is a 'hump' solution centered at the point $x = -4.5$. Thus, we can envision the solution \[ f(x-ct) = \exp\left[-(x + 4.5 - ct)^2\right], \]

as the profile $f(x)$ but now travelling rightwards at a speed $c$. But what do we make of $-f(-x-ct)$? A second key to this question is to expand your domain to the entire real line. For this, we ask whether the posited form of $y$ is an odd or even function. If we set $x \to -x$, then \[ y(-x,t) = f(-x-ct) - f(x-ct) = -y(x,t), \]

and thus we expect $y$ to be an odd function, and the addition of $-f(-c-ct)$ creates a second, identical, but downward facing bump initially set at $x = 4.5$ and then moving leftwards. Thus, when we add the two pieces together, \[ y(x,t) \equiv f(x-ct) - f(-x-ct). \]

we should have an upwards facing hump moving from left to right, and a downwards facing hump moving from right to left. This is shown in the below figure:

Now let us return to the original question and ask ourselves what happens strictly in $(-\infty, 0]$. We know from the boundary conditions (and from oddness of the function) that $y(0, t) = 0$. Had you *only* been looking on $x < 0$, you would have seen a wave moving in from left-to-right, shrink once it reaches $x = 0$, and then 'bounce' back (the reflected wave), but now inverted about the x-axis (try and hide the region $x > 0$ in the above figure with your hand).