C1 January 2010 Q9

EdexcelOld spec13 marksCo-ordinate Geometry Sketching & Transforming Graphs

(a) Factorise completely \(x^3 - 4x\) (3)

(b) Sketch the curve \(C\) with equation\[y = x^3 - 4x,\]showing the coordinates of the points at which the curve meets the \(x\)-axis. (3)

The point \(A\) with \(x\)-coordinate \(-1\) and the point \(B\) with \(x\)-coordinate 3 lie on the curve \(C\).

(c) Find an equation of the line which passes through \(A\) and \(B\), giving your answer in the form \(y = mx + c\), where \(m\) and \(c\) are constants. (5)

(d) Show that the length of \(AB\) is \(k\sqrt{10}\), where \(k\) is a constant to be found. (2)

Mark scheme (a)

Scheme	Marks
\(x(x^2 - 4)\) Factor \(x\) seen in a correct factorised form of the expression.	B1
\(= x(x - 2)(x + 2)\) M: Attempt to factorise quadratic (general principles). Accept \((x - 0)\) or \((x + 0)\) instead of \(x\) at any stage. Factorisation must be seen in part (a) to score marks.	M1 A1
	(3)

Notes

\(x^3 - 4x \to x(x^2 - 4) \to (x - 2)(x + 2)\) scores B1 M1 A0.
\(x^3 - 4x \to x^2 - 4 \to (x - 2)(x + 2)\) scores B0 M1 A0 (dividing by \(x\)).
\(x^3 - 4x \to x(x^2 - 4x) \to x^2(x - 4)\) scores B0 M1 A0.
\(x^3 - 4x \to x(x^2 - 4) \to x(x - 2)^2\) scores B1 M1 A0

Special cases: \(x^3 - 4x \to (x - 2)(x^2 + 2x)\) scores B0 M1 A0.
\(x^3 - 4x \to x(x - 2)^2\) (with no intermediate step seen) scores B0 M1 A0

Sketch: positive cubic through (-2, 0), the origin and (2, 0) — Mark scheme (b)

Scheme	Marks
Shape (2 turning points required)	B1
Through (or touching) origin	B1
Crossing \(x\)-axis or “stopping at \(x\)-axis” (not a turning point) at \((-2, 0)\) and \((2, 0)\).	B1
Allow \(-2\) and 2 on \(x\)-axis. Also allow \((0, -2)\) and \((0, 2)\) if marked on \(x\)-axis. Ignore extra intersections with \(x\)-axis.
	(3)

Mark scheme (c)

Scheme	Marks
Either \(\ y = 3\) (at \(x = -1\)) or \(\ y = 15\) (at \(x = 3\)) Allow if seen elsewhere.	B1
Gradient \(= \dfrac{\text{"}15 - 3\text{"}}{3 - (-1)}\ (= 3)\) Attempt correct grad. formula with their \(y\) values. For gradient M mark, if correct formula not seen, allow one slip, e.g. \(\dfrac{\text{"}15 - 3\text{"}}{3 - 1}\)	M1
\(y - \text{"}15\text{"} = m(x - 3)\) or \(y - \text{"}3\text{"} = m(x - (-1))\), with any value for \(m\).	M1
\(y - 15 = 3(x - 3)\) or the correct equation in any form, e.g. \(y - 3 = \dfrac{15 - 3}{3 - (-1)}(x - (-1)),\ \ \dfrac{y - 3}{x + 1} = \dfrac{15 - 3}{3 + 1}\)	A1
\(y = 3x + 6\)	A1
	(5)

Notes

1st M: May be implicit in the equation of the line, e.g. \(\dfrac{y - \text{"}15\text{"}}{3 - \text{"}15\text{"}} = \dfrac{x - \text{"}3\text{"}}{-1 - \text{"}3\text{"}}\)

2nd M: An equation of a line through \((3, \text{"}15\text{"})\) or \((-1, \text{"}3\text{"})\) in any form, with any gradient (except 0 or \(\infty\)).

2nd M: Alternative is to use one of the points in \(y = mx + c\) to find a value for \(c\), in which case \(y = 3x + c\) leading to \(c = 6\) is sufficient for both A marks.

1st A1: Correct equation in any form.

Mark scheme (d)

Scheme	Marks
\(AB = \sqrt{\left(\text{"}15 - 3\text{"}\right)^2 + \left(3 - (-1)\right)^2}\) (With their non-zero \(y\) values)... Square root is required.	M1
\(= \sqrt{160}\ \left(= \sqrt{16}\sqrt{10}\right) = 4\sqrt{10}\) (Ignore \(\pm\) if seen) (\(\sqrt{16}\sqrt{10}\) need not be seen).	A1
	(2)
	(13 marks)