June 2022 Paper 1 Q5
5 Find an equation of the tangent to the curve
\[y = (x - 2)^4\]at the point where \(x = 0\) [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Differentiates to obtain a correct derivative either \(4(x - 2)^3\) OE or \(4x^3 - 24x^2 + 48x - 32\) PI by \(-32\) obtained with no errors seen in evaluating \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) | B1 | 1.1b |
| Substitutes \(x\) = 0 into their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) to obtain a numerical value or PI by constant from their \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) or PI by \(-32\) obtained with no errors seen in evaluating \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) | M1 | 1.1a |
| Obtains \(y = -32x + 16\) ACF Award the mark at the first opportunity and ISW any incorrect rearrangement No errors seen | A1 | 1.1b |
| (3 marks) |
Typical solution
\[\frac{\mathrm{d}y}{\mathrm{d}x} = 4(x - 2)^3\]When \(x\) = 0
\[\frac{\mathrm{d}y}{\mathrm{d}x} = -32\]\[y = 16\]\[y = -32x + 16\]