Every Grade 12 topic, explained in plain language and worked through using real DBE exam questions from 2023 to 2026. Every line of working is shown, one tap at a time, so you never get lost in a wall of algebra. Missed something in Grade 10 or 11? Start with the gap finder and we patch it first.
Start by fixing Grade 10 and 11 gaps, then work through every Paper 1 and Paper 2 topic. Each lesson teaches the idea, lets you see it in a live simulation, shows how it is examined with a marked memo, and checks that it stuck.
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Built for brains that wander
Short bursts. Clear next steps. No walls of text.
If your attention jumps around, you are not bad at maths. Most textbooks are simply built for one long, quiet sitting. This page is built for 15-minute sprints instead.
ChunkOne topic at a time
Open a topic and switch on Focus in the bottom bar. Everything else fades so your eyes stay on one card.
RevealTap “Next step”
Worked examples show one line at a time. Try the next line in your head before you tap. That guess is where the learning happens.
RewardQuick checks earn XP
Every topic ends with short questions and instant feedback. A wrong answer tells you exactly which slip you made.
RestSprint, then break
Start the timer in the bottom bar. When it rings, take the break. Your brain files what you learnt while you rest.
Stop 1 · Gap finder
Matric maths is built on Grade 10 and 11 bricks.
Most people who “can’t do” matric maths are missing two or three older skills, and every new topic falls on top of those gaps. Ten questions find yours.
No marks, no pressure. Pick an answer and keep moving. Guessing is fine.
Each question tests one skill. A wrong answer points to the exact foundation card to read.
About 4 minutes. Scrap paper helps.
Question 1 of 10Skill
Stop 2 · Foundations
Eight skills every matric question quietly assumes.
Read only the cards the gap finder flagged. Each one takes about three minutes.
Signs and order of operations F1
Half of all lost marks are sign slips. Fix this and you gain marks in every single topic.
Order: Brackets, Of/powers, Divide and Multiply (left to right), Add and Subtract (left to right).
Same signs multiply to positive, different signs to negative: \((-3)(-4)=12\), \((-3)(4)=-12\).
\(-3^2=-9\) because the power touches only the 3. But \((-3)^2=9\).
Subtracting a negative is adding: \(5-(-2)=7\).
Example: \(-3^2+2(-4)=-9-8=-17\)
Brackets around a negative number are like a seatbelt. Put them in whenever you substitute a negative: \(f(-2)\) for \(x^2\) is \((-2)^2=4\).
Fractions and percentages F2
Finance, probability and calculus all lean on fractions.
Add or subtract: get a common denominator first. \(\frac{2}{3}+\frac14\)\({}=\frac{8}{12}+\frac{3}{12}\)\({}=\frac{11}{12}\)
Multiply straight across: \(\frac{2}{3}\times\frac{5}{7}\)\({}=\frac{10}{21}\). Divide: flip the second and multiply.
A percentage is a fraction of 100: \(15\%=0{,}15\).
Increase R400 by 15%: \(400\times1{,}15=R460\). Decrease by 15%: \(400\times0{,}85=R340\).
“Increase by \(r\%\)” always means multiply by \(1+\frac{r}{100}\). That one idea is the whole of compound interest.
Exponent laws F3
These five laws unlock exponential equations, logs and every derivative.
Special angles: \(\sin30^\circ=\frac12\), \(\cos60^\circ=\frac12\), \(\tan45^\circ=1\), \(\sin60^\circ\)\({}=\frac{\sqrt3}{2}\), \(\cos30^\circ\)\({}=\frac{\sqrt3}{2}\)
SOH CAH TOA. For the special angles, draw two triangles: the half-square (sides \(1;1;\sqrt2\), angles 45°) and the half-equilateral (sides \(1;\sqrt3;2\), angles 30° and 60°). You can read every special value off them.
Where the marks are
Spend your time where the marks live.
These are the official CAPS mark weightings (each can move by about 3 marks either way). Tap a topic to jump to it. Your progress shows on the right of each row.
Bars are drawn to one scale per paper: the longest bar is the heaviest topic on that paper. Trigonometry alone is a third of Paper 2.
Stop 3 · Paper 1
Paper 1: algebra, functions and calculus.
Seven topics. Open one, read the big idea, then work the examples one step at a time. Mark the topic done when the quick check feels easy.
Solving an equation means finding the number that makes both sides equal. In Question 1 you meet the same six disguises every year.
Quadratic that factorises: \(x^2-9=0\)Factorise, then set each bracket to zero.
Quadratic that won’t factorise, or “TWO decimal places”Use the quadratic formula. The decimal instruction is a hint.
Inequality: \(x^2-2x-3\gt0\)Find the roots, sketch the parabola, read off the region.
Exponential in disguise: \(2\cdot2^{2x}-9\cdot2^x+4=0\)Let \(k=2^x\). It becomes an ordinary quadratic.
Surd equation: \(\sqrt{-2x+4}-x=2\)Isolate the root, square both sides, then check every answer.
Simultaneous (one linear, one quadratic)Make \(y\) the subject of the linear one and substitute.
Zero is magic. If two things multiply to give zero, one of them must be zero. That is why we always get everything onto one side first: \(ax^2+bx+c=0\).
Quadratic formula\(x\)\({}=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\) Given on the formula sheet. Write \(a\), \(b\), \(c\) down first.
Discriminant \(\Delta=b^2-4ac\)\(\Delta\lt0\): no real roots \(\Delta=0\): equal real roots \(\Delta\gt0\) and a perfect square: rational roots \(\Delta\gt0\), not a square: irrational roots
Inequality signs\((x-a)(x-b)\gt0\) with \(a\lt b\): \(x\lt a\) or \(x\gt b\) \((x-a)(x-b)\lt0\): \(a\lt x\lt b\)
Exponential facts\(2^x\) is always positive, so \(2^x=-4\) has no solution. Same base: \(2^x=2^3\Rightarrow x=3\)
For \(8x^2-2x\gt0\): the parabola sits above the axis outside the roots.
“Greater than zero” means above the \(x\)-axis. “Less than zero” means below it. Sketch the happy or sad parabola and the answer draws itself.
Solve for \(x\): \(5x^2+2=-9x\) (correct to TWO decimal places)
Get everything on one side, equal to zero. Add \(9x\) to both sides: \(5x^2+9x+2=0\)The formula only works on the standard form.
Read off the values. \(a=5,\ b=9,\ c=2\)
Substitute into the formula. \[x=\frac{-9\pm\sqrt{9^2-4(5)(2)}}{2(5)}=\frac{-9\pm\sqrt{41}}{10}\]
Split into the + and the − answer. \(x\)\({}=\frac{-9+\sqrt{41}}{10}\)\({}=-0{,}26\) or \(x\)\({}=\frac{-9-\sqrt{41}}{10}\)\({}=-1{,}54\)
Answer: \(x=-0{,}26\) or \(x=-1{,}54\)
Solve for \(x\): \(8x^2\gt2x\)
Move everything to one side. \(8x^2-2x\gt0\)Never divide by \(x\). You don’t know if \(x\) is positive or negative, and you would lose half the answer.
Factorise. Common factor \(2x\): \(2x(4x-1)\gt0\)
Find the critical values (where it equals zero): \(x=0\) or \(x=\frac14\)
Sketch. The \(x^2\) term is positive, so the parabola smiles. It is above the axis (greater than zero) to the left of 0 and to the right of \(\frac14\).
