The curve ${{C}_{1}}$ has equation $y=\mathrm{f}\left( x \right)$, where \(\mathrm{f}\left(

Question

The curve ${{C}_{1}}$ has equation $y=\mathrm{f}\left( x \right)$, where $\mathrm{f}\left( x \right)={{x}^{3}}-4x+1$. A second curve ${{C}_{2}}$ undergoes, in succession, the following transformations to get the resulting curve ${{C}_{1}}$: $A$: A translation of $1$ unit in the negative direction of the $y$-axis $B$: A reflection in the $y$-axis $C$: A scaling of factor $\frac{1}{4}$ parallel to the $x$-axis Determine the equation of ${{C}_{2}}$, showing your workings clearly. The diagram shows the curve $y=\mathrm{f}\left( x \right)$. The curve passes through $\left( 4,0 \right)$ and $\left( 0,-2 \right)$, and has a maximum point $\left( 6,a \right)$ where $a>0$. It has asymptotes $x=2$, $y=\frac{1}{2}a$ and $y=0$. Sketch the graph of $y=\frac{1}{\mathrm{f}\left( x \right)}$,[3] $y=\mathrm{f}\left( \left| x \right| \right)$,[3] showing clearly the stationary points, axial intercepts and equations of all asymptotes.

Accepted Answer

(a) $C_2:y=-\dfrac{x^3}{64}+x+2$ (b)(i) Vertical asymptote $x=4$, horizontal asymptote $y=\dfrac2a$, hole at $(2,0)$, minimum $\left(6,\dfrac1a\right)$, $y$-intercept $\left(0,-\dfrac12\right)$, no $x$-intercepts (b)(ii) Vertical asymptotes $x=\pm2$, horizontal asymptote $y=\dfrac a2$, stationary points $(\pm6,a)$, axial intercepts $(-4,0)$, $(0,-2)$ and $(4,0)$

2018 AJC Promo Q6

2018 AJC Promo Q6

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