It is given that $\mathrm{f}(x) = 6kx + x^2$, where $k 0$. Sketch the graph of \(

Question

It is given that $\mathrm{f}(x) = 6kx + x^2$, where $k > 0$. Sketch the graph of $y = \mathrm{f}(x)$, indicating clearly the coordinates of the points where the curve crosses the axes and of the turning points, if any.[2] Describe the transformation that maps the graph of $y = \mathrm{f}(x)$ onto the graph of $y = x^2 - 9k^2$.[2] It is given that $\mathrm{g}(x) = \begin{cases} 6kx + x^2 & \text{for}\hspace{0.5em} -3k \leq x < k \ \quad 7k^2 & \text{for}\hspace{0.5em} k \leq x < 2k \end{cases}$ $\quad$where $k > 0$ and that $\mathrm{g}(x) = \mathrm{g}(x + 5k)$ for all real values of $x$. Sketch the graph of $y = \mathrm{g}(x)$ for $-3k \leq x < 5k$.[4] Sketch the graph of $y = \frac{1}{\mathrm{g}(x)}$ for $-3k \leq x < 2k$.[3]

Accepted Answer

(a) intercepts $(-6k,0)$, $(0,0)$, turning point $(-3k,-9k^2)$ (b) translate $3k$ units right (c) period $5k$, with the stated parabola and constant segments (d) vertical asymptote $x=0$, reciprocal branches through $\left(-3k,-\dfrac1{9k^2}\right)$ and $\left(k,\dfrac1{7k^2}\right)$, then $y=\dfrac1{7k^2}$ for $k\le x<2k$

2024 CJC P2 Q2

2024 CJC P2 Q2

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