Higher June 2024 Paper 1 Q25
25 \(\mathrm{f}(x) = 17 - 3x^2 + 12x\)
Write \(\mathrm{f}(x)\) in the form \(\;a - b(x - c)^2\;\) where \(a\), \(b\) and \(c\) are constants.
(4)
| Scheme | Marks |
|---|---|
| \(\pm 3(x^2 \pm 4x)\)……… or \(\pm 3(x^2 \pm 4x\ldots\ldots\ldots)\) or \(b = 3\) | M1 |
| \(-3\big[(x - 2)^2 \ldots\ldots\ldots\big]\) or \(-3(x - 2)^2\)……… | M1 |
| \(-3\big[(x - 2)^2 - (2)^2\big]\)…….. oe or \(-3(x - 2)^2 + 12\)…….. or \(-3\big[(x - 2)^2 - (2)^2 \ldots\ldots\ldots\big]\) oe | M1 |
| Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: \(29 - 3(x - 2)^2\) | A1 |
| (4) | |
| (4 marks) |
Notes
M1: for factorising \(-3x^2 + 12x\)
or
stating the correct value of \(b\) or \(b = 3\) embedded in an incorrect final answer in the form \(a - 3(x - c)^2\)
M1: for a correct first step to complete the square
M1: for a correct second step to complete the square
A1: oe eg \(-3(x - 2)^2 + 29\)
25 ALT
| Scheme | Marks |
|---|---|
| \(-bx^2 + 2bcx - bc^2 + a\) oe or \(b = 3\) | M1 |
| \(2bc = 12\) or \(a - bc^2 = 17\) oe | M1 |
| \(2 \times \text{``}{3}\text{''} \times c = 12\) or \(a - \text{``}{3}\text{''} \times \text{``}{2}\text{''}^2 = 17\) oe | M1 |
| Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: \(29 - 3(x - 2)^2\) | A1 |
| (4 marks) |
Notes
M1: for multiplying out \(a - b(x - c)^2\)
or
stating the correct value of \(b\) or \(b = 3\) embedded in an incorrect final answer in the form \(a - 3(x - c)^2\)
M1: for equating coefficients
M1: for finding at least 2 from \(a\) or \(b\) or \(c\)
A1: oe eg \(-3(x - 2)^2 + 29\)