NESA HSC-style-style practice: Draw up a table of signs for y = (x - 1)(x - 3)(x - 5), and hence solve (x - 1)(x - 3)(x - 5) 0.

The factored form already gives the zeroes: x = 1, x = 3 and x = 5. A polynomial has no discontinuities, so these are the only places the sign can change. They cut the number line into four intervals; pick one test value in each. $y(0) = (-1)(-3)(-5) = -15, y(2) = (1)(-1)(-3) = 3, y(4) = (3)(1)(-1) = -3, y(6) = (5)(3)(1) = 15.$ So the sign row is -, +, -, + across x 5. We want y 0, so keep the intervals where y is negative and, because the inequality is weak, include the zeroes where y = 0: $x 1 or 3 x 5.$ Markers reward computing a test value in every interval, the correct sign row, and including the zeroes for the weak inequality.

NESA HSC-style-style practice: Find the zero and the discontinuity of y = x - 1x - 4, and state where the function is positive and where it is negative.

The function is zero where the numerator is zero, at x = 1, and is undefined (discontinuous) where the denominator is zero, at x = 4. These are the only two places the sign can change, so they split the line into three intervals. Test one value in each: $y(0) = (-1)/(-4) = (1)/(4) > 0, y(2) = (1)/(-2) = -(1)/(2) 0.$ So the sign row is +, -, + across x 4. Hence y is positive for x 4, and negative for 1 < x < 4. The sign genuinely changes at both the zero x = 1 and the discontinuity x = 4. Markers reward identifying both the zero and the discontinuity, testing each interval, and stating the sign on each piece.

NSWMaths Extension 1Syllabus dot point

Where can a function change sign, and how do we use that to solve inequations and to begin a sketch?

Examine the sign of a function by building a table of test values that dodge around its zeroes and discontinuities, and use the resulting sign pattern to solve inequations and to start a curve sketch

The Year 11 Extension 1 dot point on the sign of a function. Why a function can only change sign at a zero or a discontinuity, how to build a table of signs with test values, the repeated-factor exception, rational functions with vertical asymptotes, and using the sign pattern both to solve inequations and to begin a sketch.

Generated by Claude Opus 4.817 min answerUpdated 2026-06-21

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

NESA wants you to work out the sign of a function, that is, the set of $x$ for which it is positive, negative or zero, by building a short table of test values, and then to use that sign pattern for two purposes: to solve inequations and to begin a sketch of the curve. It is the engine room behind a surprising amount of the course.

The whole technique rests on one idea, and it is worth stating before anything else: a function can only change sign at a zero or at a discontinuity. Everywhere else its sign is locked. So if you know the zeroes and the discontinuities, the sign of the function is settled on each interval between them, and a single test value reveals it.

A quick note on language, the same one made on the companion inequations page: we solve an inequation such as $x^2 - x > 0$ , and we prove an inequality such as $(x - 2)^2 \ge 0$ . NESA uses the word "inequality" for both, so do not be thrown if a question says "solve the inequality".

The answer

Where can a function change sign?

Picture tracing a continuous curve from left to right. To get from below the $x$ -axis to above it, the pen must cross the axis, that is, pass through a zero. It cannot jump from negative to positive without either touching zero or lifting off the page. "Lifting off the page" is a discontinuity: a break in the graph, which for the functions in this course happens only where a denominator is zero. This is the intuition behind the Intermediate Value Theorem, and for Year 11 the picture is all the justification you need.

So the candidate places for a sign change are exactly the zeroes and the discontinuities. The word "candidate" matters: the sign changes at some of them but not necessarily all. The figure below shows a curve with a simple zero (sign flips) and a double zero (sign holds).

Building a table of signs

The method turns that idea into a routine you can lay out cleanly for a marker.

Factor the expression completely so the zeroes and any denominator are visible.
List the critical values: every zero (of the whole expression) and every discontinuity (every zero of a denominator). Mark each discontinuity as already excluded.
Order them on a number line; they carve it into intervals.
Test one value in each interval, dodging around the critical values, and record the sign.
To solve an inequation, keep the intervals with the sign you want; to start a sketch, use the signs to know which side of the axis the curve sits on in each interval.

A small but real efficiency: you can find the sign of the whole expression either by substituting the test value into the function directly, or by reading the sign of each factor and multiplying. Both are examined; the factor method scales better when there are several factors, and is what the stage-by-stage figures below use.

