Higher November 2023 Paper 2 Q18
18 By completing the square, prove that \(\quad x^2 + 6x + 13 \quad\) is always positive. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((x + 3)^2\) … | M1 | |
| \((x + 3)^2 + 4\) | A1 | |
| \((x + 3)^2 + 4\) and valid argument | A1 | eg \((x + 3)^2 + 4\) and \((x + 3)^2 \geqslant 0\) and adding 4 or \((x + 3)^2 + 4\) and this is \(\geqslant 4\) or correct reference to a minimum point and its position above the \(x\)-axis |
Additional guidance
| \((x + 3)^2 + 4\) and Even if \(x\) is negative it is squared so will be positive so the expression is always positive (no reference to the \(+ 4\)) | M1A1A0 |
| \((x + 3)^2 + 4\) and Turning point is \((-3, 4)\) which is positive on the \(y\)-axis and as \(x^2\) coefficient it is a U-shape therefore always positive | M1A1A1 |
| Incorrect work after \((x + 3)^2 + 4\) seen, eg \((x + 3)^2 + 4 = 0\) | M1A1A0 |
| Condone \(\gt\) for \(\geqslant\) |