<p>The diagram shows the curve $y=\mathrm{f}\left( x \right)$. The curve has maximum points at $\left( -5,4 \right)$ and the origin, and crosses the $x$-axis at $\left( -4,0 \right)$. The lines $y=0$, $x=-3$ and $y=-x+2$ are the horizontal, vertical and oblique asymptotes to the curve respectively.</p>
<figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/public/question-images/converted/1759132149707-cciqofelaba.jpg" width="400" class="aligncenter" /></figure>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">$y=\frac{1}{\mathrm{f}\left( x \right)}$</span></div><p style="text-align: right;">[3]</p></li><li><div class="list-item-content"><span class="list-item-text">$y=\mathrm{f}\left( \frac{x+1}{2} \right)$</span></div><p style="text-align: right;">[3]</p></li></ol>
<p>labelling clearly the equation(s) of any asymptote(s), coordinates of any axial intercept(s) and turning point(s) where applicable.</p>

The diagram shows the curve $y=\mathrm{f}\left( x \right)$. The curve has maximum points at $\left( -5,4 \right)$ and the origin, and crosses the $x$-axis at $\left( -4,0 \right)$. The lines $y=0$, $x=-3$ and $y=-x+2$ are the horizontal, vertical and oblique asymptotes to the curve respectively.

$y=\mathrm{f}\left( \frac{x+1}{2} \right)$

labelling clearly the equation(s) of any asymptote(s), coordinates of any axial intercept(s) and turning point(s) where applicable.

2017 RI P1 Q5

2017 RI P1 Q5

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