Polynomials: a worked sign table

Polynomials are the easy case because they have no discontinuities at all, so the test values only have to dodge around the zeroes. Take $y = (x - 1)(x - 3)(x - 5)$ , with zeroes at $x = 1, 3, 5$ . Testing $x = 0, 2, 4, 6$ gives

y(0) = -15, \quad y(2) = 3, \quad y(4) = -3, \quad y(6) = 15,

so the sign row is $-, +, -, +$ . The graph below is the same information drawn as a curve: below the axis, above, below, above. The $y$ -intercept is $y(0) = -15$ .

To solve $(x - 1)(x - 3)(x - 5) \le 0$ , keep the intervals where the sign is negative and, because the inequality is weak, include the zeroes: $x \le 1$ or $3 \le x \le 5$ . Notice that the sketch is incomplete, we have not found the turning points, but it is complete enough to confirm the inequality. That dual use, enough for the inequation, a head start on the sketch, is the point of the method.

The repeated-factor exception

There is one trap worth isolating. A factor raised to an even power, like $(x - 2)^2$ , is never negative, so it does not change the sign of the product as $x$ passes its zero. A factor raised to an odd power behaves, for sign purposes, like a single factor and does change the sign. The shortcut: count each multiplicity modulo $2$ , so even powers can be ignored when you track sign changes.

Concretely, for $y = (x + 1)(x - 2)^2$ the sign changes at the simple zero $x = -1$ but not at the double zero $x = 2$ , which is exactly the curve drawn in the first figure: it crosses at $x = -1$ and merely touches at $x = 2$ . This connects sign analysis to the graphing of polynomials by multiplicity: a simple zero is a crossing, an even zero is a touch.

Functions with discontinuities

When the function has a denominator, the test values must dodge around the discontinuities (zeroes of the denominator) as well as the zeroes. The method is otherwise identical. Take $y = \dfrac{x - 1}{x - 4}$ : a zero at $x = 1$ (numerator zero) and a discontinuity at $x = 4$ (denominator zero). Testing $x = 0, 2, 5$ gives $\tfrac{1}{4}, -\tfrac{1}{2}, 4$ , so the sign row is $+, -, +$ . The sign genuinely flips at both the zero and the discontinuity.

The dashed lines show the vertical asymptote $x = 4$ (the function shoots off to $\pm\infty$ at the discontinuity) and a horizontal asymptote $y = 1$ . Finding the vertical asymptote at a denominator zero is in your Year 11 toolkit; the horizontal asymptote (here $y = 1$ , found by writing $\tfrac{x - 1}{x - 4} = 1 + \tfrac{3}{x - 4}$ ) is a Year 12 refinement and is shown only to make the sketch honest. For the sign, all you need is the table.

Not every candidate point produces a change. The function $y = \dfrac{1}{1 + x^2}$ has denominator $1 + x^2 \ge 1$ , so it has no zeroes and no discontinuities at all; a single test value $y(0) = 1 > 0$ settles that it is positive everywhere. Always check whether the table is even needed.

Two routes for a rational inequation

For an inequation with the unknown in the denominator you now have two complete methods, and the HSC examines both. The companion page develops the first; this page develops the second.

Multiply by the square of the denominator. Since a square is never negative, the inequality sign is preserved, and you reduce the problem to a polynomial inequation. Fast for a single fraction against a constant.
Collect on one side as a single fraction, then sign-table. No multiplying by anything of unknown sign; you analyse the sign of one quotient directly. This is the more general method and the one that scales to the harder rational inequations met in the Year 12 polynomial and rational inequalities page.

The worked centrepiece below solves the same inequation, $\dfrac{3}{x + 2} \le x$ , by the sign-table route; the final worked example solves it again by multiplying by the square, so you can compare them side by side and pick the right tool under exam pressure.

Solve a rational inequation by sign table, stage by stage

Take $\dfrac{3}{x + 2} \le x$ . Collect everything on the left over the common denominator $x + 2$ :

\frac{3}{x + 2} - x \le 0 \implies \frac{3 - x(x + 2)}{x + 2} \le 0 \implies \frac{3 - 2x - x^2}{x + 2} \le 0.

