Higher January 2023 Paper 1R Q23
23 \(G\) is the point on the curve with equation \(\;y = 8x^2 - 14x - 6\;\) where the gradient is 10
The straight line Q passes through the point \(G\) and is perpendicular to the tangent at \(G\)
Find an equation for Q
Give your answer in the form \(\;ax + by + c = 0\;\) where \(a\), \(b\) and \(c\) are integers.
(5)
| Scheme | Marks |
|---|---|
| \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x} =\right) 16x - 14\) | M1 |
| \(16x - 14 = 10\) | M1 |
| (1.5, −9) or \(x = 1.5\), \(y = -9\) | A1 |
eg \(y - {-9} = -\dfrac{1}{10}\left(x - \dfrac{3}{2}\right)\) oe or eg \(-9 = -\dfrac{1}{10} \times 1.5 + c\) oe | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: \(2x + 20y + 177 = 0\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: Differentiation to obtain 2 terms with at least 1 correct
M1: their \(\mathrm{d}y/\mathrm{d}x = 10\) dep on M1
A1: coordinates of point on curve at which gradient is 10 – allow given as coordinates or as \(x\) worked out and \(y\) worked out if meaning is clear
M1: A correct method to find the equation for line Q using (1.5, −9)
A1: oe where \(a\), \(b\), \(c\) are integers eg \(10x + 100y + 885 = 0\)