Higher November 2019 Paper 1 Q21
21 Sketch the graph of
\[y = 2x^2 - 8x - 5\]
showing the coordinates of the turning point and the exact coordinates of any intercepts with the coordinate axes. (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| Sketch graph with TP at \((2, -13)\) and intercepts at \((0, -5)\), \(\left(2 + \sqrt{\frac{13}{2}}, 0\right)\) and \(\left(2 - \sqrt{\frac{13}{2}}, 0\right)\) | B1 | for a parabola drawn with intercept at the point \((0, -5)\) |
| M1 | for the start of a method to find the roots of \(y = 0\), eg. \(2(x - 2)^2 - 13\ (= 0)\) oe or (\(x =\)) \(\dfrac{--8 \pm \sqrt{(-8)^2 - 4 \times 2 \times -5}}{2 \times 2}\) | |
| M1 | (dep) for method to find the roots, eg. \(2 \pm \sqrt{\dfrac{13}{2}}\) oe | |
| B1 | for turning point at \((2, -13)\) | |
| C1 | for a fully correct parabola drawn with turning point at \((2, -13)\) and intercepts at \((0, -5)\), \(\left(2 + \sqrt{\dfrac{13}{2}}, 0\right)\) oe and \(\left(2 - \sqrt{\dfrac{13}{2}}, 0\right)\) oe clearly shown |
Additional guidance
Turning point may be just seen and labelled on the sketch