The numerator factors as $3 - 2x - x^2 = -(x^2 + 2x - 3) = -(x + 3)(x - 1)$ , so the inequation is $\dfrac{-(x + 3)(x - 1)}{x + 2} \le 0$ , equivalently $\dfrac{(x + 3)(x - 1)}{x + 2} \ge 0$ after multiplying both sides by $-1$ (which flips the sign). Now sign-table $\dfrac{(x + 3)(x - 1)}{x + 2}$ .

Stage 1, mark the zeroes and the discontinuity. The numerator gives zeroes at $x = -3$ and $x = 1$ ; the denominator gives a discontinuity at $x = -2$ , which is excluded from the outset. These three values cut the line into four intervals.

Stage 2, find the sign of each factor in every interval. Test $x = -4, -\tfrac{5}{2}, 0, 2$ , one inside each interval, and record the sign of each of $(x + 3)$ , $(x - 1)$ and $(x + 2)$ . Each linear factor is negative to the left of its own zero and positive to the right, so every row is a clean run of minuses then pluses.

Stage 3, combine the signs into the quotient row. In each interval, multiply the two numerator signs and divide by the denominator sign: $\tfrac{(-)(-)}{(-)} = -$ , then $\tfrac{(+)(-)}{(-)} = +$ , then $\tfrac{(+)(-)}{(+)} = -$ , then $\tfrac{(+)(+)}{(+)} = +$ . So the quotient row is $-, +, -, +$ . A quotient sign behaves just like a product sign, since dividing by a negative flips the result exactly as multiplying by one would.

Stage 4, read off the solution and fix the endpoints. We need the quotient $\ge 0$ , so keep the intervals where it is positive, $(-3, -2)$ and $(1, \infty)$ . Fix the endpoints: include the numerator zeroes $x = -3$ and $x = 1$ (there the quotient equals $0$ , allowed by $\ge$ ), and exclude the discontinuity $x = -2$ . The solution is $-3 \le x < -2$ or $x \ge 1$ .

How exam questions ask about the sign of a function

The wording tells you what the table is for:

"Draw up a table of signs for $y = \ldots$ " or "examine the sign of $\ldots$ ". Factor, list zeroes and discontinuities, test one value per interval, and state the sign on each piece. This is the direct task.
"Hence solve $\ldots \le 0$ (or $\ge 0$ , $< 0$ , $> 0$ )". Read the inequation straight off the sign row you just built: keep the intervals with the matching sign and fix the endpoints by the rules below.
"Find the zeroes and discontinuities, and hence sketch the curve". Use the sign table to know which side of the axis the curve is on in each interval, then add the intercepts and asymptotes. The sketch will be incomplete without turning points, but the sign pattern is the skeleton.
"For what values of $x$ is $f(x) > g(x)$ ?" Rewrite as $f(x) - g(x) > 0$ and sign-table the difference. "Above" means the difference is positive.
"Solve $\dfrac{ax + b}{cx + d} \ge m$ (a fraction against a constant)". Subtract $m$ first, combine into one fraction, then sign-table; or multiply by the square of the denominator. Never multiply by the bare denominator.

Exam technique

A clean table of signs is the presentation markers most want to see, and it earns method marks even if you slip on a single endpoint.

Factor fully and list every critical value first, zeroes and discontinuities together. A missed discontinuity silently merges two intervals and corrupts the sign row.
Show the table or sign diagram, not just the final intervals. Write the test value you used in each interval; that is the evidence the marker rewards.
Mark each discontinuity as excluded the moment you write it down. A value that makes a denominator zero is never a solution, even for $\ge$ or $\le$ .
Decide each endpoint deliberately. Include a numerator zero for a weak inequality ( $\le$ or $\ge$ ); exclude it for a strict one; always exclude a denominator zero. Use open and closed circles on a number line.
Sanity-check with one test value from your final solution set, substituted back into the original. One quick check catches a flipped sign before it costs the answer mark.

Worked example

Solve a cubic inequation with a table of signs

Solve $x^3 + 1 \le x^2 + x$ using a table of signs.

Collect everything on one side. Move all terms left: $x^3 - x^2 - x + 1 \le 0$ .

Factor by grouping. Group in pairs: $x^2(x - 1) - (x - 1) \le 0$ , so $(x - 1)(x^2 - 1) \le 0$ , and since $x^2 - 1 = (x - 1)(x + 1)$ this is

(x + 1)(x - 1)^2 \le 0.

List the zeroes. $x = -1$ (simple) and $x = 1$ (double). No discontinuities. They cut the line into three intervals.

