Higher November 2019 Paper 1 Q27
27 A curve has the equation \(\quad y = x^2 - 6x + 17\)
The turning point of the curve is at \((a, 8)\)
(a) By completing the square, or otherwise, work out the value of \(a\). [2 marks]
(b) The turning point of the curve \(\quad y = x^2 + 4x + b \quad\) also has \(y\)-coordinate 8
Work out the value of \(b\). [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \((x - 3)^2\) | M1 | may be preceded by \(y =\) |
| 3 | A1 | |
| Alternative method 2 | ||
| \((8 = x^2 - 6x + 17\) and\()\) \(x^2 - 6x + 9\ (= 0)\) | M1 | |
| 3 | A1 | |
| Answer | Mark | Comments |
|---|---|---|
| \((x + 2)^2 - 4 + b\) or \(-4 + b = 8\) | M1 | |
| 12 | A1 | SC1 12 from \((x - 2)^2 - 4 + b\) |