Higher November 2020 Paper 6 Q19
19 The graph of \(y = 2x^2 + 3x - 9\) is drawn below.
(a) Use the graph to solve \(2x^2 + 3x - 9 = 0\). [2]
(b) The equation \(2x^2 + x - 4 = 0\) can be solved by finding the intersection of the graph of \(y = 2x^2 + 3x - 9\) and the line \(y = ax + b\).
(i) Find the value of \(a\) and the value of \(b\). [2]
(ii) Hence use the graph to solve the equation \(2x^2 + x - 4 = 0\). [3]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [x =] -3, 1.5 | 2 | B1 for 1 correct | |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) [a =] 2 [b =] -5 | 2 | B1 for each or for \(2x - 5\) seen | |
| (ii) \(y = 2x - 5\) or FT \(y = \textit{their}\ ax + \textit{their}\ b\) ruled on grid | M2 | M2 and M1 apply to \(y = 2x - 5\) or FT \(y = \textit{their}\ ax + \textit{their}\ b\) M1 for ‘correct’ \(y\)-intercept or for ‘correct’ gradient or for freehand or broken ‘correct’ line or for at least 3 ‘correct’ plots and no ‘incorrect’ plots | For M2 line must cross curve For M2 and M1, accuracy 1 small square at y-intercept (extended if necessary provided it fits on the grid) and gradient ±1 small square vertically for a run of 1 unit horizontally Do not FT if a = 0 or b = 0 |
| 1.1 to 1.3 and -1.8 to -1.6 | A1 | Only award if M2 scored | |