Test one value in each interval. Take $x = -2, 0, 2$ :

f(-2) = (-1)(-3)^2 = -9, \quad f(0) = (1)(-1)^2 = 1, \quad f(2) = (3)(1)^2 = 3.

So the sign row is $-, +, +$ . The sign does not change at the double zero $x = 1$ , as expected.

Keep the right intervals and fix the endpoints. We want $\le 0$ . The function is negative on $x < -1$ , and equals $0$ at the two zeroes. Include $x = -1$ . At $x = 1$ the function is also $0$ , which satisfies $\le 0$ , so include the isolated point $x = 1$ even though the curve only touches there.

Final answer: $x \le -1$ or $x = 1$ .

Examine the sign of a rational function

Find the zeroes and discontinuities of $y = \dfrac{x - 1}{x - 4}$ , and examine its sign.

Identify the critical values. The numerator is zero at $x = 1$ , so that is a zero of the function. The denominator is zero at $x = 4$ , so that is a discontinuity. These split the line into three intervals.

Test one value in each interval. Take $x = 0, 2, 5$ :

y(0) = \frac{-1}{-4} = \frac{1}{4} > 0, \quad y(2) = \frac{1}{-2} = -\frac{1}{2} < 0, \quad y(5) = \frac{4}{1} = 4 > 0.

So the sign row is $+, -, +$ across $x < 1$ , $1 < x < 4$ , $x > 4$ .

State the sign. $y$ is positive for $x < 1$ or $x > 4$ , and negative for $1 < x < 4$ . The sign changes at the zero $x = 1$ and again at the discontinuity $x = 4$ , the behaviour drawn in the rational-function figure above.

Final answer: positive for $x < 1$ or $x > 4$ ; negative for $1 < x < 4$ ; zero at $x = 1$ ; undefined at $x = 4$ .

Solve the same inequation by multiplying by the square of the denominator

Solve $\dfrac{3}{x + 2} \le x$ by first multiplying through by the square of the denominator, then compare with the sign-table solution above.

State the excluded value. The left side is undefined at $x = -2$ , so $x \ne -2$ throughout.

Multiply by the square. $(x + 2)^2 > 0$ for all $x \ne -2$ , so the inequality sign is preserved:

3(x + 2) \le x(x + 2)^2.

Collect on one side and factor without expanding. Bring everything to the right; the factor $(x + 2)$ is common:

x(x + 2)^2 - 3(x + 2) \ge 0 \implies (x + 2)\big(x(x + 2) - 3\big) \ge 0 \implies (x + 2)(x^2 + 2x - 3) \ge 0,

and since $x^2 + 2x - 3 = (x + 3)(x - 1)$ this is $(x + 2)(x + 3)(x - 1) \ge 0$ , with $x \ne -2$ .

Sign-table the cubic. Zeroes at $x = -3, -2, 1$ . Testing $x = -4, -\tfrac{5}{2}, 0, 2$ gives

h(-4) = -10, \quad h\!\left(-\tfrac{5}{2}\right) = \tfrac{7}{8} = 0.875, \quad h(0) = -6, \quad h(2) = 20,

so the sign row is $-, +, -, +$ . We need $\ge 0$ , giving $-3 \le x \le -2$ or $x \ge 1$ ; but $x = -2$ is excluded as a discontinuity of the original, so it becomes $-3 \le x < -2$ or $x \ge 1$ .

Final answer: $-3 \le x < -2$ or $x \ge 1$ , the same solution reached by the sign-table route.

Common traps

Assuming the sign changes at every zero: A zero of even multiplicity, like $(x - 2)^2$ , does not flip the sign; the curve touches the axis and turns back. Count multiplicities modulo $2$ .
Forgetting to dodge around discontinuities: A denominator zero is a candidate sign change just like a numerator zero. Leaving it out merges two intervals and gives the wrong sign row.
Including a denominator zero in the answer: The expression is undefined there, so it is never a solution, even under $\ge$ or $\le$ . State the exclusion when you first write the critical value.
Multiplying an inequation by the bare denominator: Multiplying $\dfrac{A}{B} \le c$ by $B$ is invalid because $B$ may be negative and would flip the sign. Subtract $c$ and combine into one fraction, or multiply by $B^2$ .
Expanding a common factor: After multiplying by $(x + 2)^2$ you get $(x + 2)$ on both sides. Expanding everything forces you to re-factor a cubic and usually introduces a sign slip; collect to one side and factor the common term out.
Testing on a critical value instead of between them: Test points must lie strictly inside the intervals, dodging around the zeroes and discontinuities; a test exactly at a critical value gives $0$ or undefined and tells you nothing about the interval.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

