(a) On the axes below, sketch the graph of $y = (x - 2)(x + 1)(3 - x)$, stating the intercepts on the coordinate axes. (b) Hence write down the values of $x$ such that $(x - 2)(x + 1)(3 - x) > 0$.

Additional Mathematics 4037/13 — October/November 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 · Straight-line graphs · Logarithmic and exponential functions · Factors of polynomials · Equations, inequalities and graphs · +4 more

Q1Equations, inequalities and graphsFree sample

(a)

On the axes below, sketch the graph of $y = (x - 2)(x + 1)(3 - x)$ , stating the intercepts on the coordinate axes.

Fig. 1

DifficultyMedium-Easy

Worked solution

Approach

To sketch the cubic curve $y = (x - 2)(x + 1)(3 - x)$ :

Find the $x$ -intercepts by setting $y = 0$ .
Find the $y$ -intercept by setting $x = 0$ .
Determine the end behaviour by finding the sign of the coefficient of $x^3$ .
Sketch a smooth curve passing through these intercepts with correct turning points and orientation.

Working

Find the $x$ -intercepts by solving $y = 0$ :

(x - 2)(x + 1)(3 - x) = 0

x = -1, \quad x = 2, \quad x = 3

Find the $y$ -intercept by substituting $x = 0$ :

y = (0 - 2)(0 + 1)(3 - 0) = (-2)(1)(3) = -6

Determine the shape and end behaviour:
Expanding the leading term gives $x \times x \times (-x) = -x^3$ .
Since the leading coefficient is negative:

As $x \to -\infty$ , $y \to +\infty$
As $x \to +\infty$ , $y \to -\infty$

The curve comes down from the second quadrant, crosses the $x$ -axis at $(-1, 0)$ , reaches a local minimum below the $x$ -axis passing through the $y$ -intercept $(0, -6)$ , crosses the $x$ -axis at $(2, 0)$ , reaches a local maximum above the $x$ -axis, crosses the $x$ -axis at $(3, 0)$ , and continues downwards into the fourth quadrant.

Answer

A negative cubic curve with $x$ -intercepts at $-1$ , $2$ , and $3$ , and $y$ -intercept at $-6$ .

Final answer

Negative cubic curve crossing the x-axis at -1, 2, 3 and the y-axis at -6

Detailed explanation

Walkthrough

Find the $x$ -intercepts: The curve crosses the $x$ -axis where $y = 0$ . Since the polynomial is already in factorised form, $(x - 2)(x + 1)(3 - x) = 0$ , setting each factor to zero gives $x = 2$ , $x = -1$ , and $x = 3$ .
Find the $y$ -intercept: Setting $x = 0$ gives $y = (0 - 2)(0 + 1)(3 - 0) = -6$ . So the curve crosses the $y$ -axis at $(0, -6)$ .
Determine the orientation (end behaviour): Look at the highest power of $x$ . Multiplying the variable terms in each bracket gives $x \times x \times (-x) = -x^3$ . A cubic with a negative coefficient of $x^3$ starts high in the top-left (as $x \to -\infty$ , $y \to +\infty$ ) and ends low in the bottom-right (as $x \to +\infty$ , $y \to -\infty$ ).
Sketch the curve: Draw a smooth cubic curve that enters from quadrant 2, passes through $(-1, 0)$ , drops to a local minimum while passing through $(0, -6)$ , turns up to cross the $x$ -axis at $(2, 0)$ , reaches a local maximum between $x = 2$ and $x = 3$ , crosses at $(3, 0)$ , and continues downwards.

Key Takeaways

For a polynomial in factored form, the roots directly give the $x$ -intercepts.
The $y$ -intercept is found by evaluating $y$ at $x = 0$ .
The sign of the leading coefficient dictates the global shape and end behaviour of the polynomial.

Common Mistakes

Incorrect orientation: Assuming a positive cubic shape because $(x-2)$ and $(x+1)$ have positive $x$ terms, while missing the $-x$ inside $(3 - x)$ .
Miscalculating the $y$ -intercept: Sign errors such as getting $+6$ instead of $-6$ .
Failing to extend arms: Stopping the graph precisely at the outer intercepts instead of extending the branches beyond $x = -1$ and $x = 3$ .

Things to Be Careful About

Ensure the turning points are placed in the correct quadrants (local minimum in the fourth quadrant below the $y$ -axis/intercept, local maximum in the first quadrant between $x = 2$ and $x = 3$ ).
Clearly label all coordinate intercepts with their values: $-1, 2, 3$ on the $x$ -axis and $-6$ on the $y$ -axis.

Techniques used

find intercepts with coordinate axesdetermine end behaviour of a negative cubic polynomialsketch a cubic curve with turning points and intercepts

(b)

Hence write down the values of $x$ such that $(x - 2)(x + 1)(3 - x) > 0$ .

DifficultyMedium-Easy

Worked solution

Approach

The inequality $(x - 2)(x + 1)(3 - x) > 0$ asks for the values of $x$ where the curve $y = (x - 2)(x + 1)(3 - x)$ lies strictly above the $x$ -axis ( $y > 0$ ). Read these intervals directly from the sketch in part (a).

Working

From the graph drawn in part (a), the curve lies above the $x$ -axis ( $y > 0$ ) in two disjoint regions:

To the left of $x = -1$ :

x < -1

Between $x = 2$ and $x = 3$ :

2 < x < 3

Answer

x < -1, \quad 2 < x < 3

Final answer

x < -1, 2 < x < 3

Detailed explanation

Walkthrough

Identify the geometric meaning of the inequality: The condition $(x - 2)(x + 1)(3 - x) > 0$ represents all $x$ -values where $y > 0$ , meaning the parts of the graph that lie strictly above the horizontal $x$ -axis.
Identify the intervals from the sketch:
- For $x < -1$ , the curve is above the $x$ -axis ( $y > 0$ ).
- For $-1 < x < 2$ , the curve is below the $x$ -axis ( $y < 0$ ).
- For $2 < x < 3$ , the curve forms an arch above the $x$ -axis ( $y > 0$ ).
- For $x > 3$ , the curve falls below the $x$ -axis ( $y < 0$ ).
State the solution set: Combining the regions where $y > 0$ gives $x < -1$ and $2 < x < 3$ .

Key Takeaways

A polynomial inequality $f(x) > 0$ corresponds directly to the intervals on the $x$ -axis where the graph $y = f(x)$ is above the $x$ -axis.
Strict inequalities ( $>$ ) require strict inequality symbols in the solution ( $<$ and $>$ rather than $\leq$ and $\geq$ ).

Common Mistakes

Including the boundary points: Writing $\leq$ or $\geq$ instead of strict inequalities ( $<$ or $>$ ).
Combining disjoint intervals incorrectly: Writing invalid compound inequalities like $2 < x < -1$ .

Things to Be Careful About

Ensure both disjoint intervals are stated clearly as separate statements ( $x < -1$ and $2 < x < 3$ ).
Note that this is a 'Hence' question, meaning the solution should be directly read from the graph in part (a).

Techniques used

read inequality solution sets from a sketched graph

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