Higher November 2024 Paper 3 Q24

24 A curve has the equation \(\quad y = x^2 + 4x - 4\)

A straight line has the equation \(\quad y = 3x - 2\)

Work out the two points of intersection of the curve and the straight line. [5 marks]

Mark scheme

Answer	Mark	Comments
Alternative method 1
\(x^2 + 4x - 4 = 3x - 2\)	M1
\(x^2 + x - 2\ (= 0)\)	A1
\((x + 2)(x - 1)\ (= 0)\) or \((x =)\dfrac{-1 \pm \sqrt{1^2 - 4 \times 1 \times -2}}{2 \times 1}\) or \((x + 0.5)^2 - 2.25\ (= 0)\)	M1	oe correct for their 3-term quadratic in the form \(ax^2 + bx + c\ (= 0)\) if the quadratic formula is used with \(+\) and \(-\) separately, both must be seen correctly for this mark
\((x =)\) \(-2\) and \((x =)\) 1	A1
\((-2, -8)\) and \((1, 1)\)	A1	SC1 for 1 correct coordinate
Alternative method 2
\(y = \left(\dfrac{y + 2}{3}\right)^2 + 4\left(\dfrac{y + 2}{3}\right) - 4\)	M1
\(y^2 + 7y - 8\ (= 0)\)	A1
\((y + 8)(y - 1)\ (= 0)\) or \((y =)\dfrac{-7 \pm \sqrt{7^2 - 4 \times 1 \times -8}}{2 \times 1}\) or \((y + 3.5)^2 - 20.25\ (= 0)\)	M1	oe correct for their 3-term quadratic in the form \(ay^2 + by + c\ (= 0)\) if the quadratic formula is used with \(+\) and \(-\) separately, both must be seen correctly for this mark
\((y =)\) \(-8\) and \((y =)\) 1	A1
\((-2, -8)\) and \((1, 1)\)	A1	SC1 for 1 correct coordinate

Trial and improvement may be awarded SC1 for one correct coordinate or full marks if both coordinates are correct