Answer: \(x\lt0\) or \(x\gt\frac14\)
Solve for \(x\): \(2\cdot2^{2x}-9\cdot2^x+4=0\)
Spot the disguise. \(2^{2x}=(2^x)^2\). So this is a quadratic in \(2^x\).
Substitute \(k=2^x\): \(2k^2-9k+4=0\)
Factorise. \((2k-1)(k-4)=0\), so \(k=\frac12\) or \(k=4\)Check: \(2k\cdot k=2k^2\), and \(-8k-k=-9k\). ✓
Swap back to \(2^x\): \(2^x=\frac12=2^{-1}\) or \(2^x=4=2^2\)
Same base, so the exponents are equal. \(x=-1\) or \(x=2\)
Answer: \(x=-1\) or \(x=2\)
Solve for \(x\): \(\sqrt{-2x+4}-x=2\)
Get the root alone. \(\sqrt{-2x+4}=x+2\)
Square both sides. \(-2x+4=(x+2)^2=x^2+4x+4\)Square the whole right side as a bracket, not term by term.
Standard form and factorise. \(0\)\({}=x^2+6x\)\({}\Rightarrow x(x+6)\)\({}=0\), so \(x=0\) or \(x=-6\)
Check both in the original. \(x=0\): \(\sqrt4-0=2\) ✓. \(x=-6\): \(\sqrt{16}+6=10\neq2\) ✗Squaring can create fake answers. The check is worth a mark.
Answer: \(x=0\)
\(x\) is the sum of 2 and \(y\). Five times the product of \(x\) and \(y\) is 6 more than the square of \(x\).Calculate the values of \(x\) and \(y\).
Translate the words into maths. \(x=y+2\) and \(5xy=x^2+6\)
Substitute the linear one into the other. \(5y(y+2)=(y+2)^2+6\)
Expand. \(5y^2+10y=y^2+4y+4+6\)
Standard form. \(4y^2+6y-10=0\). Divide by 2: \(2y^2+3y-5=0\)
Factorise. \((2y+5)(y-1)=0\), so \(y=1\) or \(y=-\frac52\)
Find each matching \(x\) from \(x=y+2\): \(y=1\Rightarrow x=3\); \(y\)\({}=-\frac52\)\({}\Rightarrow x\)\({}=-\frac12\)
Answer: \(x=3,\ y=1\) or \(x=-\frac12,\ y=-\frac52\)
✗ Dividing \(8x^2\gt2x\) by \(x\) to get \(x\gt\frac14\). ✓ Factorise instead. You keep the \(x\lt0\) part.
✗ Forgetting to check surd answers. ✓ Substitute every answer back into the original equation.
✗ Writing \(2^x=-\frac12\) as an answer. ✓ \(2^x\) is never negative, so write “no solution” for that factor.
✗ Rounding \(\sqrt{41}\) to 6,4 halfway through. ✓ Keep the full value in your calculator; round only the final answer.
✗ Giving \(x\) without \(y\) in simultaneous equations. ✓ Every \(x\) needs its partner \(y\).
Nature of rootsWhat is the nature of the roots of \(2x^2-3x+5=0\)?
\(\Delta\)\({}=(-3)^2-4(2)(5)\)\({}=9-40\)\({}=-31\). A negative discriminant means no real roots: the parabola never touches the \(x\)-axis.
DBE Nov 2024 · Q1.1.3Solve \(x^2-2x-3\gt0\)
\((x-3)(x+1)\gt0\). Critical values \(-1\) and \(3\). Smiling parabola, so “greater than zero” is outside the roots.
DBE Nov 2024 · Q1.1.4Solve \(2^{2x}-2^{x+2}-32=0\)
\(2^{x+2}=4\cdot2^x\). Let \(k=2^x\): \(k^2-4k-32\)\({}=0\)\({}\Rightarrow(k-8)(k+4)\)\({}=0\). \(2^x=8\Rightarrow x=3\). \(2^x=-4\) has no solution.
Comfortable with all three quick checks? Mark it done and take a short break.
A pattern is a recipe for the next number. Your first job is always to identify the recipe by testing the gaps between terms.
First differences are equal (6; 10; 14 …)Arithmetic. \(d\) is the common difference.
Ratios are equal (8; −4; 2 …)Geometric. \(r=\frac{T_2}{T_1}\).
Second differences are equal (36; 71; 104 …)Quadratic. \(T_n=an^2+bn+c\).
A sequence is the list of terms. A series is the sum of the terms. \(T_n\) is one term; \(S_n\) is the total of the first \(n\) terms.
Think of a taxi rank. Arithmetic: the fare goes up by R5 each stop. Geometric: a rumour doubles every hour. Quadratic: the first differences themselves form an arithmetic pattern.
Sum to infinity\(S_\infty=\dfrac{a}{1-r}\) only if \(-1\lt r\lt1\) The terms shrink towards zero, so the total settles.
Quadratic\(2a=\) second difference \(3a+b=T_2-T_1\) \(a+b+c=T_1\)
Sigma notation\(\sum_{n=1}^{5}(2n+1)\)\({}=3+5+7+9+11\) Number of terms \(=\) top \(-\) bottom \(+1\)
Useful link\(T_n=S_n-S_{n-1}\)
A torpedo’s depth forms a quadratic pattern: 36 m after 1 s, 71 m after 2 s, 104 m after 3 s.Find the depth after 5 seconds, and show that \(T_n=-n^2+38n-1\).
First differences: \(71-36=35\), \(104-71=33\). Second difference: \(33-35=-2\)
Keep the pattern going. First differences drop by 2: 35; 33; 31; 29. So \(T_4=104+31=135\) and \(T_5=135+29=164\) m.
The \(t^2\) cancels: \(-18t=36\Rightarrow t=-2\) ✓
The terms are now 8; −4; 2. So \(a=8\), \(r=\frac{-4}{8}=-\frac12\)
\(T_{25}\): \(8\left(-\frac12\right)^{24}\)\({}=2^3\cdot2^{-24}\)\({}=2^{-21}\)An even power kills the minus sign. Write 8 as \(2^3\) so the bases match.
Sum to infinity (allowed because \(-1\lt-\frac12\lt1\)): \(S_\infty\)\({}=\dfrac{8}{1-(-\frac12)}\)\({}=\dfrac{8}{\frac32}\)\({}=\dfrac{16}{3}\)
Standard form: \(2k^2-3k-819\)\({}=0\)\({}\Rightarrow k\)\({}=\frac{3\pm\sqrt{9+6552}}{4}\)\({}=\frac{3\pm81}{4}\)
Choose the sensible one: \(k=21\) or \(k=-19{,}5\). \(k\) must be a whole number, so \(k=21\).
Answer: \(k=21\)
Clock A runs correctly from 09:00. Clock B measures half of clock A’s time in the first five minutes (2,5 min), and then, for every five minutes, measures 50% of what it measured in the previous five minutes.What does clock B show when clock A reaches 10:00? Will clock B ever reach 09:06?
Spot the pattern. Clock B’s five-minute amounts: 2,5; 1,25; 0,625 … Geometric with \(a=2{,}5\), \(r=\frac12\).
Count the terms. One hour on clock A is 12 five-minute blocks, so \(n=12\).
Add them: \(S_{12}\)\({}=\dfrac{2{,}5\left(1-(\frac12)^{12}\right)}{1-\frac12}\)\({}=4{,}9988\) minutes, just under 5 minutes.
So clock B shows about 09:05 (4 min 59,9 s after 09:00).