HSC-style3 marksDraw up a table of signs for

y = (x - 1)(x - 3)(x - 5)

, and hence solve

(x - 1)(x - 3)(x - 5) \le 0

Show worked answer →

The factored form already gives the zeroes: $x = 1$ , $x = 3$ and $x = 5$ . A polynomial has no discontinuities, so these are the only places the sign can change. They cut the number line into four intervals; pick one test value in each.

y(0) = (-1)(-3)(-5) = -15, \quad y(2) = (1)(-1)(-3) = 3, \quad y(4) = (3)(1)(-1) = -3, \quad y(6) = (5)(3)(1) = 15.

So the sign row is $-, +, -, +$ across $x < 1$ , $1 < x < 3$ , $3 < x < 5$ , $x > 5$ .

We want $y \le 0$ , so keep the intervals where $y$ is negative and, because the inequality is weak, include the zeroes where $y = 0$ :

x \le 1 \quad \text{or} \quad 3 \le x \le 5.

Markers reward computing a test value in every interval, the correct sign row, and including the zeroes for the weak inequality.

HSC-style3 marksFind the zero and the discontinuity of

y = \dfrac{x - 1}{x - 4}

, and state where the function is positive and where it is negative.

Show worked answer →

The function is zero where the numerator is zero, at $x = 1$ , and is undefined (discontinuous) where the denominator is zero, at $x = 4$ . These are the only two places the sign can change, so they split the line into three intervals. Test one value in each:

y(0) = \frac{-1}{-4} = \frac{1}{4} > 0, \quad y(2) = \frac{1}{-2} = -\frac{1}{2} < 0, \quad y(5) = \frac{4}{1} = 4 > 0.

So the sign row is $+, -, +$ across $x < 1$ , $1 < x < 4$ , $x > 4$ .

Hence $y$ is positive for $x < 1$ or $x > 4$ , and negative for $1 < x < 4$ . The sign genuinely changes at both the zero $x = 1$ and the discontinuity $x = 4$ .

Markers reward identifying both the zero and the discontinuity, testing each interval, and stating the sign on each piece.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksSolve

x(x - 2)(x + 3) > 0

using a table of signs.

Show worked solution →

List the zeroes. The factored form gives zeroes at $x = -3$ , $x = 0$ and $x = 2$ . A polynomial has no discontinuities, so these are the only places the sign can change. They cut the line into four intervals.

Test one value in each interval. Take $x = -4, -1, 1, 3$ :

f(-4) = (-4)(-6)(-1) = -24, \quad f(-1) = (-1)(-3)(2) = 6,

f(1) = (1)(-1)(4) = -4, \quad f(3) = (3)(1)(6) = 18.

So the sign row is $-, +, -, +$ across $x < -3$ , $-3 < x < 0$ , $0 < x < 2$ , $x > 2$ .

Keep the positive intervals. We want $f(x) > 0$ , and the inequality is strict, so exclude the zeroes.

Answer: $-3 < x < 0$ or $x > 2$ .

core3 marksSolve

(x - 2)^2 (x + 1) < 0

, taking care over the repeated factor.

Show worked solution →

Find the zeroes and their multiplicities. The zeroes are $x = -1$ (from the single factor $x + 1$ ) and $x = 2$ (from the squared factor $(x - 2)^2$ , a double zero). They cut the line into three intervals.

Test one value in each interval. Take $x = -2, 0, 3$ :

f(-2) = (-4)^2(-1) = -16, \quad f(0) = (-2)^2(1) = 4, \quad f(3) = (1)^2(4) = 4.

So the sign row is $-, +, +$ across $x < -1$ , $-1 < x < 2$ , $x > 2$ .

Read the repeated factor: The sign does not change across the double zero $x = 2$ , because $(x - 2)^2 \ge 0$ on both sides; the run of pluses confirms it. The only sign change is at the simple zero $x = -1$ .
Keep the negative interval: We want $f(x) < 0$ , which holds only on $x < -1$ . The strict inequality excludes $x = 2$ (where $f = 0$ ) anyway.
Answer: $x < -1$ .

core3 marksFind the zero and the discontinuity of

\dfrac{x + 4}{x - 1}

, and hence solve

\dfrac{x + 4}{x - 1} \ge 0

Show worked solution →

Locate the critical values. The expression is zero where the numerator is zero, at $x = -4$ , and undefined where the denominator is zero, at $x = 1$ . These split the line into three intervals.

