Higher June 2019 Paper 1 Q26
26 The turning point of the graph \(\quad y = (x + a)^2 + b \quad\) has \(x\)-coordinate \(-2\)
(3, 1) is another point on the graph.
Work out the \(y\)-coordinate of the turning point. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((x - -2)^2\) or \((x + 2)^2\) or \(a = 2\) | M1 | oe implied by \(x^2 + 2x + 2x + 4\ (+ b)\) or \(x^2 + 4x + 4\ (+ b)\) |
| \(1 = (3 + 2)^2 + b\) | M1dep | oe |
| \(-24\) | A1 | accept \((-2, -24)\) |
Additional guidance
| \((x - 2)^2\) \(1 = (3 - 2)^2 + b\) | M0 M0 |