For 09:06, use the sum to infinity: \(S_\infty\)\({}=\dfrac{2{,}5}{1-\frac12}\)\({}=5\) minutes. Even if it runs forever, clock B never measures more than 5 minutes, so it never reaches 09:06.
Answer: About 09:05. It never reaches 09:06, because the total time is capped at \(S_\infty=5\) minutes.
✗ Counting \(\sum_{p=21}^{117}\) as 96 terms. ✓ \(117-21+1=97\) terms.
✗ Using \(S_\infty\) when \(r=2\). ✓ Only when \(-1\lt r\lt1\). State the condition for the mark.
✗ \(r=\frac{T_1}{T_2}\). ✓ \(r=\frac{T_2}{T_1}\): a term divided by the one before it.
✗ Answering \(n=-19{,}5\) for a position. ✓ \(n\) is a counting number. Reject negatives and fractions, and say why.
DBE Nov 2024 · Q2.1.1An arithmetic series has \(a=7\), \(d=5\) and 20 terms. What is its sum?
DBE Nov 2024 · Q3.1Circles are drawn inside each other: radius 6 cm, then each new circle has half the radius of the one before. What is the radius of the 3rd circle?
Geometric, \(a=6\), \(r=\frac12\): \(T_3\)\({}=6\left(\frac12\right)^2\)\({}=1{,}5\) cm.
SigmaHow many terms are in \(\sum_{n=100}^{150}(4n+2)\)?
Top − bottom + 1 \(=150-100+1=51\). (DBE May/June 2026 asked for this sum: \(S_{150}-S_{99}\)\({}=45\,600-19\,998\)\({}=25\,602\).)
Patterns are reliable marks. Most papers give 25 for this topic.
A function is a machine: put in \(x\), get out exactly one \(y\). A graph is just a picture of every input and output at once.
Each family has its own shape, and the numbers in the equation move that shape around. Play with the sliders below and watch what each letter does.
Inside the bracket, things go backwards: \((x-2)\) moves the graph 2 to the right. Outside the bracket, things go the way they look: \(+3\) moves it 3 up.
Hyperbola\(y=\dfrac{a}{x+p}+q\) Asymptotes: \(x=-p\) and \(y=q\) Axes of symmetry: \(y=(x+p)+q\) and \(y=-(x+p)+q\)
Straight line\(y=mx+c\). Find it from two points: gradient first, then substitute one point.
Reading two graphs\(f(x)\ge g(x)\): where \(f\) is on or above \(g\) Vertical distance \(=\) top graph \(-\) bottom graph
Domain and rangeDomain: the \(x\)-values allowed. Range: the \(y\)-values produced. Parabola with max \(q\): range \(y\le q\)
Intercepts\(y\)-intercept: put \(x=0\) \(x\)-intercepts: put \(y=0\) and solve
\(f(x)=ax^2+bx+c\) and \(g(x)=\frac{-4}{x-3}+8\). P is the turning point of \(f\) and also where the asymptotes of \(g\) meet. The graphs intersect at D(5 ; 6). M and T are the \(x\)-intercepts of \(f\) and \(g\).Show that \(f(x)=-\frac12x^2+3x+\frac72\), then calculate the length of MT.
Read P from \(g\). Asymptotes of \(g\): \(x=3\) and \(y=8\), so P(3 ; 8).
M: set \(f(x)=0\) and multiply by \(-2\): \(x^2-6x-7\)\({}=0\)\({}\Rightarrow(x-7)(x+1)\)\({}=0\). M is on the left, so M(−1 ; 0).
T: set \(g(x)=0\): \(\frac{-4}{x-3}\)\({}=-8\)\({}\Rightarrow x-3\)\({}=\frac12\)\({}\Rightarrow x\)\({}=3{,}5\). T(3,5 ; 0).
Length: \(MT=3{,}5-(-1)=4{,}5\) units
Answer: \(MT=4{,}5\) units
\(f(x)=\frac{a}{x+p}+q\) has domain \((-\infty;1)\cup(1;\infty)\), cuts the \(y\)-axis at (0 ; 1), and one axis of symmetry is \(g(x)=x-3\).Find \(p\), \(q\) and \(a\), then solve \(f(x)\ge0\).
The domain skips \(x=1\), so the vertical asymptote is \(x=1\). That means \(x+p=0\) at \(x=1\): \(p=-1\).
Axes of symmetry pass through the asymptotes’ meeting point. At \(x=1\): \(g(1)=1-3=-2\). So \(q=-2\).
Use (0 ; 1): \(1\)\({}=\frac{a}{0-1}-2\)\({}\Rightarrow3\)\({}=-a\)\({}\Rightarrow a\)\({}=-3\). So \(f(x)=\frac{-3}{x-1}-2\).
Find the \(x\)-intercept: \(0\)\({}=\frac{-3}{x-1}-2\)\({}\Rightarrow x-1\)\({}=-\frac32\)\({}\Rightarrow x\)\({}=-\frac12\)
Read the graph. \(f\) is on or above the \(x\)-axis from the intercept up to the asymptote (the asymptote itself is excluded).
Answer: \(p=-1\), \(q=-2\), \(a=-3\); \(f(x)\ge0\) for \(-\frac12\le x\lt1\)
\(f(x)=-x^2+4x+5\) and a straight line \(g\) meet at A (on the \(x\)-axis) and C(3 ; 8). EH is a vertical line with E on \(f\) and H on \(g\).Find the turning point B, show \(g(x)=2x+2\), and find the maximum length of EH where \(f\gt g\).
Find A: \(-x^2+4x+5\)\({}=0\)\({}\Rightarrow x^2-4x-5\)\({}=0\)\({}\Rightarrow(x-5)(x+1)\)\({}=0\). A is on the left: A(−1 ; 0).
Line through A and C: \(m=\frac{8-0}{3-(-1)}=2\); \(y=2(x+1)=2x+2\) ✓
Vertical distance = top − bottom: \(EH\)\({}=(-x^2+4x+5)-(2x+2)\)\({}=-x^2+2x+3\)
This is a sad parabola, so its maximum is at \(x=-\frac{2}{2(-1)}=1\): \(EH=-1+2+3=4\)
Answer: B(2 ; 9); maximum EH \(=4\) units
Given \(f(x)=\frac{-6}{x+2}-1\). Write down the domain, find the intercepts A (on the \(x\)-axis) and B (on the \(y\)-axis), and find the axis of symmetry with a positive gradient.
Domain: the denominator can’t be zero, so \(x\neq-2\) (\(x\in\mathbb{R}\), \(x\neq-2\)).
B (\(x=0\)): \(f(0)=\frac{-6}{2}-1=-4\). B(0 ; −4).
A (\(y=0\)): \(\frac{-6}{x+2}\)\({}=1\)\({}\Rightarrow x+2\)\({}=-6\)\({}\Rightarrow x\)\({}=-8\). A(−8 ; 0).
Axis of symmetry passes through (−2 ; −1) with gradient 1: \(y+1=x+2\Rightarrow h(x)=x+1\)
✗ \(y=\frac{3}{x+2}\) has asymptote \(x=2\). ✓ Set the denominator to zero: \(x=-2\).
✗ Including the asymptote in an interval. ✓ Use a round bracket or a strict sign at an asymptote: \(-\frac12\le x\lt1\).
✗ Vertical distance \(=g-f\) when \(f\) is on top. ✓ Always top minus bottom.
✗ Range of a sad parabola written as \(y\ge q\). ✓ Its highest point is \(q\), so \(y\le q\).
AsymptotesWhat are the asymptotes of \(y=\frac{3}{x-4}+1\)?
