June 2019 Paper 1 Q7
7 A function is given by
\[\begin{aligned} \mathrm{f}(x) &= -2x &\qquad& -1 \leqslant x \lt 0 \\ &= x(4 - x) && 0 \leqslant x \lt 3 \\ &= 2x - 3 && 3 \leqslant x \leqslant 4 \end{aligned}\]Draw the graph of \(\quad y = \mathrm{f}(x) \quad\) on the grid. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
 | B4 | B1 for a straight line from \((-1,\ 2)\) to \((0,\ 0)\). This should be drawn with a ruler (give BOD) B1 for a quadratic style curve through \((0,\ 0)\), \((1,\ 3)\), \((2,\ 4)\) and \((3,\ 3)\) within tolerance. Condone one straight line between only one of the sections between \((0,\ 0)\) and \((1,\ 3)\), \((1,\ 3)\) and \((2,\ 4)\) or \((2,\ 4)\) and \((3,\ 3)\) B1 for any quadratic graph drawn with correct curvature and no straight lines (see examples) with clear vertices at \((0,0)\) and \((3,3)\) and within tolerance for these points. There needs to be evidence of a maximum turning point drawn. B1 for a straight line from \((3,\ 3)\) to \((4,\ 5)\). This should be drawn with a ruler (give BOD) SC1 for all six stated points plotted correctly and clearly defined (don’t need to be joined up or could be joined up incorrectly). If any incorrect points plotted then no marks can be awarded |
Additional guidance
| Tolerance of plot \(\pm\) 2mm for each stated point (these are 1cm squares) | |
| For \(\mathrm{f}(x) = -2x\) extending to the left of \(x = -1\) or \(\mathrm{f}(x) = 2x - 3\) extending to the right of \(x = 4\) greater than 2mm, award maximum 1 mark from B1 B1 only for a section that would have otherwise scored ie. If both lines extended then B0 B1 | |
| If more than one line drawn in any section then choice so loss of marks | |
| Ignore shading under or over the lines as long as the graph is clear | |
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