Mathematics 9709/13 — October/November 2020

Cambridge AS Level · Pure Mathematics 1 · worked solutions for every part, with the mark scheme

questions

marks

110

minutes

Topics Quadratics · Trigonometry · Integration · Functions · Series · Differentiation · +2 more

Q1QuadraticsFunctionsFree sample

(a)

Express $x^2 + 6x + 5$ in the form $(x + a)^2 + b$ , where $a$ and $b$ are constants.

DifficultyEasy

Worked solution

Approach

Complete the square by halving the coefficient of $x$ , adding and subtracting its square, and combining the constants.

Working

Halve the coefficient of $x$ : half of $6$ is $3$ , so the squared bracket is $(x + 3)^2$ .

(x + 3)^2 = x^2 + 6x + 9

Therefore:

\begin{aligned} x^2 + 6x + 5 &= (x + 3)^2 - 9 + 5 \\ &= (x + 3)^2 - 4 \end{aligned}

So $a = 3$ and $b = -4$ .

Answer

x^2 + 6x + 5 = (x + 3)^2 - 4

with $a = 3$ and $b = -4$ .

Final answer

a = 3, b = -4

Detailed explanation

Walkthrough

This is a straightforward completing-the-square question. The expression is $x^2 + 6x + 5$ . Start with the coefficient of $x$ , which is $6$ . Half of $6$ is $3$ , so the bracket will be $(x + 3)^2$ . Expanding $(x + 3)^2$ gives $x^2 + 6x + 9$ , which is $4$ more than $x^2 + 6x + 5$ . Therefore subtract $4$ to compensate, giving $(x + 3)^2 - 4$ . Hence $a = 3$ and $b = -4$ .

Key Takeaways

Completing the square rewrites a quadratic in a form that directly shows its vertex or turning point. The coefficient of $x$ determines the number inside the bracket, and the constant term has to be adjusted by subtracting the square of half that coefficient.

Common Mistakes

A common mistake is forgetting to subtract the square of half the $x$ -coefficient, giving $x^2 + 6x + 5 = (x + 3)^2 + 5$ instead of $(x + 3)^2 - 4$ . Another common mistake is getting the sign of $b$ wrong when writing the final answer.

Things to Be Careful About

In the form $(x + a)^2 + b$ , $a$ is the number inside the bracket with the correct sign. Here $x^2 + 6x$ becomes $(x + 3)^2$ , so $a = 3$ , not $-3$ . The value of $b$ is the constant left after adjusting, here $-4$ . These values will be used in part (b), so care with signs is important.

Techniques used

complete the square for a quadratic expressionidentify the constants a and b in completed-square form

(b)

The curve with equation $y = x^2$ is transformed to the curve with equation $y = x^2 + 6x + 5$ .

Describe fully the transformation(s) involved.

DifficultyMedium-Easy

Worked solution

Approach

Rewrite the transformed equation in completed-square form so that the translations from $y = x^2$ can be read directly. The transformation is a translation.

Working

From part (a):

y = x^2 + 6x + 5 = (x + 3)^2 - 4

Compare this with $y = x^2$ .

Replacing $x$ with $x + 3$ moves the graph $3$ units to the left.

Subtracting $4$ moves the graph $4$ units down.

So the transformation is:

\begin{pmatrix} -3 \\ -4 \end{pmatrix}

Answer

A translation (or shift) of $\begin{pmatrix} -3 \\ -4 \end{pmatrix}$ , i.e. $3$ units in the negative $x$ -direction (left) and $4$ units in the negative $y$ -direction (down).

Final answer

Translation by vector (-3, -4): 3 units left and 4 units down.

Detailed explanation

Walkthrough

The curve $y = x^2$ is being changed into $y = x^2 + 6x + 5$ . To see the transformation clearly, write the new equation in completed-square form. From part (a), $x^2 + 6x + 5 = (x + 3)^2 - 4$ . Compare this with $y = x^2$ . Replacing $x$ by $x + 3$ means the graph moves left by $3$ units, and subtracting $4$ means the graph moves down by $4$ units. Combined, this is a translation by the column vector $\begin{pmatrix} -3 \\ -4 \end{pmatrix}$ . The word "translation" is important because it tells us that every point moves by the same vector and the shape is unchanged.

Key Takeaways

A quadratic can be written as $(x + a)^2 + b$ to reveal its graph as a translation of $y = x^2$ . In general, $y = (x + a)^2 + b$ is the graph of $y = x^2$ translated $\begin{pmatrix} -a \\ b \end{pmatrix}$ . Replacing $x$ with $x + a$ shifts the graph horizontally in the negative direction; adding or subtracting a constant shifts it vertically.

Common Mistakes

A common mistake is describing the horizontal translation as $3$ units to the right instead of $3$ units to the left. Another mistake is to state only one of the two translations, or to call the transformation a stretch or reflection. The mark scheme requires both the translation and its direction/components.

Things to Be Careful About

Use the word "translation" or "shift" explicitly. State both components: $3$ units left and $4$ units down. If giving a vector, ensure the signs are correct: $\begin{pmatrix} -3 \\ -4 \end{pmatrix}$ . The mark scheme also allows the form "translation by $-3$ units in the $x$ -direction and $-4$ units in the $y$ -direction".

Techniques used

rewrite a quadratic in completed-square form to identify transformationsdescribe translations of a quadratic graph in vector form

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