Vertical: \(x-4=0\Rightarrow x=4\). Horizontal: the number added at the end, \(y=1\).
Turning pointWhat is the turning point of \(f(x)=-x^2+4x+5\)?
\(x=-\frac{4}{2(-1)}=2\), and \(f(2)=-4+8+5=9\).
RangeWhat is the range of \(f(x)=-\frac12x^2+3x+\frac72\)? (Turning point (3 ; 8).)
\(a\lt0\), so the parabola frowns and 8 is the highest \(y\)-value.
Next up is topic 4: inverses, exponentials and logs, which finishes the functions marks.
If \(f\) turns 2 into 9, then \(f^{-1}\) turns 9 back into 2. Every point \((a;b)\) becomes \((b;a)\), so the graph is mirrored in the line \(y=x\).
The exponential graph \(y=b^x\) and the log graph \(y=\log_bx\) are inverses of each other. A log is simply the question “what power?”: \(\log_28=3\) because \(2^3=8\).
\(\log_b x=y\) means \(b^y=x\). Say it as: “log base \(b\) of \(x\) is the power I put on \(b\) to get \(x\).”
Finding \(f^{-1}\)1. Swap \(x\) and \(y\). 2. Make \(y\) the subject. 3. Write \(f^{-1}(x)=\dots\) or \(y=\dots\)
Log \(y=\log_bx\)Passes through (1 ; 0). Asymptote \(x=0\). Domain \(x\gt0\). Range: all real numbers.
Is the inverse a function?Only if \(f\) is one-to-one. A parabola is not, so restrict its domain to one side: \(x\ge0\) or \(x\le0\).
\(f(x)=\log_{\frac13}x\). A is the \(x\)-intercept and \((3;t)\) lies on \(f\). Find \(t\), A, \(f^{-1}\), its asymptote, and the \(y\)-values of \(h\) (which is \(f^{-1}\) shifted 5 units right) for \(x\gt4\).
\(t\): \(t=\log_{\frac13}3\). Ask: \((\frac13)^{?}=3\). The answer is \(-1\). So \(t=-1\).
A: every log graph passes through (1 ; 0), so A(1 ; 0).
Inverse: swap: \(x=\log_{\frac13}y\). Rewrite as a power: \(y=\left(\frac13\right)^x\)
Asymptote of \(f^{-1}\): \(y=0\)
Shift 5 right: \(h(x)\)\({}=\left(\frac13\right)^{x-5}\). It is decreasing, so for \(x\gt4\): \(h(x)\lt h(4)\)\({}=\left(\frac13\right)^{-1}\)\({}=3\), and it stays above the asymptote.
\(f(x)=a^x-1\), \(a\gt0\), passes through \(B\left(2;-\frac59\right)\). Find \(a\), the range of \(f\), and the image of C (on \(f\) at \(y=\frac{19}{8}\)) after reflection in \(y=x\).
Range: the asymptote is \(y=-1\), and \(a^x\) is always positive, so \(y\gt-1\).
Find C: \(\left(\frac23\right)^x-1\)\({}=\frac{19}{8}\)\({}\Rightarrow\left(\frac23\right)^x\)\({}=\frac{27}{8}\)\({}=\left(\frac32\right)^3\)\({}=\left(\frac23\right)^{-3}\), so \(x=-3\).
Reflect in \(y=x\): swap the coordinates. C\(\left(-3;\frac{19}{8}\right)\) becomes \(C'\left(\frac{19}{8};-3\right)\).
Given \(f(x)=-6x^2\). How would you restrict the domain so that \(f^{-1}\) is a function? Then find \(f^{-1}\) for \(f^{-1}(x)\le0\).
Restrict: a parabola fails the horizontal line test, so keep one half: \(x\le0\) (or \(x\ge0\)).
Swap: \(x=-6y^2\)
Make \(y\) the subject: \(y^2\)\({}=-\frac{x}{6}\)\({}\Rightarrow y\)\({}=\pm\sqrt{-\frac{x}{6}}\)
Pick the sign using \(f^{-1}(x)\le0\): take the negative root.
Answer: \(x\le0\); \(y=-\sqrt{-\frac{x}{6}}\) for \(x\le0\)
✗ \(f^{-1}(x)=\frac{1}{f(x)}\). ✓ The inverse undoes \(f\). It is not the reciprocal.
✗ Shifting 5 right as \(x+5\). ✓ Right is \(x-5\): replace \(x\) with \((x-5)\).
✗ Log graph domain “all real numbers”. ✓ You can only take the log of a positive number: \(x\gt0\).
InverseFind the inverse of \(y=2x+6\).
Swap: \(x=2y+6\). Then \(2y=x-6\), so \(y=\frac{x-6}{2}\).
LogsWhat is \(\log_2 8\)?
“What power on 2 gives 8?” \(2^3=8\), so 3.
Reflection(3 ; −1) lies on \(f\). Which point lies on \(f^{-1}\)?
The inverse swaps \(x\) and \(y\): \((a;b)\to(b;a)\).
With topics 3 and 4 done, you have covered the full 35 marks of functions.
Compound interest is interest on interest. Every period, the balance is multiplied by \((1+i)\). Everything else in this topic is bookkeeping around that one idea.
One amount, simple interest or straight-line depreciation\(A=P(1+ni)\) or \(A=P(1-ni)\)
One amount, compound growth or reducing-balance depreciation\(A=P(1+i)^n\) or \(A=P(1-i)^n\)
Equal payments that build up savings (a sinking fund, a retirement annuity)Future value \(F=\dfrac{x[(1+i)^n-1]}{i}\)
Equal payments that pay off a loan (a bond, a car loan)Present value \(P\)\({}=\dfrac{x[1-(1+i)^{-n}]}{i}\)
F is for the Future pile you are saving towards. P is for the Present money the bank lends you today. If the bank hands you money now, it’s P.
Converting the rate\(i=\dfrac{r}{100}\div\) periods per year Monthly: \(\div12\), quarterly: \(\div4\), half-yearly: \(\div2\)
Number of periods\(n=\) years \(\times\) periods per year 16 years quarterly: \(n=64\)
Payment at the start of each periodEach payment gets one extra period of interest: multiply \(F\) by \((1+i)\).
Draw a timeline for every annuity question. Mark “now”, each payment, and the date the question asks about. Counting payments on a timeline is where most marks are won.
Mary’s grandparents deposited R5 000 on the day she was born, at 6,8% p.a. compounded quarterly. How much is in the account on her 16th birthday?
Which formula? One lump sum, compound growth: \(A=P(1+i)^n\)
Convert: \(i=\frac{0{,}068}{4}=0{,}017\) and \(n=16\times4=64\)
After 4 years a printer was worth half its original value. Find the straight-line depreciation rate.
Straight-line: \(A=P(1-ni)\). Half the value means \(A=\frac12P\).
Substitute: \(\frac12P=P(1-4i)\). Divide by \(P\): \(\frac12\)\({}=1-4i\)\({}\Rightarrow i\)\({}=0{,}125\)
Answer: 12,5% per year
Sarah deposited R2 300 on 1 January 2020 and at the beginning of every quarter after that, at 5,8% p.a. compounded quarterly. Her last deposit was on 1 October 2025. How much was in the account on 1 January 2026?
Count the deposits on a timeline. Four per year for 2020 to 2025 is 6 years \(\times\) 4 \(=24\) deposits.