Test one value in each interval. Take $x = -5, 0, 2$ :

\frac{-5 + 4}{-5 - 1} = \frac{-1}{-6} = \frac{1}{6} > 0, \quad \frac{0 + 4}{0 - 1} = \frac{4}{-1} = -4 < 0, \quad \frac{2 + 4}{2 - 1} = \frac{6}{1} = 6 > 0.

So the sign row is $+, -, +$ across $x < -4$ , $-4 < x < 1$ , $x > 1$ .

Fix the endpoints. We want the quotient $\ge 0$ , so keep the positive intervals. Include $x = -4$ , where the numerator is zero so the quotient equals $0$ (allowed by $\ge$ ). Exclude $x = 1$ , where the expression is undefined.

Answer: $x \le -4$ or $x > 1$ .

exam4 marksSolve

\dfrac{x - 1}{x + 2} \le 1

by collecting everything on one side as a single fraction, then using a table of signs.

Show worked solution →

Move the constant across first. Do not multiply by $x + 2$ , whose sign is unknown. Subtract $1$ from both sides and combine over the common denominator $x + 2$ :

\frac{x - 1}{x + 2} - 1 \le 0 \implies \frac{(x - 1) - (x + 2)}{x + 2} \le 0 \implies \frac{-3}{x + 2} \le 0.

Simplify the sign question: The numerator is the constant $-3 < 0$ , so $\dfrac{-3}{x + 2} \le 0$ is equivalent to $\dfrac{1}{x + 2} \ge 0$ , which (since the numerator $1$ is positive) holds exactly when $x + 2 > 0$ .
Watch the discontinuity: The original expression is undefined at $x = -2$ , and the rearranged fraction can never equal $0$ (its numerator is $-3 \ne 0$ ), so $x = -2$ is excluded and no endpoint is gained there.
Solve: $x + 2 > 0$ gives $x > -2$ .
Check: At $x = 0$ , $\dfrac{0 - 1}{0 + 2} = -\tfrac{1}{2} \le 1$ , true. At $x = -3$ , $\dfrac{-3 - 1}{-3 + 2} = \dfrac{-4}{-1} = 4 \le 1$ , false, so $x = -3$ is correctly excluded.
Answer: $x > -2$ .

exam4 marksFor

y = (x + 2)(x - 1)^2

, draw up a table of signs, state the

y

-intercept, and hence solve

(x + 2)(x - 1)^2 \ge 0

. Use the table to describe how the curve meets the

x

-axis at each zero.

Show worked solution →

Find the zeroes and multiplicities. The zeroes are $x = -2$ (simple) and $x = 1$ (double, from $(x - 1)^2$ ). They cut the line into three intervals.

Test one value in each interval. Take $x = -3, 0, 2$ :

f(-3) = (-1)(-4)^2 = -16, \quad f(0) = (2)(-1)^2 = 2, \quad f(2) = (4)(1)^2 = 4.

So the sign row is $-, +, +$ across $x < -2$ , $-2 < x < 1$ , $x > 1$ .

Read the y-intercept: $f(0) = (2)(-1)^2 = 2$ , so the curve cuts the $y$ -axis at $(0, 2)$ .
Describe each zero from the signs: At $x = -2$ the sign changes from $-$ to $+$ , so the curve crosses the $x$ -axis there. At $x = 1$ the sign does not change (it stays $+$ ), so the curve touches the $x$ -axis and turns back, the hallmark of a double zero.
Solve the inequation: We want $f(x) \ge 0$ . The function is positive on $-2 < x < 1$ and $x > 1$ , zero at $x = -2$ and $x = 1$ , and these two intervals plus the touching point join into one piece. Including the zeroes (weak inequality):
Answer: $x \ge -2$ .

What this dot point is asking

The answer

Where can a function change sign?

Building a table of signs

Polynomials: a worked sign table

The repeated-factor exception

Functions with discontinuities

Two routes for a rational inequation

Solve a rational inequation by sign table, stage by stage

How exam questions ask about the sign of a function

Exam-style practice questions

Practice questions

Related dot points