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.