Rajesh got a loan of R900 000 on 28 February 2024 at 6,8% p.a. compounded monthly. He skipped the first three payments and paid R10 000 on 30 June 2024, then R10 000 at the end of each month. How long (in completed months) to repay the loan from when it was granted, and what is the final payment?
Rate: \(i=\frac{0{,}068}{12}\)
The loan grows for 3 months untouched (March, April, May): \(900\,000\left(1+\frac{0{,}068}{12}\right)^3\)\({}=R915\,386{,}86\)The annuity starts one month after this balance, so this is the “present value” of the payments.
Set up \(P\): \(915\,386{,}86\)\({}=\dfrac{10\,000\left[1-\left(1+\frac{0{,}068}{12}\right)^{-n}\right]}{\frac{0{,}068}{12}}\)
Solve for \(n\): \(\left(1+\frac{0{,}068}{12}\right)^{-n}\)\({}=1-\frac{915\,386{,}86\times0{,}068/12}{10\,000}\), so \(n=129{,}42\)
Interpret: 129 full payments plus one smaller payment = 130 payments. Add the 3 skipped months: \(3+130=133\) months.
Balance after 129 payments: \(915\,386{,}86(1+i)^{129}\)\(-\dfrac{10\,000[(1+i)^{129}-1]}{i}\)\(=R4\,173{,}56\)
Final payment is a month later, so add one month of interest: \(4\,173{,}56\left(1+\frac{0{,}068}{12}\right)\)\({}=R4\,197{,}21\)
Answer: 133 months; final payment R4 197,21
Tshepo borrowed R100 000 on 1 March 2022 at 13,5% p.a. compounded monthly, repaying R2 300,98 monthly for 5 years from 1 April 2022. (a) Find the total interest. (b) On 1 March 2024 he paid R22 300,98 (his instalment plus R20 000 extra), then carried on as normal. How many months earlier does he finish?
(a) Total paid minus loan: \(60\times2\,300{,}98-100\,000\)\({}=R38\,058{,}80\)
(b) Count payments to 1 March 2024: April 2022 to March 2024 is 24 payments.
Balance after 24 payments: \(100\,000(1+i)^{24}\)\(-\dfrac{2\,300{,}98[(1+i)^{24}-1]}{i}\)\(=R67\,805{,}20\), with \(i=\frac{0{,}135}{12}\)
Take off the extra R20 000: \(R47\,805{,}20\) still owed.
Payments still needed: \(47\,805{,}20\)\({}=\dfrac{2\,300{,}98[1-(1+i)^{-n}]}{i}\)\({}\Rightarrow n\)\({}=23{,}8\), so 24 more payments.
Compare: originally \(60-24=36\) payments were left. Now 24. \(36-24=12\).
Answer: (a) R38 058,80 (b) 12 months earlier
✗ Using 6,8 instead of 0,068, or forgetting to divide by 12. ✓ Write \(i\) as a fraction in your working: \(\frac{0{,}068}{12}\).
✗ Counting 23 deposits from January 2020 to October 2025. ✓ Use a timeline. Count both ends.
✗ Rounding \(n=129{,}42\) down to 129 months. ✓ The 0,42 is a real (smaller) payment, so it counts as a month.
✗ Rounding \(i\) to 0,0057. ✓ Keep \(\frac{0{,}068}{12}\) exact in the calculator. Early rounding changes the rand answer.
DBE Nov 2025 · Q7.1A holiday costs R40 000 now and rises by 7,8% per year. What will it cost in 5 years?
\(A\)\({}=40\,000(1{,}078)^5\)\({}=R58\,230{,}94\). R55 600 is simple interest, which is wrong for compound growth.
Effective rateWhat is the effective annual rate of 15% p.a. compounded monthly?
\(\left(1+\frac{0{,}15}{12}\right)^{12}-1\)\({}=0{,}1608\), so 16,08%.
Choose the formulaThandi takes out a car loan and repays it monthly. Which formula finds the loan amount?
The bank gives her money now and she pays it off. That is present value, P.
Finance is about 15 marks on Paper 1. Timelines make it predictable.
Imagine walking along a graph. The derivative tells you how steep the path is under your feet. At the top of a hill or the bottom of a valley, the path is flat: the derivative is zero.
That single idea answers most calculus questions: tangents (use the gradient), turning points (gradient \(=0\)), increasing and decreasing (gradient positive or negative), and real-life maximum or minimum problems.
Power rule: bring the power down, then take one off. \(x^5\to5x^4\). A constant on its own disappears, because a flat line has zero gradient.
First principles\(f'(x)\)\({}=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}{h}\) Given on the sheet. Keep writing “lim” until you substitute \(h=0\).
Rules\(\frac{d}{dx}(ax^n)\)\({}=anx^{n-1}\) \(\frac{d}{dx}(k)=0\) Differentiate term by term.
Tangent at \(x=a\)Point: \((a;f(a))\). Gradient: \(m=f'(a)\). Then \(y-y_1=m(x-x_1)\)
Cubic graphsTurning points: \(f'(x)=0\) Point of inflection: \(f''(x)=0\) Concave up: \(f''(x)\gt0\); concave down: \(f''(x)\lt0\)
Optimisation1. Write the quantity in one variable. 2. Differentiate, set equal to 0. 3. Solve and check it makes sense.
A cubic with \(a\gt0\): gradient positive, zero at the turning points, negative between them. The point of inflection sits halfway between the turning points.
Determine \(f'(x)\) from first principles if \(f(x)=x^2+2x\).
Find \(f(x+h)\) by replacing every \(x\) with \((x+h)\): \(f(x+h)\)\({}=(x+h)^2+2(x+h)\)\({}=x^2+2xh+h^2+2x+2h\)
Subtract \(f(x)\): \(f(x+h)-f(x)=2xh+h^2+2h\)Everything without an \(h\) cancels. If it doesn’t, check your expansion.
Write the limit: \(f'(x)\)\({}=\displaystyle\lim_{h\to0}\frac{2xh+h^2+2h}{h}\)
Factor out \(h\) and cancel: \(=\displaystyle\lim_{h\to0}\frac{h(2x+h+2)}{h}\)\({}=\lim_{h\to0}(2x+h+2)\)
Now let \(h=0\): \(f'(x)=2x+2\)
Answer: \(f'(x)=2x+2\)
Find (a) \(\dfrac{dy}{dx}\) if \(y=\dfrac{2x^4+1}{x^2}\) and (b) \(g'(x)\) if \(g(x)\)\({}=\dfrac{2}{x^2}-\sqrt[3]{x^7}\).
(a) Split the fraction over the single denominator: \(y\)\({}=\frac{2x^4}{x^2}+\frac{1}{x^2}\)\({}=2x^2+x^{-2}\)
(a) Differentiate: \(\frac{dy}{dx}=4x-2x^{-3}\)
(b) Rewrite as powers: \(g(x)=2x^{-2}-x^{\frac73}\)
Answer: E(5 ; −36); concave down for \(x\lt\frac83\)
\(h(x)=-x^3+px^2+qx+r\) cuts the \(x\)-axis at C(2 ; 0) and D(5 ; 0). C is also a turning point.Show that \(p=9\), \(q=-24\), \(r=20\). Find the point of inflection, and the tangent with an angle of inclination of 96,34° that touches \(h\) between B (the \(y\)-intercept) and C.
A turning point on the axis is a double root. So \(h(x)=-(x-2)^2(x-5)\)
Expand: \(-(x^2-4x+4)(x-5)\)\({}=-(x^3-9x^2+24x-20)\)\({}=-x^3+9x^2-24x+20\). So \(p=9\), \(q=-24\), \(r=20\) ✓
\(y\)-value: \(h(3)=-27+81-72+20=2\). Point of inflection (3 ; 2).
