Specimen 2020 Paper 1 Q15
15 Point \(A\) lies on the curve
\[y = x^2 + 5x + 8\]The \(x\)-coordinate of \(A\) is \(-4\)
(a) Show that the equation of the normal to the curve at \(A\) is \(\quad 3y = x + 16\) [5 marks]
(b) The normal at \(A\) also intersects the curve at \(B\).
Work out the \(x\)-coordinate of \(B\). [4 marks]
| Answer | Mark | Comments |
|---|---|---|
| \((-4)^2 + 5 \times -4 + 8\) or 4 | M1 | oe |
| \(2x + 5\) | M1 | \(\dfrac{\mathrm{d}y}{\mathrm{d}x}\) |
| \(2 \times -4 + 5\) or \(-3\) | M1dep | gradient of tangent |
| \(-\dfrac{1}{\text{their } -3}\) or \(\dfrac{1}{3}\) | M1dep | dep on 2nd and 3rd M1 |
| \(y - 4 = \dfrac{1}{3}(x + 4)\) and \(3y = x + 16\) | A1 | must see correct working leading to \(3y = x + 16\) |
| Answer | Mark | Comments |
|---|---|---|
| \(x + 16 = 3(x^2 + 5x + 8)\) | M1 | oe |
| \(3x^2 + 14x + 8\ (= 0)\) | A1 | |
| \((3x + 2)(x + 4)\ (= 0)\) or \(\dfrac{-14 \pm \sqrt{14^2 - 4 \times 3 \times 8}}{2 \times 3}\) or \(-\dfrac{7}{3} \pm \sqrt{\dfrac{25}{9}}\) | M1 | oe correct attempt to solve their 3-term quadratic |
| \(-\dfrac{2}{3}\) | A1 |