Higher November 2021 Paper 4 Q14
14 Find the coordinates of the turning point of the graph of \(y = x^2 + 6x + 17\). [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| −3, 8 | 4 | B1 for \((x + 3)^2\) or \(-6 \div 2\) B2FT for +8, correct or ft their \((x + 3)^2\) or M1 for \((\textit{their } {-3})^2 + 6 \times (\textit{their } {-3}) + 17\) B1FT for \((-a, b)\) FT their \(\{(x + a)^2 + b\}\) to a maximum of 3 marks If no working B2 for either ordinate correct | accept any correct method (see appendix) B3 implied by \((x + 3)^2 + 8\) |
Appendix: Question 14
The curve does not cross the \(x\)-axis so solving \(y = 0\) does not help.
Alternative method : find the line of symmetry
Find the two points where e.g. \(y = 17\)
\(x^2 + 6x + 17 = 17\) …..M1
\(x(x + 6) = 0\) so \(x = 0\) or \(-6\) …A1
So line of symmetry is \(x = -3\) and by substitution \(y = 8\) …B1 each to maximum of 3 marks