Gradient from the angle: \(m=\tan96{,}34^\circ=-9\)
Where is the gradient \(-9\)? \(-3x^2+18x-24\)\({}=-9\)\({}\Rightarrow x^2-6x+5\)\({}=0\)\({}\Rightarrow x\)\({}=1\) or \(x=5\). Between B and C means \(0\lt x\lt2\), so \(x=1\).
Point and line: \(h(1)=-1+9-24+20=4\). \(y-4\)\({}=-9(x-1)\)\({}\Rightarrow y\)\({}=-9x+13\)
Answer: Inflection (3 ; 2); tangent \(y=-9x+13\)
A rectangular sheet \(x\) by \(h\) (\(x\gt h\)) has a perimeter of 50 units. It is rolled into an open cylinder of height \(h\). Show that \(V\)\({}=\frac{25x^2}{4\pi}-\frac{x^3}{4\pi}\), and find the \(x\) that maximises the volume.
Choose: \(x=0\) gives no cylinder, so \(x=\frac{50}{3}=16{,}67\). Check: \(h=25-16{,}67=8{,}33\lt x\) ✓
Answer: \(x\)\({}=\frac{50}{3}\approx16{,}67\) units
✗ Dropping “lim” halfway through first principles. ✓ Write \(\lim_{h\to0}\) on every line until \(h\) is replaced by 0. Examiners take a mark for it.
✗ Substituting the \(x\) of a turning point into \(f'\) to get the \(y\)-value. ✓ \(y\)-values always come from the original \(f\).
✗ Differentiating \(\frac{2}{x^2}\) as \(\frac{2}{2x}\). ✓ Rewrite as \(2x^{-2}\) first.
✗ Using \(f'(x)\) notation when asked for \(\frac{dy}{dx}\). ✓ Answer in the notation the question uses.
DBE Nov 2024 · Q8.1.1\(\frac{d}{dx}\left[3x-5x^2\right]\)\({}=\)
\(3x\to3\) and \(-5x^2\to-10x\).
DBE Nov 2024 · Q10.1A cyclist’s speed is \(s'(t)=-3t^2+18t\) km/h. What is the maximum speed?
Maximum where the derivative of speed is zero: \(-6t+18=0\Rightarrow t=3\). Then \(s'(3)=-27+54=27\) km/h.
DBE Nov 2024 · Q9A cubic has turning points A(1 ; 9) and B\(\left(\frac52;8\right)\). Where is it concave up?
The point of inflection is halfway between the turning points: \(\frac{1+2{,}5}{2}=\frac74\). The minimum is on the right, so it is concave up to the right of \(\frac74\).
Calculus is 35 marks. First principles and the rules alone are usually worth 10 to 12 of them.
\(P(\text{event})\)\({}=\dfrac{\text{ways it can happen}}{\text{all possible outcomes}}\). The hard part is never the division. It is counting correctly, so draw a Venn diagram, a tree, or a row of boxes.
Overlapping groups (learners taking Maths, Tourism, Geography)Venn diagram. Fill the middle first.
Events happening one after the other (rain, then coffee)Tree diagram. Multiply along branches, add between branches.
Arranging things or building codesRow of boxes. Fill restricted boxes first, then multiply.
Along a branch, multiply (“and”). Between branches, add (“or”).
Addition rule\(P(A\text{ or }B)\)\({}=P(A)+P(B)-P(A\text{ and }B)\)
Mutually exclusiveCan’t happen together: \(P(A\text{ and }B)=0\)
IndependentOne doesn’t affect the other: \(P(A\text{ and }B)\)\({}=P(A)\times P(B)\)
Complement\(P(\text{not }A)=1-P(A)\). “At least one” \(=1-P(\text{none})\)
Arrangements\(n\) different objects: \(n!\) With repeats: \(\dfrac{n!}{a!\,b!}\) (e.g. BOOK: \(\frac{4!}{2!}=12\))
Kept togetherGlue them into one block, arrange the blocks, then arrange inside the block.
Every learner writes at least one of Maths (M, 22 learners), Tourism (T, 16), Geography (G, 18). 5 write M and T but not G; 4 write M and G but not T; 3 write T and G but not M; 6 write only T.Draw the Venn diagram, find P(at least two subjects), and test whether M and T are independent.
Let \(x\) = learners writing all three. Tourism total: \(6+5+3+x=16\Rightarrow x=2\)
Only M: \(22-5-4-2=11\). Only G: \(18-4-3-2=9\)
Total learners: \(11+5+4+2+6+3+9=40\)
At least two subjects: \(5+4+3+2=14\), so \(P\)\({}=\frac{14}{40}\)\({}=\frac{7}{20}\)
Independence test. \(P(M)=\frac{22}{40}\), \(P(T)=\frac{16}{40}\), so \(P(M)\times P(T)=0{,}22\)
Compare with the actual overlap: \(P(M\text{ and }T)\)\({}=\frac{5+2}{40}\)\({}=0{,}175\). Since \(0{,}175\neq0{,}22\), they are not independent.
Answer: \(\frac{7}{20}\); not independent
A code has the form letter, digit, letter, digit, using 26 letters and the digits 0–9.(a) How many codes if repeats are allowed? (b) How many if D, F, I, Q, U, V may not be used, the code may not start with W or Z, nothing repeats, and it ends in an odd digit? (c) What is the percentage increase if D, F, I, Q, U, V are allowed again?
(a) Four boxes: \(26\times10\times26\times10\)\({}=67\,600\)
(b) Restricted boxes first. Last digit is odd: 5 choices. First letter: \(26-6=20\) letters, minus W and Z \(=18\).
(b) Then the others. Second letter: 20 allowed letters minus the one used \(=19\). First digit: 10 minus the odd one used \(=9\).
At a kiosk, 120 people buy either coffee or water. The chance of rain is 75%. On a rainy day the chance of buying coffee is three times the chance on a dry day. Overall, P(coffee) \(=\frac{7}{12}\). How many cups are sold on a non-rainy day?
Let \(p\) = P(coffee on a dry day). Then P(coffee on a rainy day) \(=3p\).
Tree: rain then coffee \(=0{,}75\times3p\); dry then coffee \(=0{,}25\times p\)
Add the branches: \(2{,}25p+0{,}25p\)\({}=\frac{7}{12}\)\({}\Rightarrow2{,}5p\)\({}=\frac{7}{12}\)\({}\Rightarrow p\)\({}=\frac{7}{30}\)
On a dry day: \(120\times\frac{7}{30}=28\) cups
Answer: 28 cups
Eight runners race (no ties), including Andrew and Bongi. (a) In how many ways can they finish if Bongi finishes immediately after Andrew? (b) What is the probability that two or more runners finish after Andrew and before Bongi?
(a) Glue “Andrew, Bongi” into one block. Now there are 7 things to arrange: \(7!=5\,040\)
(b) Total outcomes: \(8!=40\,320\)
(b) Count positions. If exactly \(k\) runners are between them (Andrew first), the pair can sit in \(7-k\) ways along the 8 places.
(b) Two or more between: \(k=2,3,4,5,6\) gives \(5+4+3+2+1=15\) position pairs. The other 6 runners fill the rest in \(6!=720\) ways.
✗ Thinking independent means mutually exclusive. ✓ Independent: multiply. Mutually exclusive: the overlap is zero. They are different tests.
