Higher June 2018 Paper 1 Q29
29 Here are sketches of two graphs.
The graph of \(\quad y = x^2 - 1 \quad\) is translated 3 units to the left to give graph A.
(a) The equation of graph A can be written in the form \(\quad y = x^2 + bx + c\)
Work out the values of \(b\) and \(c\). [3 marks]
(b) The graph of \(\quad y = x^2 - 1 \quad\) is reflected in the \(x\)-axis to give graph B.
Work out the equation of graph B. [1 mark]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \((x + 3)^2 - 1\) | M1 | |
| \(x^2 + 3x + 3x + 9 - 1\) or \(x^2 + 6x + 8\) | M1 | oe |
| \(b = 6\) and \(c = 8\) | A1 | SC1 \(b = 6\) or \(c = 8\) |
| Alternative method 2 | ||
| \((x - 3)^2 + b(x - 3) + c = x^2 - 1\) | M1 | |
| \(x^2 - 6x + 9 + bx - 3b + c = x^2 - 1\) | M1 | |
| \(b = 6\) and \(c = 8\) | A1 | SC1 \(b = 6\) or \(c = 8\) |
| Alternative method 3 | ||
| \((x + 3 + 1)(x + 3 - 1)\) or \((x - -4)(x - -2)\) or \((x + 4)(x + 2)\) | M1 | difference of two squares from the original roots |
| \(x^2 + 4x + 2x + 8\) or \(x^2 + 6x + 8\) | M1 | |
| \(b = 6\) and \(c = 8\) | A1 | SC1 \(b = 6\) or \(c = 8\) |
Additional guidance
Working out the roots of the original curve or the translated curve is not enough for M1 in alt 3
| Answer | Mark | Comments |
|---|---|---|
| \(y = 1 - x^2\) or \(y = -x^2 + 1\) | B1 | oe equation |
Additional guidance
| \(-y = x^2 - 1\) | B1 |
| \(y = -(x^2 - 1)\) | B1 |
| \(y = -(x - 1)(x + 1)\) | B1 |
| \(y = 1 - (-x)^2\) | B1 |
| (\(y = 1 - x^2\) in working with answer) \(1 - x^2\) | B0 |
| \(y = (-x)^2 + 1\) | B0 |
| \(\mathrm{f}(x) = 1 - x^2\) | B0 |