Pearson Edexcel International Advanced Level Mathematics: Inequalities, polynomials and graphs (YMA01)

This guide is aligned to the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification (YMA01), Issue 3, April 2019. It covers Unit P1: Pure Mathematics 1, section 1 (Algebra and functions), topics 1.6 to 1.12. According to the specification, P1 and P2 are compulsory in IAS Mathematics, in IAL Mathematics and in IAL Pure Mathematics; no tier labels apply.

Use it with the revision notes and the practice questions. Course pages: the Edexcel A Level Mathematics hub, the printable checklist and the free 10-minute diagnostics. Other P1 and P2 pages: Pure Mathematics 1, Pure Mathematics 2 and the IAL exam preparation page. Topics 1.1 to 1.5 (indices, surds, quadratic functions, discriminant, completing the square) are not taught here.

What this unit covers

Topic	You must be able to
1.6	Solve simultaneous equations by substitution
1.7	Interpret linear and quadratic inequalities graphically, including brackets and fractions that reduce to linear or quadratic form
1.8	Represent inequalities graphically, with shading and dotted or solid lines
1.9	Solve linear and quadratic inequalities
1.10	Manipulate and factorise polynomials of degree up to 3; use f(x) notation
1.11	Sketch simple cubics, y = k/x and y = k/x² (x ≠ 0), and trigonometric graphs; use asymptotes; use intersections to solve equations
1.12	Apply y = af(x), y = f(x) + a, y = f(x + a) and y = f(ax) to quadratics, cubics, reciprocals, sine, cosine and tangent, and to any given graph

1.6 Simultaneous equations by substitution

When one equation is linear, make one letter the subject of it, substitute into the other equation, solve the resulting equation, then go back to the linear equation for the second letter. Always pair the answers, because the final answer is a set of coordinate pairs.

Example 1. Solve y = x − 3 and x² + xy = 14.

x² + x(x − 3) = 14
2x² − 3x − 14 = 0
(2x − 7)(x + 2) = 0
x = 7/2 or x = −2

Substitute into y = x − 3: x = 7/2 gives y = 1/2, and x = −2 gives y = −5.

Answer: (7/2, 1/2) and (−2, −5).

Each pair gives the coordinates of one intersection of the two graphs. If the substituted quadratic has no real solutions, the graphs do not meet.

Example 2. y = 2x − 7 and y = x² − 3x + 1 give x² − 5x + 8 = 0. Here b² − 4ac = 25 − 32 = −7, which is negative, so the line and curve never meet and there are no solutions.

1.7 and 1.9 Solving inequalities

Linear. Solve as an equation, but reverse the inequality sign when you multiply or divide by a negative number.

Example 3. Solve 3(x − 2) ≥ 5x + 4. Expand: 3x − 6 ≥ 5x + 4. Subtract 5x and add 6: −2x ≥ 10. Divide by −2 and reverse: x ≤ −5.

Quadratic. Rearrange to have 0 on one side, factorise (or use the formula), sketch the curve, and read off the range. For a positive x² coefficient, “below the axis” gives the interval between the roots and “above the axis” gives two outer regions.

Example 4. Solve 3x² − 2x − 8 < 0. Factorise: (3x + 4)(x − 2) < 0, with roots x = −4/3 and x = 2. The U-shaped curve is below the axis between the roots, so −4/3 < x < 2.

Fractions. Never multiply by x alone, because x may be negative. Multiply by x², which is positive for x ≠ 0, and state x ≠ 0.

Example 5. Solve 6/x < 5.

6x < 5x²                (multiplied by x², x ≠ 0)
5x² − 6x > 0
x(5x − 6) > 0
x < 0  or  x > 6/5

Check x = −1: −6 < 5, true. Check x = 1: 6 < 5, false. Check x = 2: 3 < 5, true. Answer: x < 0 or x > 6/5.

