Parabolas again

This problem was solved by Ali Abu Hijleh, age 13, Alex Liao, age 14 and Prateek Mehrotra, age 14 from Riccarton High School, Christchurch, New Zealand and Andrei Lazanu, age: 12 from School No. 205, Bucharest, Romania. Well done all of you. This is Andrei's solution:

I observed that the equations of the parabolas given in the problem are of the form: $y = a(x + b)^2 + c$

where positive corresponds to parabolas with the opening towards the top, and negative to parabolas with the opening towards the bottom; are the coordinates of the peak. I identified the parabolas in your pattern and I verified them with the help of a small program in Matlab:

x=-8:0.1:8

y1=-2*(x+6).*(x+6)-2

y2=-2*(x+4).*(x+4)-1

y3=-2*(x+2).*(x+2);

y4=-2*x.*x+1;

y5=-2*(x-2).*(x-2)+2;

y6=-2*(x-4).*(x-4)+3;

y7=-2*(x-6).*(x-6)+4;

y8=2*(x+6).*(x+6)+2;

y9=2*(x+4).*(x+4)+1;

y10=2*(x+2).*(x+2);

y11=2*x.*x-1;

y12=2*(x-2).*(x-2)-2;

y13=2*(x-4).*(x-4)-3;

y14=2*(x-6).*(x-6)-4;

plot(x,y1,x,y2,x,y3,x,y4,x,y5,x,y6,x,y7,x,y8,x,y9,x,y10,x,y11,x,y12,x,y13,x,y14)

hold

axis([-8 8 -4 4]);

grid

Consequently, the equations of the 14 parabolas in your pattern are:

$y = -2(x + 6)^2 - 2$

$y = -2(x + 4)^2 - 1$

$y = -2(x + 2)^2$

$y = -2x^2 + 1$

$y = -2(x - 2)^2 + 2$

$y = -2(x - 4)^2 + 3$

$y = -2(x - 6)^2 + 4$

$y = 2(x + 6)^2 + 2$

$y = 2(x + 4)^2 + 1$

$y = 2(x + 2)^2$

$y = 2x^2 - 1$

$y = 2(x - 2)^2 - 2$

$y = 2(x - 4)^2 - 3$

$y = 2(x - 6)^2 - 4$

Image

Andrei supplied the Matlab program but in order to challenge you to find his equations it is not reproduced here