The diagram shows the graph of a cubic curve $y = \mathrm{f}(x)$. (a) Find an expression for $\mathrm{f}(x)$. (b) Solve $\mathrm{f}(x) \leq 0$.

(a) f(x) = -(1/2)(x + 2)(x + 1)(x - 5) · (b) -2 = 5

Additional Mathematics 4037/11 — May/June 2020

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

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120

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Topics Calculus · Trigonometry · [Legacy] Indices and surds · Straight-line graphs · Factors of polynomials · Equations, inequalities and graphs · +5 more

Q1Factors of polynomialsEquations, inequalities and graphsFree sample

The diagram shows the graph of a cubic curve $y = \mathrm{f}(x)$ .

Fig. 1

(a)

Find an expression for $\mathrm{f}(x)$ .

DifficultyMedium-Easy

Worked solution

Approach

The graph crosses or touches the $x$ -axis at $x = -2$ , $x = -1$ and $x = 5$ , so these are the three roots. Write $\mathrm{f}(x) = k(x+2)(x+1)(x-5)$ and find $k$ from the $y$ -intercept $(0, 5)$ .

Working

\mathrm{f}(x) = k(x+2)(x+1)(x-5)

Using $\mathrm{f}(0) = 5$ :

k(2)(1)(-5) = 5 \implies -10k = 5 \implies k = -\frac{1}{2}

Answer

\mathrm{f}(x) = -\frac{1}{2}(x+2)(x+1)(x-5)

Final answer

f(x) = -(1/2)(x + 2)(x + 1)(x - 5)

Detailed explanation

Walkthrough

The curve meets the $x$ -axis at $x = -2$ and $x = 5$ and touches it at $x = -1$ , so each of $(x+2)$ , $(x+1)$ and $(x-5)$ is a factor. The general cubic with these roots is $k(x+2)(x+1)(x-5)$ for some constant $k$ . Substituting $x = 0$ , where the graph shows $y = 5$ , gives $k(2)(1)(-5) = 5$ , so $k = -\frac{1}{2}$ . The negative sign is consistent with the shape: the curve falls to $-\infty$ as $x \to \infty$ .

Key Takeaways

A cubic's $x$ -intercepts give its factors directly; a touch point is still a root.
The vertical stretch factor is fixed by any one other known point, usually the $y$ -intercept.

Common Mistakes

Writing $(x-2)$ instead of $(x+2)$ for the root $x = -2$ (sign slips in the brackets).
Forgetting the constant $k$ entirely, which would give $y$ -intercept $-10$ rather than $5$ .
Missing the negative sign on $k$ — the mark scheme awards B1 specifically for $-\frac{1}{2}$ .

Things to Be Careful About

Both B marks are accuracy-only (B1 for the bracket product, B1 for $-\frac{1}{2}$ ), so both must be exactly right — no follow-through.
Check your answer by substituting $x = 0$ : it should return $5$ .

Techniques used

read the roots of a cubic from its graphwrite the cubic in factorised formuse the y-intercept to find the constant multiplier

(b)

Solve $\mathrm{f}(x) \leq 0$ .

DifficultyEasy

Worked solution

Approach

$\mathrm{f}(x) \leq 0$ means the parts of the graph on or below the $x$ -axis. Read those intervals straight off the sketch.

Working

From the graph, the curve is below the $x$ -axis between $x = -2$ and $x = -1$ (including the endpoints, since equality is allowed), and again for all $x$ beyond $x = 5$ .

-2 \leq x \leq -1 \quad \text{or} \quad x \geq 5

Answer

-2 \leq x \leq -1 \;\text{ or }\; x \geq 5

Final answer

-2 <= x <= -1 or x >= 5

Detailed explanation

Walkthrough

Solving $\mathrm{f}(x) \leq 0$ graphically means finding every $x$ for which the curve sits on or below the $x$ -axis. Between $x = -2$ and $x = -1$ the curve dips below the axis (it touches at $x = -1$ ); after crossing at $x = 5$ it stays below forever. Because the inequality is $\leq$ (not strict), the boundary points $x = -2$ , $x = -1$ and $x = 5$ are included.

Key Takeaways

An inequality like $\mathrm{f}(x) \leq 0$ is answered by reading intervals from the graph, not by algebra here.
$\leq$ includes the endpoints; $<$ would exclude them.

Common Mistakes

Using strict inequalities and losing the endpoint marks.
Including the region between $-1$ and $5$ , where the curve is above the axis.
Giving only one of the two intervals — each interval carries its own B1.

Things to Be Careful About

Each interval is a separate B1 mark, so both must be stated with correct inclusive signs.

Techniques used

read where a curve lies below the x-axis from its graphexpress the solution set of an inequality using interval notation

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