4MA0 Higher June 2015 Paper 4R Q15
15
(a) Complete the table of values for \(y = x^3 - 3x^2 + 5\) (1)
| \(x\) | \(-2\) | \(-1\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|
| \(y\) | \(-15\) | 1 | 5 | 3 |
(b) On the grid, complete the graph of \(y = x^3 - 3x^2 + 5\) for \(-2 \leqslant x \leqslant 4\) (1)
(c) Use the graph to find an estimate for the solution of the equation \(x^3 - 3x^2 + 5 = 0\) (1)
(d) By drawing a suitable straight line on the grid, find an estimate for the solution of the equation \(x^3 - 3x^2 + 2x + 4 = 0\) (3)
| Scheme | Marks |
|---|---|
| 1, 5, 21 | B1 |
| (1) |
| Scheme | Marks |
|---|---|
| correct curve | B1 |
| (1) |
Notes
B1: Correct curve through (1, 3), (2, 1), (3, 5), and (4, 21).
| Scheme | Marks |
|---|---|
| −1.1 | B1 |
| (1) |
Notes
B1: Accept \(-1.15 \leqslant x \leqslant -1.05\)
| Scheme | Marks |
|---|---|
| \(y = 1 - 2x\) drawn | M2 |
| −0.8 | A1 |
| (3) |
Notes
M2: Line must be long enough to cross curve and verify accuracy.
M1 for \(x^3 - 3x^2 + 5 = -2x + 1\) or \(y = -2x + 1\) oe
A1: dep on M2
Accept \(-0.9 \leqslant x \leqslant -0.7\)