Additional Mathematics 4037/12 — October/November 2010

Cambridge O-Level · Calculator · worked solutions for every part, with the mark scheme

questions

marks

120

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Topics Trigonometry · Calculus · Straight-line graphs · Quadratic functions · Factors of polynomials · Series · +3 more

Q1Quadratic functionsFree sample

The equation of a curve is given by $y = 2x^2 + ax + 14$ , where $a$ is a constant.

Given that this equation can also be written as $y = 2(x - 3)^2 + b$ , where $b$ is a constant, find

(i)

the value of $a$ and of $b$ ,

DifficultyMedium-Easy

Worked solution

Approach

Expand $y = 2(x - 3)^2 + b$ and compare coefficients with $y = 2x^2 + ax + 14$ .

Working

2(x - 3)^2 + b = 2\left(x^2 - 6x + 9\right) + b = 2x^2 - 12x + (18 + b)

Comparing with $y = 2x^2 + ax + 14$ :

a = -12

and equating the constant terms:

18 + b = 14 \quad\Rightarrow\quad b = -4

Answer

a = -12, \qquad b = -4

Final answer

a = -12, b = -4

Detailed explanation

Walkthrough

The two forms describe the same curve, so expanding the completed-square form must reproduce the original expression. Expanding $(x-3)^2 = x^2 - 6x + 9$ and multiplying by 2 gives $2x^2 - 12x + 18$ ; adding $b$ gives $2x^2 - 12x + (18 + b)$ . Since two polynomials are equal only when their matching coefficients are equal, the coefficient of $x$ gives $a = -12$ , and the constant term gives $18 + b = 14$ , so $b = -4$ .

Key Takeaways

A quadratic written as $a(x - h)^2 + k$ can always be expanded back to standard form by multiplying out.
Equating coefficients is the reliable way to find unknown constants when two forms of the same polynomial are given.

Common Mistakes

Sign slips when expanding $(x - 3)^2$ : forgetting that $(-3)^2 = +9$ or writing $-6x$ incorrectly.
Forgetting to multiply the whole bracket by 2 before comparing constants.
Comparing only one coefficient and guessing the other.

Things to Be Careful About

Both values are required for the two B1 marks; each value earns its own mark.
Check the constant equation carefully: it is $18 + b = 14$ , not $b = 14$ .

Techniques used

expand the completed-square formequate coefficients of x and of the constant term

(ii)

the minimum value of $y$ .

DifficultyEasy

Worked solution

Approach

Use the completed-square form $y = 2(x - 3)^2 + b$ from part (i): since $2(x-3)^2 \geq 0$ , the least value of $y$ occurs at $x = 3$ .

Working

y_{\min} = 2(3 - 3)^2 + (-4) = -4

Answer

y_{\min} = -4

Final answer

-4

Detailed explanation

Walkthrough

In the form $y = 2(x - 3)^2 - 4$ , the squared term $2(x-3)^2$ can never be negative, and it equals zero exactly when $x = 3$ . So the smallest possible value of $y$ is just the constant term, $-4$ . This is why completing the square is useful: the minimum is read off directly without differentiating.

Key Takeaways

For a positive quadratic $a(x-h)^2 + k$ with $a > 0$ , the minimum value is $k$ , occurring at $x = h$ .
The completed-square form makes turning points visible immediately.

Common Mistakes

Giving $x = 3$ instead of the minimum value of $y$ ; the question asks for the value of $y$ .
Sign errors carrying $b$ through from part (i).

Things to Be Careful About

The mark scheme allows follow-through on the candidate's own value of $b$ , but the correct answer here is $-4$ .
State the minimum value of $y$ , not the $x$ -coordinate where it occurs.

Techniques used

read the minimum from the completed-square form

The rest of this paper

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