June 2024 Paper 1 Q20
20 Here is the graph of \(\quad y = -x^2 + 3x + 6 \quad\) for \(\quad -2 \leqslant x \leqslant 5\)
By drawing a suitable straight line, work out approximate solutions to
\[x^2 - \frac{7}{2}x - 3 = 0\][4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((y =)\ \dfrac{-x}{2} + 3\) | B2 | oe eg \(2y = -x + 6\) can be implied from correct line B1 for \((y =)\ \dfrac{x}{2} - 3\) can be implied from drawing this line or \(m = -\dfrac{1}{2}\) or \(c = +3\) not from incorrect working for either \(m\) or \(c\) |
| \(y = \dfrac{-x}{2} + 3\) accurately drawn on diagram | M1 | ft their stated linear function in \(x\) accuracy is half a square at \(x = 0\) and \(x = 4\) needs to be a straight line (mark intention) |
| \([-0.6, -0.8]\) and [4.1, 4.3] | A1 | dependent on B2M1 |
Additional guidance
| Factorising by use of the quadratic formula alone | B0M0A0 |
| Ignore any y values for A1 | |
| Do not allow M mark for \(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\) linear function drawn | |
| Graph of \(x^2 - \dfrac{7}{2}x - 3 = 0\) | B0M0A0 |