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Converse

Clearly if a, b and c are the lengths of the sides of a triangle and the triangle is equilateral then a^2 + b^2 + c^2 = ab + bc + ca. Is the converse true, and if so can you prove it? That is if a^2 + b^2 + c^2 = ab + bc + ca is the triangle with side lengths a, b and c necessarily equilateral?

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Consecutive Squares

The squares of any 8 consecutive numbers can be arranged into two sets of four numbers with the same sum. True of false?

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Parabolic Patterns

The illustration shows the graphs of fifteen functions. Two of them have equations y=x^2 and y=-(x-4)^2. Find the equations of all the other graphs.

More Quadratic Transformations

Stage: 4 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3

Muntej, from Wilson's School, sent us screenshots showing that for the first pair of functions, using the same input gave an output from $g$ which was twice the output from $f$.

Josh explained this by using  $f(x)$.
 
$f(x) = x^2 + 3x - 4$
$g(x) = 2x^2 +6x - 8 = 2(x^2 + 3x - 4)$
so $g(x) = 2f(x)$
 
He then went on to describe what happens to the graphs.
 
 $2f(x)$ will be twice as large as $f(x)$ for any value of x including negative numbers, for example if $f(x) = -4$ then $2f(x) = -8$. However, the solutions for $y = 2f(x)$ will be the same as $y = f(x)$ because $2x0 = 0$. Thus, the graph $y = g(x)$ is $y=f(x)$ stretched in the $y$ axis by scale factor $2$.
 
Henry from Latymer School did the same for the second set of functions.
 
$f(x) = x^2 + 3x - 4 $
$g(x) = (2x)^2 + 3(2x) - 4 $
$x$ has essentially been doubled before being put into the function so $g(x) = f(2x)$.
$y=g(x)$ is $y=f(x)$ stretched by scale factor $0.5$ parallel to the $x$-axis.
 
Josh told us that you can work out the input numbers for which both machines give the same results by calculating $f(x) = g(x)$, which is the same as finding where the functions cross on a graph.He used algbra to find the solutions to the second functions. 
 
The $x$ value for which $g(x) = f(x)$ is:
 
$4x^2 + 6x - 4 = x^2 + 3x - 4$
$3x^2 + 3x = 0$
$x^2 + x = 0$
$x(x+1) = 0$
so $x = 0$ or $x = -1$ because $0(0 +1) = 0$ and $-1(-1+1) = 0$
 
Ellen tried her own set of equations and used graphs to show the transformations.
 
$f(x) = x^2 +2, g(x) = 2f(x) = 2x^2 +4$
 
 2f(x)
This is a stretch in the y-axis because the y coordinates are being affected.

$f(x) = x^2 - 4, g(x) = f(2x) = 4x^2 - 4$ f(2x)
This is a stretch in the x-axis because the x coordinates are being affected.
 
Now see if you can explain these findings further.