The graph of $y=\mathrm{f}(x)$ has turning points at $A(-a, 0)$, $B(a, b)$ and $C(c, -b)$ an

Question

The graph of $y=\mathrm{f}(x)$ has turning points at $A(-a, 0)$, $B(a, b)$ and $C(c, -b)$ and intersects the $x$-axis at the points $D(d, 0)$ and $E(e, 0)$. The graph of $y=\mathrm{f}(x)$ is as shown in the diagram below.[3] Sketch the graph of $y=\mathrm{f}(a-x)$, labelling the coordinates of the corresponding points of $A$, $B$, $C$, $D$, and $E$ clearly.[3] The graph of $y=\mathrm{g}(x)$ has asymptotes $y=0$, $y=-m$ and $x=-n$, and intersects the $y$-axis at $P(0, p)$, where $p < m$. The graph of $y=\mathrm{g}(x)$ is as shown in the diagram below. Sketch on separate diagrams the graphs of $y=|\mathrm{g}(x)|$ and[3] $y=\mathrm{g}(|x|)$,[2] labelling clearly the asymptote(s) and coordinates of $P$ on each graph.

Accepted Answer

(a) $A'(2a,0),\ B'(0,b),\ C'(a-c,-b),\ D'(a-d,0),\ E'(a-e,0)$ (b)(i) Asymptotes $x=-n$, $y=0$, and $y=m$, with $P(0,p)$ (b)(ii) Asymptote $y=0$, with $P(0,p)$

2025 NYJC P1 Q4

2025 NYJC P1 Q4

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