4MA0 Higher January 2015 Paper 4 Q15
15 Here is the graph of \(y = x^2 - 2x - 1\)
(a) Use the graph to solve the equation \(x^2 - 2x - 1 = 2\) (2)
The equation \(x^2 + 5x - 7 = 0\) can be solved by finding the points of intersection of the line \(y = ax + b\) with the graph of \(y = x^2 - 2x - 1\)
(b) Find the value of \(a\) and the value of \(b\). (2)
| Scheme | Marks |
|---|---|
| M1 | |
| −1, 3 | A1 |
| (2) |
Notes
M1: line \(y = 2\) drawn
A1: SC: B1 for (−1, 2) and (3, 2)
| Scheme | Marks |
|---|---|
| \(x^2 + 5x - 7 + 6 = 6\) or \(x^2 + 5x - 7 - 7x = -7x\) or \(x^2 - 2x - 1 = -7x + 6\) | M1 |
| \(a = -7\), \(b = 6\) | A1 |
| (2) | |
| (4 marks) |
Notes
M1: addition of 6 to both sides or subtraction of \(7x\) from both sides or \(a = -7\) or \(b = 6\)
A1: SC : B1 for \(a = 7\) and \(b = -6\)