Did you know? Two curves can cross at the right points on a graph and still lead a learner to the wrong inequality interval. Finding the intersections tells you where two functions are equal. Solving an inequality asks a different question: where is one function above or below the other?
For parents looking for a JC1 H1 Mathematics tutor in Bukit Timah, graphical inequalities reveal whether a pupil can move between pictures, algebra and practical meaning. A confident learner identifies intersections, checks the order of the curves in each interval, includes or excludes endpoints correctly and knows when a graphing calculator needs a better viewing window.
Quick answer: to solve f(x) > g(x), draw or inspect both graphs, find or estimate their intersections, then take the x-values where the graph of f is higher than the graph of g. For f(x) ≥ g(x), include the intersections themselves when the functions are defined there. An equivalent algebraic route is to solve f(x) − g(x) > 0 or ≥ 0.
Quick reading route: three-step method · line and parabola · two quadratics · exponential threshold · FAQs.
The three-step graphical inequality method
- Draw or identify the two functions. State which is f and which is g; do not confuse the variables and vertical axes.
- Find their intersections. Solve f(x) = g(x) exactly when practical, or use the permitted graphical method with appropriate accuracy.
- Read the vertical order. In each interval, decide where f(x) − g(x) is positive, zero or negative and apply the strict or inclusive boundary symbols.
The crucial word is vertical. At a particular x-coordinate, compare the heights of the graphs. Whichever function has the larger y-value is the higher curve. That comparison answers f(x) > g(x) without needing to name a special kind of equation.
Graphical inequalities can often be checked algebraically. Define h(x) = f(x) − g(x). Then f(x) > g(x) means h(x) > 0, so a graph of h above the horizontal axis describes exactly the same permitted x-values.
- Crossing: two functions have equal output at an intersection.
- Above: f(x) > g(x).
- Below: f(x) < g(x).
- On or above: f(x) ≥ g(x).
- On or below: f(x) ≤ g(x).
- Strict inequality: equality points are excluded.
- Inclusive inequality: equality points are included if they belong to the permitted domain.
Where this fits the official H1 syllabus
The SEAB H1 Mathematics 8865 syllabus for the 2027 examination includes solving inequalities by graphical methods, solving quadratic inequalities analytically, simultaneous equations involving one linear and one quadratic equation, and using a graphing calculator or graphing software for numerical solutions.
It also includes graph features such as intercepts, symmetry, turning points and asymptotes. These skills reinforce one another. A graph of a parabola can provide root locations, sign intervals and an estimate of the relevant region, but a pupil still needs to interpret the inequality symbols.
The 2027 H1 paper has one three-hour, 100-mark assessment, with Pure Mathematics comprising 40 marks. A JC1 learner whose eventual A-Level year is later should follow that year’s official syllabus rather than assume every paper has unchanged scope.
Our existing quadratic inequalities tutorial focuses on one quadratic and its sign. This article asks a distinct practical question: which of two different curves is higher in a stated x-interval, and what does that mean?
Worked example one: a line above a parabola
Consider the line y = 2x and parabola y = x². Find the values of x where the line is strictly above the parabola. This is the inequality 2x > x².
- Find intersections by solving 2x = x².
- Rearrange: x² − 2x = x(x − 2) = 0.
- Thus the graphs meet at x = 0 and x = 2.
- Test a value between the roots, such as x = 1: the line gives 2 and the parabola gives 1.
- Therefore 0 < x < 2 is the interval where the line is higher.
Outside the two roots, the parabola sits above the line. At x = 3, the parabola’s output is nine while the line’s output is six. At x = −1, the parabola gives one and the line gives −2. These tests confirm the vertical ordering on either side.
The strict inequality excludes x = 0 and x = 2, where the functions have equal output. If the question instead asks where 2x ≥ x², the answer is 0 ≤ x ≤ 2.
What does the correct sketch contain?
Draw the upward-opening parabola through the origin and the point (2,4). Draw the straight line of gradient two through those same intersections. Between x = 0 and x = 2, shade or label the region where the line sits above the curve.
The diagram should convey an interval, not merely the two intersection coordinates. A pair of correct roots without the inequality interval would answer ‘where are they equal?’ rather than ‘where is one greater?’
The same answer from a sign chart
Subtract the two expressions: 2x − x² = x(2 − x). The product is positive when 0 < x < 2 because both factors are positive. On x > 2 the second factor becomes negative, and on x < 0 the first factor is negative while the second is positive.
Thus the product sign method and graph comparison agree. This kind of cross-check is valuable because the learner is not dependent on a perfect screen resolution or an accurately drawn hand sketch.
A tutor can ask the pupil to explain the algebra after the picture, then reverse the sequence on a new example. The goal is flexible understanding, not memorising one visual layout.
