1MA0 Higher June 2013 Paper 2 Q15
15
(a) Complete the table of values for \(y = x^2 - 2x\)
(2)
| \(x\) | \(-2\) | \(-1\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|---|---|
| \(y\) | 3 | 0 | 3 |
(b) On the grid, draw the graph of \(y = x^2 - 2x\) for values of \(x\) from \(-2\) to 4
(c) Solve \(x^2 - 2x - 2 = 1\) (2)
| Answer | Mark | Notes |
|---|---|---|
| \(x\): \(-2\), \(-1\), 0, 1, 2, 3, 4 \(y\): 8, 3, 0, \(-1\), 0, 3, 8 | 2 | B2 for 8, \(-1\), 0, 8 (B1 for at least two of 8, \(-1\), 0, 8) |
| Answer | Mark | Notes |
|---|---|---|
| Correct curve | 2 | M1 (ft) for at least 5 points plotted correctly A1 for a fully correct curve |
| Working | Answer | Mark | Notes |
|---|---|---|---|
| \(x^2 - 2x - 3 = 0\) OR \((x - 3)(x + 1) = 0\) | 3 and \(-1\) | 2 | M1 for the straight line \(y = 3\) drawn to intersect the “graph” from (a) A1 for both solutions OR M1 for identifying \(y = 3\) from the table A1 for both solutions OR M1 for \((x \pm 3)(x \pm 1)\) A1 for both solutions |