June 2025 Paper 1 Q5
5 Here is a sketch of \(\quad y = x^2 + bx + c\)
The graph passes through \((-1, 0)\) and \((5, 0)\)
Work out the values of \(b\) and \(c\). [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1: by using the roots | ||
| \((x - -1)(x - 5)\) or \((x + 1)(x - 5)\) | M1 | |
| \(x^2 - 5x + x - 5\) or \(x^2 - 4x - 5\) or \(b = -4\) or \(c = -5\) with correct factors | M1dep | oe eg could be in a grid |
| \(b = -4,\ c = -5\) | A1 | |
| Alternative method 2: by substitution | ||
| \(0 = 1 - b + c\) and \(0 = 25 + 5b + c\) or \(1 - b + c = 25 + 5b + c\) | M1 | oe do not allow \(0 = (-1)^2 + (-1)b + c\) or \(0 = (5)^2 + (5)b + c\) until processed correctly |
| \(0 = 24 + 6b\) or \(0 = 30 + 6c\) | M1dep | oe eliminating either \(b\) or \(c\) |
| \(b = -4,\ c = -5\) | A1 | |
| Alternative method 3: by using the minimum from completing the square | ||
| \((x - 2)^2 + k\) or \(\left(x + \dfrac{b}{2}\right)^2 + k\) | M1 | oe \(k\) could be any letter or number or combination or 0 |
| \(0 = (3)^2 + k\) or \(k = -9\) or \(\dfrac{-b}{2} = 2\) | M1dep | oe by substituting in one pair of coordinates 3 could be \(-3\) or \(+3\) or \(\pm 3\) |
| \(b = -4,\ c = -5\) | A1 | |
| Alternative method 4: by using the minimum from differentiation | ||
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = 2x + b\) | M1 | |
| \(2(2) + b = 0\) | M1dep | oe |
| \(b = -4,\ c = -5\) | A1 | |
Additional guidance
| Up to M1 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| Both coefficients correct but no working | M2A1 |
| One coefficient correct and the other incorrect or missing and no working | M0A0 |