4400 Higher November 2004 Paper 3 Q17
17 A curve has equation \(y = x^2 - 4x + 1\).
(a) For this curve find
(i) \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\),
(ii) the coordinates of the turning point. (4)
(b) State, with a reason, whether the turning point is a maximum or a minimum. (2)
(c) Find the equation of the line of symmetry of the curve \(y = x^2 - 4x + 1\) (2)
| Scheme | Marks |
|---|---|
| (i) Answer: \(2x - 4\) | B1 |
| (ii) his \(2x - 4 = 0\) | M1 |
| \(x = 2\) | A1ft |
| \((2, -3)\) | A1ft |
| (4) |
Notes
A1ft: Follow her linear \(y'\)
A1ft: Follow her \(x\)
| Scheme | Marks |
|---|---|
| Coeff of \(x^2\) +ve or shape is “U” oe | B1 |
| Min | B1dep |
| (2) |
Notes
B1: or any correct method
B1dep: B1
| Scheme | Marks |
|---|---|
| \(x =\) constant | M1 |
| \(x = 2\) | A1 |
| (2) | |
| (8 marks) |