Find the \(x\)-coordinates of all the points on the curve \(y = 2{x^4} + 6{x^3} + \frac{7}{2}{x^2} - 5x + \frac{3}{2}\) at which

the tangent to the curve is parallel to the tangent at \(( - 1,{\text{ }}6)\).

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = 8{x^3} + 18{x^2} + 7x - 5\)    A1

when \(x =  - 1,{\text{ }}\frac{{{\text{d}}y}}{{{\text{d}}x}} =  - 2\)     A1

\(8{x^3} + 18{x^2} + 7x - 5 =  - 2\)    M1

\(8{x^3} + 18{x^2} + 7x - 3 = 0\)

\((x + 1)\) is a factor     A1

\(8{x^3} + 18{x^2} + 7x - 3 = (x + 1)(8{x^2} + 10x - 3)\)    (M1)

Note:     M1 is for attempting to find the quadratic factor.

\((x + 1)(4x - 1)(2x + 3) = 0\)

\((x =  - 1),{\text{ }}x = 0.25,{\text{ }}x =  - 1.5\)    (M1)A1

Note:     M1 is for an attempt to solve their quadratic factor.

[7 marks]

The first half of the question was accessible to all the candidates. Some though saw the word ‘tangent’ and lost time calculating the equation of this. It was a pity that so many failed to spot that \(x + 1\) was a factor of the cubic and so did not make much progress with the final part of this question.