Their idea was on the right track, it's just that it worked backwards.
For \( y=f(x^2) \), you square the \(x\)-coordinate before applying the function to it. Since the square of any number is negative, effectively speaking anything that was to the left of the \(y\)-axis on the original graph gets discarded, because for something like \(f(x^2)\) you'll never end up plugging a number like -1 into \(f\).
You can see that the graph of \( y = f(x)\) passes through \( (0,0)\), so as you'd expect \(y=f(x^2)\) also passes through \( (0,0) \). Simple reason is because \(0^2 = 0\).
However, \(y=f(x)\) also passes through \( (4,0) \). Since \(4 = 2^2\), for \(y=f(x^2)\) you'd observe the point \( (2,0) \) to lie on there instead. Reason being if you plug \(x=2\) in, you get \(y = f(2^2)\), which becomes \(f(4)\) and equals to \(0\).
In a similar way, \( (-2,0) \) lies on the curve.
Noting that the pattern with the negatives always occurs, you also expect everything on the left of the \(y\)-axis to be a reflection of everything on the right. This can clearly be observed in their solution.
Also, quite visibly in their solution the graph looks more stretched inwards towards the \(y\)-axis. This can be thought of intuitively, as if \(x\) is increasing, \(x^2\) is increasing at a much faster rate. The 'squeezing in' of the graph reflects how what we're plugging into the function, i.e. \(x^2\), just increases at a faster rate than \(x\) itself does.