If you meant the two asymptotes at \(x=\pm 1\) for \(y=f(x)\) (which then become 'zeroes' at \(x=\pm 1\) for \( y=\frac{1}{f(x)} \), you're right to suspect that technically speaking we don't know. Because in theory we can't just stare at a graph and always deduce things that relate to vertical asymptotes.
However, the HSC usually rigs its examples so that they should be smooth. Essentially, for ones like that they purposely give you things like \( y= \frac{1}{ax^2+bx+c} \), so that when you reciprocate it you're always bound to get a parabola.
To be honest, I don't think they can penalise you for it either per se.
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Concavity will certainly not be required to obtain all the marks (as opposed to purposely omitting a stationary point). They never mark based off concavity because there's always a risk of too much ambiguity. That solution was generated under the assumption we already knew it was \( y=-\frac{1}{(x-1)(x+1)}\) so that we could reverse engineer it.
(For me, this is more of an advanced "intuition" thing. It's obvious that for the most part the graph is concave down, because we need to approach \(y=1\) from below. But as \(x\to 1^+\), \(f(x) \to -\infty\). As \(f(x) \to -\infty\), \( e^{-f(x)} \to 0^+ \), however it doesn't just spike to 0. Instead, it slowly decays to 0 (with a monotone, concave up behaviour) as we go to progressively large negative values.
So if we left it concave down at 0, we don't really mirror this 'decaying' behaviour. Rather, we seem to be reflecting some kind of spike in the values at \(x=1\) instead.)
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