Patrick from Woodbridge school sent us this solution:

I plotted the first few equations with graphing software, and they all seemed to have the turning point at $(-2,0)$. I will now try and show this for all cases.

Taking $y=ax^2+2bx+c$, we can replace $a$, $b$ and $c$ by consective numbers from the geometic progression, say, $$\eqalign{a&=m\cr b&=2m\cr c&=2^2 m = 4m}$$ This gives us $$\eqalign{

y&=mx^2+2\times 2mx+4m\cr &= mx^2 + 4mx + 4m}$$

Now we differentiate to get $$\eqalign{\frac{dy}{dx}&= 2mx+4a \cr &= 2m(x+2)}$$

The turning point is given by $\frac{dy}{dx}= 0$, so $2m(x+2) = 0$ imples $x=-2$.

Substituting $x=-2$ into our equation for $y$ gives: $y=4m-8m+4m = 0$.

Therefore, all these curves have turning point at $(-2,0)$.

We can now generalise these ideas for any common ratio $n$.

As before we have: $$\eqalign{y&=ax^2 + 2bx + c\cr &= mx^2 + 2mnx + mn^2\cr &= m(x^2+2nx+n^2)}$$

Differentiating gives: $$dy/dx = 2mx + 2mn = 2m(x+n)$$

The x-coordinate of the turning point will thus be $-n$, and the y-coordinate is $(-n)^2 + 2n(-n) + n^2 = n^2 - 2n^2 + n^2 = 0$.

Therefore, for a quadratic in the form $y=ax^2+2bx+c$ where the coefficients are consequtive numbers in a geometric progression, the turning point of the curve will be $(-n,0)$. This will work regardless of whether n is natural or real.