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O-Level E-Math · 2018 · P1 Q23 Functions & graphs (quadratics, tangents) · Line of symmetry from the roots 6 marks: 1 + 2 + 1 + 2 · number & algebra (quadratic graph) difficulty 4 of 5

O-Level E-Math 2018 Paper 1, Question 23: Line of symmetry from the roots

The answer

(a) \(x = \dfrac{1}{2}\)
(b) \(x \approx 2.56\) or \(x \approx -1.56\)
(c)(ii) \(y = -\dfrac{3}{2}x + 2\)

O-Level E-Math 2018 Paper 1 Question 23 · Verified worked solution by the Genius Plus Academy teaching team

What this question tests

This is Question 23 of the O-Level E-Math 2018 Paper 1. It tests line of symmetry from the roots, in the Functions & graphs (quadratics, tangents) area. It is worth 6 marks: 1 + 2 + 1 + 2. It is a worded / diagram-based question, so open your Ten-Year Series (TYS) or the official paper at this question, then follow our full worked solution below.

Step-by-step solution

(a) The curve \(y = \dfrac12(x - x^2) = -\dfrac12 x^2 + \dfrac12 x\) is a parabola; its line of symmetry runs through the vertex. The roots of \(x - x^2 = 0\) are \(x = 0\) and \(x = 1\), so the axis of symmetry is midway between them: \(x = \dfrac{1}{2}\).

(b) \(x - x^2 = -4\) is the same as \(\dfrac12(x - x^2) = -2\), i.e. \(y = -2\) on this graph. Read the \(x\)-coordinates where the curve meets the line \(y = -2\). Algebraically \(x^2 - x - 4 = 0\) gives \(x = \dfrac{1 \pm \sqrt{17}}{2}\), so \(x \approx 2.56\) or \(x \approx -1.56\). *(Read from the graph; accept roughly \(x \approx 2.6\) and \(x \approx -1.6\).)*

(c)(i) Draw the straight line through \(P(-2, 5)\) that just touches the curve (it touches near \((2, -1)\)).

(ii) Read two points off the tangent you drew in (i). The tangent from \(P(-2, 5)\) touches the curve at about \((2, -1)\), so use the two points \(P(-2, 5)\) and the point of contact \((2, -1)\). Gradient \(= \dfrac{-1 - 5}{2 - (-2)} = \dfrac{-6}{4} = -\dfrac32\). The line through \(P(-2, 5)\) with gradient \(-\dfrac32\) is \(y - 5 = -\dfrac32(x + 2)\), so \(y = -\dfrac{3}{2}x + 2\). (The point of contact is read from the graph; any carefully drawn tangent from \(P\) should give a gradient close to \(-1.5\) and an equation close to \(y = -1.5x + 2\).)

Answer: (a) \(x = \dfrac{1}{2}\)
(b) \(x \approx 2.56\) or \(x \approx -1.56\)
(c)(ii) \(y = -\dfrac{3}{2}x + 2\)

Same structure, different numbers

A question is hard because of its structure, not its surface.

Swap the constants, dress a quadratic as a length, hide a derivative inside an integral, and a student sees a brand new problem. The structure underneath is the same, and so is the method. Once a student can name the structure, a whole row of questions that look different start to open the same way.

That is where marks really leak: in choosing the method, not in the algebra that follows. We call it Lock and Key, name the lock, then the key follows.

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Our O-Level E-Math tuition trains the same recognise-the-structure method these worked solutions show, taught by a team that has marked these papers for years. It runs within our weekly Secondary Math programme, Sec 1 to 4 and IP.

Questions students ask

What does O-Level E-Math 2018 Paper 1 Question 23 test?

It is a line of symmetry from the roots question from Functions & graphs (quadratics, tangents), worth 6 marks: 1 + 2 + 1 + 2.

Is this the same as IP Math?

Yes. IP (Integrated Programme) schools teach the same O-Level Mathematics content; they just sequence it differently and set their own internal exams, so these worked solutions apply to IP students too.

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