Worked example two: a different line and parabola
Let f(x) = x² − 4x + 3 and g(x) = x − 1. Find where f(x) is lower than g(x). Form the inequality x² − 4x + 3 < x − 1.
Move everything to the left: x² − 5x + 4 < 0. Factorise to (x − 1)(x − 4) < 0. The upward-opening difference expression is negative between its roots, so 1 < x < 4.
Check x = 2: f(2) = 4 − 8 + 3 = −1, while g(2) = 1. Therefore f is indeed below g at x = 2. Check x = 5: f(5) = 25 − 20 + 3 = 8, while g(5) = 4, so the order is reversed outside the interval.
This example illustrates an important equivalence: graphing f and g and observing which is higher is the same as graphing their difference h(x) = x² − 5x + 4 and checking where h lies below the x-axis.
Our simultaneous linear and quadratic equations guide shows how to recover full intersection coordinates. The inequality task here adds the decision of which side of each intersection belongs in the solution.
Why sign can change at one intersection but not another
A simple crossing of two curves generally changes their vertical order, but a touching intersection need not. Consider f(x) = (x − 1)² and g(x) = 0. They meet at x = 1, yet the squared curve is never below the horizontal axis.
The inequality (x − 1)² ≥ 0 is true for every real x. The strict inequality (x − 1)² > 0 is true for every real x except x = 1.
A pupil who automatically assumes the sign alternates at every root could get this wrong. The repeated root produces a touching point rather than a crossing. A graph makes that behaviour visible.
A useful rule is to inspect each interval rather than blindly switch signs at every root. A test input or factor-sign reasoning provides a reliable second check.
Worked example three: two quadratic curves
Let f(x) = x² − 2x + 1 and g(x) = −x² + 4x − 1. Find the x-values for which g is greater than f. First find their difference g(x) − f(x) = −2x² + 6x − 2.
We want −2x² + 6x − 2 > 0. Divide by negative two and reverse the inequality to get x² − 3x + 1 < 0. The quadratic equation x² − 3x + 1 = 0 has roots (3 − √5)/2 and (3 + √5)/2.
Because the upward-opening difference x² − 3x + 1 is negative between its roots, the answer is (3 − √5)/2 < x < (3 + √5)/2, approximately 0.382 < x < 2.618.
Check x = 1: f(1) = 0 and g(1) = −1 + 4 − 1 = 2, so g is higher. Check x = 0: f(0) = 1 while g(0) = −1, so g is lower. The computed interval agrees with the picture.
This example shows why graphical inequalities do not have to involve a line and parabola only. Comparing two quadratics can still produce one quadratic sign problem, because subtracting them yields another quadratic.
The inequality reversal is not optional
When the expression −2x² + 6x − 2 is divided by −2, the direction of the inequality changes from greater than zero to less than zero. A student who forgets that reversal will select the outside intervals instead of the middle region.
The best repair is not merely memorising ‘flip the sign’. Show one numeric example: 2 > 1 becomes −2 < −1 after multiplication by −1. The reversed order is a property of negative scaling.
Worked example four: an always-positive difference
Let f(x) = x² + 1 and g(x) = 2x. Their difference is f(x) − g(x) = x² − 2x + 1 = (x − 1)².
Since any real square is non-negative, f(x) ≥ g(x) for all real x, and equality occurs only at x = 1. There is no interval where f is strictly lower than g.
At x = 1 both functions equal two. A sketch shows the parabola touching the line at (1,2) and lying above it everywhere else. This is a powerful counterexample to the assumption that two graphs must cross whenever they meet.
A calculator graph with an unfortunate pixel scale might make the two curves appear to cross near x = 1. The algebraic square proves that their order does not reverse.
A graphing-calculator comparison needs a sensible window
Suppose the calculator displays f(x) = x² − 4x + 3 and g(x) = x − 1 only over x from 5 to 10. Both relevant intersections x = 1 and x = 4 are outside the window. A user might incorrectly conclude that the curves do not meet.
The graphing calculator did not perform incorrect arithmetic. It showed only the interval requested. A student should use known intercepts or an algebraic estimate to select a window where the question’s important features can be seen.
The graphing calculator accuracy guide covers this issue. When graphs are used to find a solution, the official 2027 syllabus instructs students to sketch the relevant graphs as part of their answer; follow the particular question’s required method and precision.
Worked example five: an exponential threshold
Compare f(x) = ex with the horizontal line g(x) = 3. We want to know where the exponential is higher than three, so solve ex > 3.
Because ex is strictly increasing and ln reverses the exponential, the boundary occurs at ex = 3, meaning x = ln 3. Therefore ex > 3 when x > ln 3.
