The graph of $y=\mathrm{f}(x)$ is shown in the diagram. It passes through the origin and has a min

Question

The graph of $y=\mathrm{f}(x)$ is shown in the diagram. It passes through the origin and has a minimum point at $\left( 2,\mathrm{ }0 \right)$. The lines $x=1$ and $y=2$ are the asymptotes of the graph. Sketch the graph of $y=2-\mathrm{f}(x)$, stating the coordinates of the point where the graph crosses the $y$-axis, the coordinates of any stationary point(s) and the equations of any asymptotes.[2] The curve $C$ has equation $y=\frac{2{{x}^{2}}+kx-2}{x+2}$, where $k$ is a constant. Find the set of values of $k$ for which $C$ has $2$ stationary points.[3] For the rest of the question, take $k=-3$. By expressing the equation $y=\frac{2{{x}^{2}}-3x-2}{x+2}$ in the form $y=ax+b+\frac{c}{x+2}$, where $a$, $b$ and $c$ are constants, write down the equations of the asymptotes of $C$.[3] Hence sketch $C$, giving the equations of any asymptotes, the coordinates of any stationary points and of the points where $C$ crosses the axes.[2] Describe a pair of transformations which transform the graph of $y=x+\frac{6}{x+2}$ to $C$.[2] For the rest of the question, take $k=-3$.

Accepted Answer

(b)(i) Set of values of $k=\{k\in \mathbb{R}:k<3\}$ (ii) $y=2x-7+\frac{12}{x+2}$ (iv) First, stretch parallel to $y$-axis by factor $2$, with $x$-axis invariant. Next, translate $7$ units in the negative $y$-direction.

2020 VJC Promo Q8

2020 VJC Promo Q8

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