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Parabolas Again

Stage: 4 and 5 Challenge Level: Challenge Level:1

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:

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

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

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

$y = -2x^2 + 1$

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

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

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

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

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

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

$y = 2x^2 - 1$

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

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

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

I created my own pattern with a Matlab program and here are the graphs:

Matlab graphs.

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