You may also like

problem icon


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?

problem icon

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?

problem icon

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.
$f(x) = x^2 +2, g(x) = 2f(x) = 2x^2 +4$
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.