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.

$f(x) = x^2 - 4, g(x) = f(2x) = 4x^2 - 4$

This is a stretch in the x-axis because the x coordinates are being affected.

Now see if you can explain these findings further.