Higher June 2017 Paper 3 Q23
23 Here is a sketch of \(\quad y = x^2 + bx + c\)
The curve intersects
the \(x\)-axis at (5, 0) and point P
the \(y\)-axis at (0, \(-10\))
Not drawn accurately
Work out the \(x\)-coordinate of the turning point of the graph. [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(0 = 5^2 + 5b + c\) or \(-10 = 0^2 + b(0) + c\) or \(c = -10\) | M1 | oe |
| \(b = -3\) or \(\;x^2 - 3x + c\) or (\(y =\)) \(x^2 - 3x - 10\) | M1dep | oe \((x - 5)(x + k)\) and \(-5k = -10\) |
| \((x - 5)(x + 2)\) or \(\dfrac{--3 \pm \sqrt{(-3)^2 - 4 \times 1 \times -10}}{2 \times 1}\) or \(\dfrac{3 \pm \sqrt{49}}{2}\) or \(\left(x - \dfrac{3}{2}\right)^2 + \ldots\) or \(2x - 3 = 0\) or \(\;x\)-coordinate of P = \(-2\) or two symmetrical coordinates eg (1, −12) and (2, −12) | M1dep | oe Correctly factorises the 3-term quadratic expression or correctly substitutes into quadratic formula for the 3-term quadratic dep on M1 M1 |
| \(1\dfrac{1}{2}\) or \(\dfrac{3}{2}\) with no incorrect working | A1 | oe Accept (1.5, −12.25) |