On the graph, y = ex crosses the horizontal line y = 3 at approximately x = 1.099 and stays higher for larger x. The graphical and algebraic solutions agree.
A pupil who writes x > 3 has confused the height of the horizontal line with the x-coordinate of the crossing. Logarithms convert the target output into the corresponding input.
A decay threshold with a time restriction
Suppose a hypothetical index follows R(t) = 100e−0.2t for t ≥ 0 years. When does it first fall to or below 40? The target inequality is 100e−0.2t ≤ 40.
Divide by 100: e−0.2t ≤ 0.4. Take natural logs: −0.2t ≤ ln 0.4. Divide by the negative number −0.2 and reverse the inequality, giving t ≥ ln(2.5)/0.2 ≈ 4.58 years.
If the question demands the first completed whole year at or below the target, report year five. The model crosses the threshold after 4.58 years but before five full years have passed.
This is a graphical inequality, logarithmic equation and context interpretation in one problem. It is exactly the kind of multi-step choice a good H1 tutorial can make understandable without teaching unrelated advanced methods.
A cost and revenue comparison
Consider an invented producer whose revenue is R(q) = 16q and cost is C(q) = q² + 4q + 20, with q a non-negative whole-number item count. The question asks when revenue exceeds cost.
Form R(q) > C(q): 16q > q² + 4q + 20, or q² − 12q + 20 < 0. Factorise to (q − 2)(q − 10) < 0. The continuous solution is 2 < q < 10.
Because q counts whole items, the feasible positive-profit quantities are 3 through 9 inclusive. The x-intersections at two and ten represent break-even quantities; between them, the revenue line lies above the cost curve.
If the student reports q = 2 or q = 10 as profitable, they have ignored that the condition is strictly greater than, not greater than or equal. At both endpoints revenue and cost are equal, so the modelled profit is zero.
This is an illustrative classroom model rather than real business advice. A correct contextual answer includes both the mathematical interval and the permitted whole-number interpretation.
When a mathematical interval is not a real-world answer
A solution such as −3 < q < 8 may arise from valid algebra. But if q represents the number of people or physical items, negative quantities are not permitted. The relevant interval must be intersected with the question’s feasible domain.
A tutor should train learners to check domains at the beginning and end. The first check prevents impossible assumptions in the model; the last ensures that the reported result answers the practical question.
For a time variable, the question might restrict t ≥ 0. For a quantity of whole items, q must be a non-negative integer. For a percentage, a feasible value may lie between zero and one hundred. The correct restriction is supplied by the context rather than guessed.
What a sign chart can reveal faster than an elaborate sketch
Once the boundary roots are known, a simple sign chart has a column for intervals and a row showing whether f(x) − g(x) is positive or negative. For x² − 5x + 4 = (x − 1)(x − 4), the signs are positive left of one, negative between one and four, and positive right of four.
A single test value from each interval validates the chart. There is no need to calculate twenty points if the polynomial’s factor structure and root locations already explain the sign.
Graphing and sign charts are complementary. The graph makes ‘which curve is higher?’ intuitive; the chart often communicates the exact algebraic solution more compactly.
Inequalities are not just equations with an extra symbol
An equation locates equality. An inequality selects every permitted input for which one quantity is larger or smaller than another. Solving only the equality finds candidate boundaries, not the final answer.
This is why a student may get both intersection points right and still fail the inequality. The necessary extra step is to inspect the intervals between and outside those points and choose the ones satisfying the comparison.
A tutor who diagnoses this early may need only a number-line explanation, not an entire new chapter of algebra.
Six common errors and the precise repair
Error one: returning roots rather than intervals
If 2x > x², the roots 0 and 2 are the boundaries. The solution is the interval 0 < x < 2. The question asks where the inequality holds.
Error two: forgetting whether equality is allowed
An inclusive inequality includes valid intersection points; a strict one excludes them. Write the symbols before interpreting the graph.
Error three: reversing upper and lower curves
At x = 1, compare exact y-values. For y = 2x and y = x², the line is at two and the parabola at one, so the line is higher.
Error four: incorrectly reversing an inequality
Multiplying or dividing by a negative number reverses the order. The two-quadratic example uses this when dividing its difference by −2.
Error five: assuming every root changes the sign
A repeated touching root can leave a curve on the same side of the comparison. Use a test value or recognise a squared factor.
Error six: trusting a clipped calculator view
A window that misses the intersections cannot establish the full solution. Predict or locate boundaries algebraically and adjust the display.
A fifteen-minute independent practice routine
- Minutes 1–3: draw y = x² and y = 2x, then write both strict and inclusive line-above-parabola intervals.
- Minutes 4–6: compare f = x² − 4x + 3 and g = x − 1; check the solution using their difference.