✗ Filling the Venn diagram from the outside. ✓ Start with the middle (all three), then the pairs, then the singles.
✗ Saying “not independent” without numbers. ✓ Show \(P(A)\times P(B)\) and \(P(A\text{ and }B)\) and compare them.
DBE May/June 2026 · Q10.1.1A wardrobe has 15 ties, 4 pairs of trousers and 16 shirts. How many outfits (tie, trousers, shirt)?
Counting principle: \(15\times4\times16=960\).
IndependenceA and B are independent, \(P(A)=0{,}4\) and \(P(B)=0{,}5\). Find \(P(A\text{ and }B)\).
Independent, so multiply: \(0{,}4\times0{,}5=0{,}2\).
ArrangementsHow many different arrangements of the letters in BOOK are there?
4 letters with O repeated twice: \(\frac{4!}{2!}=12\).
That’s Paper 1 covered. Take a proper break before Paper 2.
Stop 4 · Paper 2
Paper 2: statistics, geometry and trigonometry.
Five topics. Trigonometry is worth a third of the paper, so give topics 10 and 11 the most sprints.
Questions 1 and 2 of Paper 2 are the friendliest marks in matric maths. You need to know which calculator buttons to press and what the answers mean in words.
CentreMean \(\bar{x}\), median (middle value), mode
SpreadRange, interquartile range (IQR \(=Q_3-Q_1\)), standard deviation \(\sigma\)
Trend between two variablesScatter plot, regression line \(\hat{y}=a+bx\), correlation \(r\)
The mean gets dragged towards the tail. Mean bigger than median means a long tail to the right: positively skewed.
Calculator (Casio fx-82ZA PLUS)MODE → STAT → A+BX. Enter \(x\) and \(y\) columns. AC, then SHIFT 1 → Reg for A, B and r; SHIFT 1 → Var for \(\bar{x}\) and \(\sigma x\). Other models: look for STAT and “linear regression”.
Regression line\(\hat{y}=a+bx\). \(b\) is the gradient: the change in \(y\) for each extra unit of \(x\).
Correlation \(r\)Between −1 and 1. Close to ±1: strong. Close to 0: no linear relationship. Sign gives the direction.
Standard deviationUse \(\sigma\) (population), not \(s\). “Within one standard deviation” means between \(\bar{x}-\sigma\) and \(\bar{x}+\sigma\).
OgivePlot cumulative frequency at the upper boundary of each class. Start at the lower boundary of the first class with 0.
Five-number summaryMinimum, \(Q_1\), median, \(Q_3\), maximum. This is what a box-and-whisker plot draws.
Ages (years) and prices (R) of 11 cars of one model: (2; 293 000), (3; 265 000), (3; 256 000), (4; 219 000), (4; 241 000), (4; 246 000), (6; 226 000), (6; 176 000), (7; 154 000), (7; 180 000), (8; 148 000).Find the least squares regression line, predict the price of a 5-year-old car, decide whether the prediction is valid, and state the average yearly decrease.
Enter the data in STAT mode (A+BX): age as \(x\), price as \(y\).
Read off: \(a=331\,397{,}20\) and \(b=-22\,988{,}32\). So \(\hat{y}\)\({}=331\,397{,}20-22\,988{,}32x\)
Predict for \(x=5\): \(\hat{y}\)\({}=331\,397{,}20-22\,988{,}32(5)\)\({}=R216\,455{,}60\)
Check validity with \(r\): \(r=-0{,}95\). This is a very strong negative correlation, and 5 years lies inside the data range (2 to 8), so the prediction is reliable.
Meaning of \(b\): for each extra year of age, the price drops by about R22 988,32.
Answer: \(\hat{y}\)\({}=331\,397{,}20-22\,988{,}32x\); about R216 455,60; valid (\(r=-0{,}95\)); about R22 988,32 per year
A basketball team of 9 players averaged 12 points each. Eight players scored 11, 14, 19, 20, 8, 10, 2, 14. How many players scored outside one standard deviation of the mean?
Find the missing score. Total \(=9\times12=108\). The eight scores add to 98, so the ninth player scored \(108-98=10\).
Standard deviation of all nine (calculator, \(\sigma x\)): \(\sigma=5{,}23\)
One SD either side: \(12-5{,}23=6{,}77\) and \(12+5{,}23=17{,}23\)
Count scores outside 6,77 to 17,23: 2, 19 and 20.
Answer: 3 players
A box-and-whisker plot has minimum \((x-2)\%\), \(Q_1=\frac{3x-3}{2}\%\), median \((2x+5)\%\), \(Q_3=(3x)\%\) and maximum 91%. The IQR is 45%. What is the minimum percentage of learners who passed?
Use the IQR: \(3x-\frac{3x-3}{2}=45\). Multiply by 2: \(6x-3x+3=90\Rightarrow x=29\)
Five-number summary: 27; 42; 63; 87; 91
Read the quartiles. \(Q_1=42\%\) is above the pass mark, so at least the top 75% of learners passed.
Answer: At least 75% passed
Cumulative frequencies for time on a website: up to 20 min: 16; up to 40: 40; up to 60: 59; up to 80: 67; up to 100: 70.How many people visited, and how many spent more than 40 and up to 80 minutes?
Total: the last cumulative frequency is everyone: 70 people.
Between 40 and 80: \(67-40=27\) peopleSubtract cumulative totals to get the count inside an interval.
Answer: 70 visitors; 27 people
✗ Using \(sx\) (sample) on the calculator. ✓ Matric uses \(\sigma x\), the population standard deviation.
✗ Plotting the ogive at class midpoints. ✓ Upper boundaries, with a starting point of 0.
✗ Writing \(\hat{y}=bx+a\) with \(a\) and \(b\) swapped. ✓ \(a\) is the intercept, \(b\) the gradient.
✗ Predicting far outside the data. ✓ Predictions are only reliable inside the data range and when \(r\) is strong.
SkewnessA data set has mean 60 and median 52. How is it skewed?
The mean is pulled towards the long tail. Mean \(\gt\) median, so the tail is on the right.
CorrelationWhat does \(r=-0{,}95\) describe?
Close to −1: points lie close to a line sloping downwards.
IQR\(Q_1=42\) and \(Q_3=87\). What is the interquartile range?
\(Q_3-Q_1=87-42=45\).
About 20 quick marks. Practise the calculator steps until they are automatic.
Analytical geometry is shapes drawn on a grid, so every length and angle can be calculated from coordinates. Make a rough sketch for every question and label everything you find.
Your formula sheet already has distance, midpoint, gradient, the line equations, \(m=\tan\theta\) and the circle. What you must bring yourself: parallel lines have equal gradients, perpendicular gradients multiply to −1, and a tangent is perpendicular to the radius.
Inclination\(\tan\theta=m\). If \(m\lt0\), add 180° to the calculator angle: \(m\)\({}=-1\)\({}\Rightarrow\theta\)\({}=135^\circ\)
Circle\((x-a)^2+(y-b)^2=r^2\), centre \((a;b)\), radius \(r\) From \(x^2+y^2+\dots\): complete the square.
Tangent to a circleGradient of radius first, then flip and change sign. Use the point of contact.
Parallelogram tricksDiagonals bisect each other: \(S=P+R-Q\) (add and subtract coordinates).
P(−1 ; 8), Q(−4 ; −6) and R(12 ; 2) are the vertices of △PQR. The inclination of QR is \(\theta\). Find QR (simplified surd), the gradient of QR, \(\theta\), and the equation of QR.