Graphical meaning (1.7). Solving ax + b > cx + d means finding where the line y = ax + b lies above the line y = cx + d. Solving px² + qx + r ≥ 0 means finding where the curve is on or above the x-axis. Solving px² + qx + r < ax + b means finding where the curve y = px² + qx + r lies below the line y = ax + b.

Example 6. Solve x² − 3x < 2x − 6. Rearrange to x² − 5x + 6 < 0, so (x − 2)(x − 3) < 0. The curve y = x² − 3x meets the line y = 2x − 6 at (2, −2) and (3, 0), and is below it between those points. Answer: 2 < x < 3.

When two conditions must hold together, solve each separately and keep the values that satisfy both, using a number line.

1.8 Regions defined by inequalities

1. Draw the boundary: y = x + c or y = ax² + bx + c.
2. A strict inequality (< or >) has a dotted boundary. An inclusive one (≤ or ≥) has a solid boundary.
3. Substitute a point that is off the boundary, and shade the side that makes the inequality true.
4. Label the region R and follow the question’s instruction about shading the wanted or the unwanted region.

Example 7. Show the region where y ≥ x² − 4x and y < x + 6.

The boundaries meet where x² − 4x = x + 6, so x² − 5x − 6 = 0 and x = −1 or 6, giving (−1, 5) and (6, 12). The parabola is solid. The line is dotted. Test (1, 0): 0 ≥ −3 is true and 0 < 7 is true. R is the region above the parabola and below the line, between the two intersection points. Both intersection points lie on the dotted line, so neither is in R.

1.10 Polynomials and f(x) notation

A polynomial is a sum of terms in whole-number powers of x. Its degree is the highest power. Expand by distributing: each term of the first bracket multiplies each term of the second, then collect like terms.

Example 8. Expand (x + 1)(x − 3)(x + 5). First (x + 1)(x − 3) = x² − 2x − 3. Then (x² − 2x − 3)(x + 5) = x³ + 5x² − 2x² − 10x − 3x − 15, which gives x³ + 3x² − 13x − 15.

Factorising, try in this order: a common factor, a difference of two squares, a quadratic factor, then grouping.

2x³ − 8x = 2x(x² − 4) = 2x(x − 2)(x + 2)

x³ + 5x² − 4x − 20 = x²(x + 5) − 4(x + 5)
                   = (x + 5)(x² − 4)
                   = (x + 5)(x − 2)(x + 2)

A cubic with no visible common factor or grouping is factorised in P2 (topic 2.1).

f(x) notation. f(x) is a rule. f(3) means replace x by 3, and f(x + 2) means replace every x by (x + 2).

Example 9. For f(x) = x² − 3x: f(−1) = 1 + 3 = 4. f(x + 2) = (x + 2)² − 3(x + 2) = x² + x − 2. f(2x) = (2x)² − 3(2x) = 4x² − 6x.

1.11 Graph sketching

Cubics. For y = ax³ + …, a positive a means the curve rises to the right. A negative a means it falls. From a factorised form read the roots, then find the y-intercept by putting x = 0. A repeated factor means the curve touches the axis at that root.

Example 10. Sketch y = x²(4 − x). Roots: x = 0 (repeated, so the curve touches the axis) and x = 4 (crosses). The y-intercept is the origin. The expanded form is 4x² − x³, so the x³ coefficient is negative. The curve comes down from the top left, touches the axis at the origin, rises to a maximum between x = 0 and x = 4, then falls through (4, 0).

Reciprocals. Both graphs have the axes as asymptotes: lines the curve approaches but never reaches.

Graph	Positive k	Negative k
y = k/x	Branches in quadrants 1 and 3	Branches in quadrants 2 and 4
y = k/x²	Both branches above the x-axis, symmetric about the y-axis	Both below

Neither graph is defined at x = 0, and neither crosses an axis.

Trigonometric graphs (degrees; replace 90° by π/2, 180° by π, 360° by 2π for radians).

