turning points.

$y = f(ax)$
This transforamtion will stretch all the x values of the function by $\frac{1}{a}$. This will therefore change the distance between the turning points of f(x).

$y=f(x-a)$
This transformation will shift the function along the x axis by a but will leave the distance between the turning points unchanged.

Part 2:

If we Select a simple cubic of the form y = $Ax^3 + Bx^2$ then one turning point will be at the origin.

$\frac{dy}{dx} = 3Ax^2 + 2Bx = x(3Ax + 2B) = 0$ at x = 0 and x = $\frac{-2B}{3A}$

therefore the turning points are $(0,0)$ and $(\frac{-2B}{3A},\frac{4B^3}{27A^2})$

$D^2 = \frac{4B^2}{9A^2}(1 + \frac{4B^4}{81A^2})$

If we let A = $\frac{1}{3}$ then $\frac{4B^2}{9A^2} (1 + \frac{4B^4}{81A^2}) = 4B^2 (1 + \frac{4B^4}{9})$

We now need $\frac{4B^4}{9}$= integer

B can therefore be any integer multiple of $\frac{\sqrt{3}}{\sqrt{2}}$or $\sqrt3$