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
Josh explained this by using
$f(x) = x^2 + 3x - 4$
$g(x) = 2x^2 +6x - 8 = 2(x^2 + 3x - 4)$
so $g(x) = 2f(x)$
He then went on to describe what happens to
$2f(x)$ 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
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) =
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
$f(x) = x^2 - 4, g(x) = f(2x) = 4x^2 - 4$
This is a stretch in the x-axis because the x coordinates are being
Now see if you can explain these