- Minutes 7–9: use a sign chart for the two-quadratic example and confirm with x = 1.
- Minutes 10–12: solve one exponential threshold with ln and explain the direction of the solution.
- Minutes 13–15: convert a continuous inequality into a permitted whole-number business or time answer.
After a short break, change the coefficients and the strictness of the inequality. The pupil should recognise what changed in the task rather than copy the previous interval.
A small-group H1 tutorial can compare methods
A tutor may ask one pupil to sketch the two curves, another to write the algebraic difference and a third to defend the endpoint choices. The class gains from seeing those approaches only if every learner then independently solves a new comparison.
In a three-student group, one student may find roots but not intervals, another may miss an inequality reversal and a third may ignore the contextual whole-number restriction. Those are different problems that require targeted mini-drills.
The established eduKateSG three-pupil teaching reference describes 90-minute weekly tutorials near Sixth Avenue MRT at 8 Fourth Avenue, Singapore 268674. That format page does not establish current JC1 H1 Mathematics vacancies or fees; confirm tutor suitability, class time and pricing directly.
Bring a marked graphing question that the pupil found difficult and ask the tutor which decision failed first: plotting, intersection calculation, interval selection or interpretation. See consultation information.
Five independent checks
One: line above parabola
Solve 2x > x². Answer: 0 < x < 2. Include neither endpoint.
Two: inclusive comparison
Solve 2x ≥ x². Answer: 0 ≤ x ≤ 2. The two intersections belong in the solution.
Three: two quadratic curves
For g − f = −2x² + 6x − 2, g is higher when (3 − √5)/2 < x < (3 + √5)/2.
Four: always non-negative
For x² + 1 ≥ 2x, the difference is (x − 1)² ≥ 0, so all real x satisfy the inequality.
Five: an exponential threshold
Solve ex > 3. Answer: x > ln 3. The logarithm gives the crossing input.
A graph-feature check before selecting an interval
Two curves should first be recognised for what they are. An upward-opening quadratic often rises on both sides of its minimum, while a straight line has constant gradient. Their possible vertical ordering depends on intersections, but their shapes can help predict what happens outside the intersection interval. Our graph intercepts, symmetry and asymptotes guide develops those basic features in a separate lesson.
For the comparison y = 2x and y = x², the parabola opens upwards and grows more rapidly than the line for large positive x, so a solution stating that the line is above the parabola for all x > 2 should look suspicious. The picture suggests the line can be higher only within a bounded region, and the algebra shows exactly which region: 0 < x < 2.
When another example involves a shifted exponential and a horizontal line, the graphical expectations differ. The exponential ex is strictly increasing and crosses a positive horizontal target once, so the greater-than inequality selects a half-line after the crossing rather than a finite middle interval. A student who predicts the rough shape of the answer before calculating is less likely to select the wrong side of the intersection.
The most useful final check is to choose one number that should satisfy the answer and one that should not. Substitute both numbers into the original inequality, not just a rearranged version. If either contradicts the claimed interval, re-examine an inequality reversal, a missed endpoint or the vertical order of the curves. This two-value check takes little time and provides strong evidence that the interval describes the intended comparison.
Frequently asked questions
Are graphical inequalities part of H1 Mathematics 8865?
Yes. The official 2027 syllabus explicitly includes solving inequalities by graphical methods, alongside analytical quadratic inequalities.
Are intersections enough to solve an inequality?
No. They supply equality boundaries. You must identify where the comparison is positive or negative between and outside those boundaries.
What is the difference between > and ≥?
Strictly greater excludes equality; greater than or equal includes any valid equality points.
Can two curves touch without crossing?
Yes. For example y = (x − 1)² touches y = 0 at x = 1 and remains on or above the axis.
Why use f(x) − g(x)?
Its sign directly tells you whether f is higher or lower than g, reducing the graph comparison to a sign problem.
Can a graphing calculator miss an intersection?
Yes. A poor window may hide it, and tracing can be imprecise. Use algebra and a suitable window where possible.
How should I answer if x counts items?
Restrict the mathematical interval to feasible non-negative whole numbers, according to the problem’s definition.
What does the tutor need to assess?
Whether the pupil can identify comparison boundaries, choose the correct intervals and transfer the method to an unseen graph without prompts.
The next time two curves appear
Do not stop when they meet. Ask where one is higher. That decision turns an intersection calculation into a graphical inequality and transforms an unfamiliar graph into an organised set of possible intervals.
For algebraic sign reasoning, see the quadratic inequalities guide. For graph intersections, use the simultaneous equations guide. The Bukit Timah H1 Mathematics overview connects the wider course, while the official 2027 syllabus sets the scope for that named examination year.