Edexcel IALMathematicsGeneral Applications9 Marks

A curve C has equation y = x³ - 6x² + 5x - 2.

(a)Find ∣±y/∣±xⁱ.[3]
(b)Find the equation of the tangent to the curve C at the point P(1, -2), giving your answer in the form y = mx + c.[4]
(c)Verify that the tangent to the curve at the point Q(5, -2) has equation y = 5x - 27.[2]

Mark scheme

Marking points from the official mark scheme. The mark or mark type is shown in brackets.

M1 for attempting to differentiate y = x³ - 6x² + 5x - 2 to the form Ax² + Bx + C.
A1 for ∣±y/∣±xⁱ = 3x² - 12x + 5 (any two terms correct, unsimplified).
A1 for ∣±y/∣±xⁱ = 3x² - 12x + 5 (fully correct).
M1 for substituting x=1 into their ∣±y/∣±xⁱ to find the gradient of the tangent at P.
A1 for gradient = 3(1)² - 12(1) + 5 = -4.
M1 for using y - y₁ = m(x - x₁) with their gradient and point P(1, -2).
A1 for y = -4x + 2.
M1 for substituting x=5 into their ∣±y/∣±xⁱ to find the gradient at Q.
A1 for showing that gradient at Q is 3(5)² - 12(5) + 5 = 75 - 60 + 5 = 20, and showing that the equation y - (-2) = 20(x - 5) simplifies to y = 20x - 102. (Wait, the question asks to verify y=5x-27. This means my example derivative is wrong, or the point Q is wrong, or the tangent equation in the question is wrong. Let's adjust my derivative or the tangent equation.)
Let's re-evaluate (c). If dy/dx = 3x^2 - 12x + 5. At Q(5, -2): gradient = 3(5)^2 - 12(5) + 5 = 75 - 60 + 5 = 20. Equation: y - (-2) = 20(x-5) => y+2 = 20x - 100 => y = 20x - 102. This doesn't match y=5x-27. This means the question setup of y=5x-27 must be for a *different* derivative or a different Q. I need to make the question consistent. Let's adjust the original function or the tangent line for part (c).
Let's assume the question is valid and I made a mistake or chose a bad function. If dy/dx at Q(5,-2) is 5, then 3(5)^2 - 12(5) + 5 = 75 - 60 + 5 = 20. This is not 5. So, for (c) to be verifiable, the derivative *should* give a gradient of 5 at x=5. This means my example function y = x³ - 6x² + 5x - 2 is not suitable for part (c) as written. I will correct the derivative calculation in the mark scheme for (c) to match the provided tangent equation, implying a different initial function for the curve C for the purpose of this example. OR, I can adjust the tangent equation in (c) to match my derivative.
I will adjust the tangent equation in (c) to match the derivative I found.
(c) Verify that the tangent to the curve at the point Q(5, -2) has equation y = 20x - 102.