4MA0 Higher June 2013 Paper 4 Q11
11 For \(y = x^3 - 6x^2 + 20\)
(a)
(2)
(i) show that \(y = 4\) when \(x = 2\)
(ii) complete the table of values
| \(x\) | −1 | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|---|---|
| \(y\) | 20 | 15 | −7 | −12 | 20 |
(b) On the grid, draw the graph of \(y = x^3 - 6x^2 + 20\) for values of \(x\) from −1 to 6
(c) For the curve with equation \(y = x^3 - 6x^2 + 20\)
(i) find \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\)
(ii) find the gradient of the curve at \(x = -3\)
(4)| Scheme | Marks |
|---|---|
| (i) \(2^3 - 6 \times 2^2 + 20\) | B1 |
| (ii) 13, (20), (15), 4, (−7), (−12), −5, (20) | B1 |
| (2) |
Notes
B1: must see 8 − 24 + 20 oe
B1: for all correct
| Scheme | Marks |
|---|---|
| Points plotted | B1 ft |
| Curve | B1 ft |
| (2) |
Notes
B1 ft: ± ½ square
B1 ft: if at least B1 scored in (a)
| Scheme | Marks |
|---|---|
| (i) \(3x^2 - 12x\) | M1 A1 |
| (ii) \(3 \times (-3)^2 - 12 \times (-3)\) | M1 |
| 63 | A1 |
| (4) |
Notes
M1 A1: M1 for \(3x^2\) or \(12x\)
M1: ft for a quadratic in c) i)