Higher June 2019 Paper 1 Q19
19 Given that \(x^2 - 6x + 1 = (x - a)^2 - b\) for all values of \(x\),
(i) find the value of \(a\) and the value of \(b\). (2)
(ii) Hence write down the coordinates of the turning point on the graph of \(y = x^2 - 6x + 1\) (1)
| Answer | Mark | Mark scheme |
|---|---|---|
| 3, 8 | M1 | for \(a = 3\), may be seen in working or as part of an expression, eg \((x - 3)^2 - 9\) |
| A1 | for \(a = 3\), \(b = 8\) |
Additional guidance
9 does not have to be seen for this mark
| Answer | Mark | Mark scheme |
|---|---|---|
| 3, \(-8\) | B1 | for 3, \(-8\) or ft (i) |