June 2023 Paper 1 Q6
6 The equation of a curve is \(\quad y = x^6 + 4x^2 - 7\)
Work out the equation of the tangent to the curve at the point \((1, -2)\)
Give your answer in the form \(\quad y = mx + c\) [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(6x^5 + 8x\) | M1 | either term correct ignore LHS |
| \(6 \times 1^5 + 8 \times 1\) or 14 | M1dep | all needs to be correct (no ft) |
| \(-2 = (\text{their } 14) \times 1 + c\) or \(y - -2 = (\text{their } 14)(x - 1)\) | M1 | from \(y = (\text{their } 14)\,x + c\) must be evidence of differentiation to get this mark |
| \(y = 14x - 16\) | A1 |
Additional guidance
| Negative reciprocal of 14 to find the normal rather than the tangent ie. \(-2 = \dfrac{-1}{14} \times 1 + c\) or \(y - -2 = \dfrac{-1}{14}(x - 1)\) | M3 A0 |
| 14 from incorrect differentiation eg \(6x^4 + 8\) in differentiation with correct answer then ft correct eg \(6x^4 + 8x\) in differentiation with correct answer then ft correct | M0 M0 M1 A0 M1 M0 M1 A0 |