Graph	Period	Key features
y = sin x	360°	Starts at 0, peaks at 1 when x is 90°, returns to 0 at 180°, bottoms out at −1 when x is 270°
y = cos x	360°	Starts at its peak of 1, is zero at 90° and 270°, has its lowest value −1 at 180°
y = tan x	180°	Zero at 0°, 180° and 360°; grows without limit towards the vertical asymptotes x = 90° and x = 270°

Using intersections. To show how many solutions an equation has, rewrite it as two curves and sketch both.

Example 11. Show that x³ − x − 5 = 0 has exactly one real root. Rearrange to x³ − x = 5, then x² − 1 = 5/x. Sketch y = x² − 1 and y = 5/x. For x > 0 the parabola rises from −1 while the reciprocal falls from very large values, so they cross once. For x < 0 the reciprocal is below −1 for −5 < x < 0, where the parabola is at least −1, and for x < −5 the parabola is positive while the reciprocal is negative. There is no crossing. Hence one real root (about 1.90).

1.12 Transformations of y = f(x)

Equation	Effect on the graph	Effect on a point (p, q)
y = af(x)	Every y-value is multiplied by a: a vertical stretch	(p, aq)
y = f(x) + a	Whole graph slides up by a (down when a is negative)	(p, q + a)
y = f(x + a)	Whole graph slides left by a (right when a is negative)	(p − a, q)
y = f(ax)	Every x-value is divided by a: a horizontal stretch of scale factor 1/a	(p/a, q)

Changes inside the bracket act on x and work “backwards”. Changes outside act on y and work as written. The case a = −1 gives a reflection: y = −f(x) in the x-axis, and y = f(−x) in the y-axis.

Example 12. The graph of f(x) = x² + 4x − 5 has a minimum at (−2, −9), crosses the y-axis at (0, −5) and crosses the x-axis at (−5, 0) and (1, 0).

Curve	Minimum	y-intercept	x-intercepts
y = 3f(x)	(−2, −27)	(0, −15)	(−5, 0), (1, 0)
y = f(x) + 1	(−2, −8)	(0, −4)	found by solving
y = f(x + 2)	(−4, −9)	(0, 7)	(−7, 0), (−1, 0)
y = f(2x)	(−1, −9)	(0, −5)	(−5/2, 0), (1/2, 0)

For y = f(x + 2) the new y-intercept is f(2) = 7. For y = f(2x) it stays at (0, −5) because x = 0 does not move.

Example 13. Sketch y = 1/(x + 2) from y = 1/x: translate 2 left. The vertical asymptote moves to x = −2, the horizontal asymptote stays at y = 0, and the y-intercept is 1/2. Sketch y = 1/x + 3: translate 3 up. The horizontal asymptote is y = 3, and the curve meets the x-axis where 1/x = −3, so x = −1/3.

For trigonometric graphs, y = 3 sin x has maximum 3 and minimum −3. The graph of y = sin 2x repeats every 180°. The graph of y = tan(x + 45°) has asymptotes at x = 45° and 225°.

Only one transformation is applied at a time in this topic. Combinations of transformations are in P3 (topic 1.4).

Common errors

• Writing f(x + 2) as a shift to the right. It is a shift to the left.
• Halving x-coordinates for f(x/2). Dividing by 1/2 doubles them.
• Multiplying an inequality by x without checking its sign.
• Giving x-values without finding the matching y-values in simultaneous equations.
• Writing “−1/2 > x > 3” for two outer ranges, or using “and” where “or” is needed.
• Solid line for a strict inequality, or dotted line for ≤ and ≥.
• Showing an asymptote as a point the curve crosses, or forgetting to move it after a translation.

Next steps

Work through the practice questions, then use the revision notes to check recall. The diagnostics show which topics need more work.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Mathematics, Further Mathematics and Pure Mathematics specification, Issue 3, April 2019, published by Pearson Education Limited. Unit P1, topics 1.6 to 1.12.