Vatsal from Wilson's School saw this function as a sin graph, and approximated its translation:

Similar to a sin graph, with a stretch by scale factor $5$ parallel to the $y$-axis

Translation of approximately $\frac{\pi}{4}$ units in the negative $x$ direction.

Hence, $y \simeq 5sin(x+ \frac{\pi}{4})$.

Louis from EMS and Timothy from Wallington Country Grammar both calculated the horizontal translation of the graph. Timothy writes:

The immediate graph that springs to mind is a sine graph, except this new

graph has two transformations: a stretch scale factor $5$ in the y direction

because the peaks are reaching $5$ and $-5$ and then the second is a shift

along the $x$-axis. This means the equation of our new graph will be of the

form:

$y = ksin(x+c)$

where $k$ is the s.f. of the stretch and $c$ is the shift along the $x$-axis.

k is 5:

$y = 5sin(x+c)$

Now we need to find $c$. From observation it is clear that the graph passes

through the point (0,4) so we can sub these values in and solve $c$.

$4 = 5sin(0+c)$

$sin(c) = \frac{4}{5}$

$c = arcsin(\frac{4}{5})$.

So adding this value to are equation we can reach the final answer of:

$y = 5sin(x+arcsin(\frac{4}{5}))$

Jack from Portsmouth Grammar School saw this function as a cosine graph, and approximated its translation:

You can solve these types of problems by firstly inspecting the maximum and

minimum values of the graph.

Since the maximum and minimum values of a

$y=cos(x)$ graph are $1$ and $-1$, the function must involve $y=5cos(x)$. The

function is the graph of $y=5cos(x)$ shifted to the right by approximately $\frac{\pi}{4}$. Therefore

you subtract $\frac{\pi}{4}$ from the $x$-coordinate, to give $y \simeq 5cos(x-\frac{pi}{4})$.

Jamie from Bentleigh Secondary College calculated the horizontal translation of $y=5cos(x)$:

$f(x-cos^{-1}(\frac{4}{5}))=5cos(x)$

This result was achieved by initially letting $f(x-z) = 5cos(x)$, and

beginning to look towards a mathematical approach to finding the offset.

The approach I used (for ease and effectiveness) was to solve for $z$ when

$5cos(x-z) = 4$ while $x = 0$. This led me to find that $z = cos^{-1}(\frac{4}{5})$.

Jamie also discussed the idea of 'new' functions:

The question 'Do we always get new functions from old ones?', is difficult

to answer. Sometimes we perceive that we have a new function from an old

one, an example being the quadratic formula's relationship with

$f(x)=ax^2+bx+c$, where we acknowledge the former being a separate one from

the latter, where it is just some  'rearrangement' of the same function.

Then in our example where the function was some transformation of the

cosine function, we are tempted to say that it is not a new function, just

some 'changing' of the original one, where widely accepted information

shows that it is.

My understanding is that every function that is 'transformed' from an old

one, is not a new one, but simply a different version of it, what that new

function does could be similar or different from its original, but

nonetheless, the same.

Thank you to everybody who submitted a solution to this problem!