Higher June 2018 Paper 5 Q18
18
(a) A straight line passes through the point (0, 6) and is perpendicular to \(y = 4x - 5\).
Find the equation of this line, giving your answer in the form \(y = mx + c\). [3]
(b) Work out the coordinates of the intersection of the graphs of \(y = 4x - 5\) and \(y = x^2 - 17\). [6]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \(y = -\dfrac{1}{4}x + 6\) | 3 | Mark final answer B2 for correct equation seen or M1 for [grad=] \(-\dfrac{1}{4}\) oe soi M1 for answer \(y = kx + 6\) oe (\(k \ne 0\)) | For 3 marks accept \(y = -0.25x + 6\) Does not have to be in form \(y = mx + c\) e.g. \(y - 6 = -\dfrac{1}{4}(x - 0)\) |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (6, 19) and (–2, –13) | 6 | M2 for \(x^2 - 4x - 12\) [= 0] or M1 for \(x^2 - 17 = 4x - 5\) or better | |
| M2 for \((x - 6)(x + 2)\) [= 0] oe or M1 for \((x + a)(x + b)\) [= 0] where \(a + b = -4\) or \(ab = -12\) | FT their 3 term quadratic equation or expression. Accept correct use of quad formula or complete the square M2 if completely correct, M1 if one error in formula or complete the square | ||
| B1 for either pair of coordinates correct or for \(x = 6\) and \(x = -2\) | See AG for alt method | ||
Alternative: Q18(b) alt algebraic method
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (6, 19) and (–2, –13) | 6 | M2 for \(y^2 - 6y - 247\) [= 0] or M1 for \(y = \left(\dfrac{y + 5}{4}\right)^2 - 17\) or better | |
| M2 for \((y + 13)(y - 19)\) [= 0] oe or M1 for \((y + a)(y + b)\) [= 0] where \(a + b = -6\) or \(ab = -247\) | FT their 3 term quadratic equation or expression. Accept correct use of quad formula or complete the square M2 if completely correct, M1 if one error in formula or complete the square | ||
| B1 for either pair of coordinates correct or for \(y = 19\) and \(y = -13\) | See AG | ||