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1989 Paper 1 Q9
D: 1516.0 B: 1500.1
curve sketching cubic turning points graph transformations scaling translation simultaneous equations

Sketch the graph of \(8y=x^{3}-12x\) for \(-4\leqslant x\leqslant4\), marking the coordinates of the turning points. Similarly marking the turning points, sketch the corresponding graphs in the \((X,Y)\)-plane, if \begin{alignat*}{3} \rm{(a)} & \quad & & X=\tfrac{1}{2}x, & \qquad & Y=y,\\ \rm{(b)} & & & X=x, & & Y=\tfrac{1}{2}y,\\ \rm{(c)} & & & X=\tfrac{1}{2}x+1, & & Y=y,\\ \rm{(d)} & & & X=x, & & Y=\tfrac{1}{2}y+1. \end{alignat*} Find values for \(a,b,c,d\) such that, if \(X=ax+b,\) \(Y=cy+d\), then the graph in the \((X,Y)\)-plane corresponding to \(8y=x^{3}-12x\) has turning points at \((X,Y)=(0,0)\) and \((X,Y)=(1,1)\).

Solution: \(8\frac{\d y}{\d x} = 3(x^2-4)\) so the turning points are at \((\pm 2, \mp 2)\)

TikZ diagram
TikZ diagram
TikZ diagram
TikZ diagram
TikZ diagram
We need either \begin{align*} && \begin{cases} -2a+b &= 0 \\ 2c+d &= 0 \\ 2a+b &= 1 \\ -2c+d &= 1 \end{cases} && \text{ or } && \begin{cases} -2a+b &= 1 \\ 2c+d &= 1 \\ 2a+b &= 0 \\ -2c+d &= 0 \end{cases} \\ \Rightarrow && \begin{cases} -2a+b &= 0 \\ 2a+b &= 1 \\ 2c+d &= 0 \\ -2c+d &= 1 \end{cases} && \text{ or } && \begin{cases} -2a+b &= 1 \\ 2a+b &= 0 \\ 2c+d &= 1 \\ -2c+d &= 0 \end{cases}\\ \Rightarrow && \begin{cases} (a,b) = (\frac14,\frac12) \\ (c,d) = (-\frac14, \frac12)\end{cases} && \text{ or } && \begin{cases} (a,b) = (-\frac14,\frac12) \\ (c,d) = (\frac14, \frac12)\end{cases} \end{align*} So either \(X = \frac14 x + \frac12, Y = -\frac14 y + \frac12\) or \(X = -\frac14x + \frac12, Y = \frac14y + \frac12\)

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