Higher June 2019 Paper 2 Q24
24 Here is a sketch of the curve \(\quad y = x^2 + 4x - 12\)
Work out the values of \(x\) for which \(\quad x^2 + 4x - 12 < 0\)
Give your answer as an inequality. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((x + 6)(x - 2)\) or \(\dfrac{-4 \pm \sqrt{4^2 - 4 \times 1 \times -12}}{2 \times 1}\) or \(-2 \pm \sqrt{16}\) | M1 | oe |
| \(-6\) and 2 | A1 | may be seen in inequalities or as intersections with \(x\)-axis on the graph must be selected if appearing in a list of values or a table |
| \(-6 < x < 2\) or \(2 > x > -6\) | A1ft | ft M1A0 and two values must be a single inequality |
Additional guidance
| To award A1ft the values must be used to give a continuous interval eg1 \((x + 6)(x - 2)\) followed by \((x =)\ 6\) and \((x =)\ {-2}\) Answer \(-2 < x < 6\) eg2 \((x + 6)(x - 2)\) followed by \((x =)\ 6\) and \((x =)\ {-2}\) Answer \(6 < x < -2\) | M1A0A1ft M1A0A0ft |
| \(x < 2\) and \(x > -6\) | M1A1A0 |
| \(-6 < x > 2\) | M1A1A0 |
| \(-6 \leqslant x < 2\) | M1A1A0 |
| \(-6 < x < 2\) in working with different answer on answer line | M1A1A0 |
| \(-6 < x < 2\) in working with integers on answer line | M1A1A0 |