Area = base × perpendicular height: \(8\sqrt5\times5\sqrt5\)\({}=40\times5\)\({}=200\) square units
Answer: S(15 ; 16); T(4 ; −2); area 200 units²
M(\(p\) ; \(q\)) is the centre of a circle cutting the \(x\)-axis at A and B. MC is parallel to the \(y\)-axis and cuts the \(x\)-axis at D and the circle at E(−6 ; −1). Tangents from C touch the circle at A and B. AD \(=(q-1)\).Find \(p\), show \(q=4\), write the circle’s equation, and find A, B, the tangent BC, point C and \(\hat{ACB}\).
\(p\): MC is vertical and passes through E, so \(p=-6\).
Radius two ways. ME goes from \(y=q\) down to \(y=-1\): \(r=q+1\). Also MD \(=q\) and AD \(=q-1\).
Pythagoras in △MDA: \((q+1)^2\)\({}=q^2+(q-1)^2\)\({}\Rightarrow q^2+2q+1\)\({}=2q^2-2q+1\)\({}\Rightarrow q^2-4q\)\({}=0\), so \(q=4\) ✓ (\(q\neq0\))
Circle: centre (−6 ; 4), \(r=5\): \((x+6)^2+(y-4)^2=25\)
A and B (\(y=0\)): \((x+6)^2+16\)\({}=25\)\({}\Rightarrow x+6\)\({}=\pm3\). A(−9 ; 0), B(−3 ; 0)
Tangent at B: \(m_{MB}\)\({}=\frac{4-0}{-6+3}\)\({}=-\frac43\), so \(m_{BC}=\frac34\). \(y\)\({}=\frac34(x+3)\)\({}=\frac34x+\frac94\)
C is on \(x=-6\): \(y\)\({}=\frac34(-6)+\frac94\)\({}=-\frac94\). C\(\left(-6;-\frac94\right)\)
Angle at C. In right-angled △BDC: \(\tan\hat{BCD}\)\({}=\frac{BD}{CD}\)\({}=\frac{3}{9/4}\)\({}=\frac43\)\({}\Rightarrow\hat{BCD}\)\({}=53{,}13^\circ\). By symmetry \(\hat{ACB}\)\({}=2\times53{,}13^\circ\).
Find the centre and radius of \(x^2-4x+y^2-8y-5=0\).
Group \(x\)’s and \(y\)’s, move the constant: \(x^2-4x+y^2-8y=5\)
Add (half the coefficient)² for each: \((\frac{-4}{2})^2=4\) and \((\frac{-8}{2})^2=16\), to both sides.
Factorise: \((x-2)^2+(y-4)^2=5+4+16=25\)
Answer: Centre (2 ; 4), radius 5
✗ \((x+6)^2\) means the centre has \(x=6\). ✓ The sign flips: centre \(x=-6\).
✗ Radius \(=25\) from \(=25\). ✓ The right side is \(r^2\), so \(r=5\).
✗ Inclination of \(m=-1\) given as −45°. ✓ Inclination is between 0° and 180°: \(-45^\circ+180^\circ\)\({}=135^\circ\).
✗ Using the gradient of the radius for the tangent. ✓ Flip it and change the sign.
CircleCentre and radius of \((x+6)^2+(y-4)^2=25\)?
Flip the signs inside the brackets for the centre; square-root 25 for the radius.
PerpendicularGradient of a line perpendicular to \(y=\frac12x-4\)?
\(\frac12\times(-2)=-1\). Flip and change sign.
InclinationA line has gradient −1. What is its angle of inclination?
Calculator gives −45°. Add 180° to land between 0° and 180°.
Analytical geometry is about 40 marks and very predictable. Sketch everything.
Trig ratios repeat every 360°. So \(\sin210^\circ\) is really \(\sin30^\circ\) wearing a minus sign. Reduction formulas are the rules for taking off the disguise.
CAST: which ratios are positive in each quadrant, and how each quadrant’s angle relates to the acute angle \(\theta\).
Start in quadrant 4 and go anticlockwise: Cos, All, Sin, Tan. “All Students Take Calculus” works too, starting in quadrant 1.
Keep the ratio\(180^\circ\pm\theta\), \(360^\circ-\theta\), \(-\theta\): same ratio, sign from CAST \(\sin(180^\circ+x)=-\sin x\)
Switch the ratio\(90^\circ\pm\theta\): sin ↔ cos, sign from CAST \(\cos(90^\circ+x)=-\sin x\), \(\sin(90^\circ-x)=\cos x\)
Identities to memorise\(\tan x\)\({}=\dfrac{\sin x}{\cos x}\) \(\sin^2x+\cos^2x=1\) Compound and double angles are on the formula sheet.
General solution\(\sin x=k\): \(x\)\({}=\text{ref}+k\cdot360^\circ\) or \(180^\circ-\text{ref}+k\cdot360^\circ\) \(\cos x=k\): \(x\)\({}=\pm\text{ref}+k\cdot360^\circ\) \(\tan x=k\): \(x\)\({}=\text{ref}+k\cdot180^\circ\) (\(k\in\mathbb{Z}\))
Proving identitiesWork on one side only. Change everything to sin and cos, then factorise or use \(\sin^2x+\cos^2x=1\).
Without a calculator, simplify \(\cos330^\circ\cos365^\circ+\sin175^\circ\sin210^\circ\) to a single ratio. If this expression equals \(p\), find \(\sin70^\circ\) in terms of \(p\).
Reduce each angle. \(\cos330^\circ\)\({}=\cos(360^\circ-30^\circ)\)\({}=\cos30^\circ\); \(\cos365^\circ=\cos5^\circ\)
Given \(\tan50^\circ=k\), express in terms of \(k\): (a) \(\cos40^\circ\) (b) \(\dfrac{2\sin25^\circ\cos25^\circ}{-2+4\sin^225^\circ}\) (c) \(\sin10^\circ\)
Draw a triangle for 50°. \(\tan50^\circ=\frac{k}{1}\): opposite \(k\), adjacent 1, hypotenuse \(\sqrt{1+k^2}\). So \(\sin50^\circ\)\({}=\frac{k}{\sqrt{1+k^2}}\), \(\cos50^\circ\)\({}=\frac{1}{\sqrt{1+k^2}}\)
Difference of cubes: \[\cos^3x-\sin^3x=(\cos x-\sin x)(\cos^2x+\sin x\cos x+\sin^2x)\]
Use \(\sin^2x+\cos^2x=1\): \(=(\cos x-\sin x)(1+\sin x\cos x)\)
Match the right side’s form by taking a minus out of the top and putting it in the bottom: \(\dfrac{(\cos x-\sin x)(1+\sin x\cos x)}{\cos x}\)\({}=\dfrac{(\sin x-\cos x)(1+\sin x\cos x)}{-\cos x}\)\({}=\) RHS ✓
Answer: LHS = RHS
Without a calculator, solve for \(x\in[0^\circ;360^\circ]\): \[\sqrt3\cos x=\sin(50^\circ+x)\cos(10^\circ-x)+\cos(50^\circ+x)\sin(10^\circ-x)\]
The right side is \(\sin A\cos B+\cos A\sin B\)\({}=\sin(A+B)\) with \(A=50^\circ+x\), \(B=10^\circ-x\).
Add the angles: \(A+B=60^\circ\), so the right side is \(\sin60^\circ\)\({}=\frac{\sqrt3}{2}\)
Cos is positive in quadrants 1 and 4: \(x=60^\circ\) or \(x\)\({}=360^\circ-60^\circ\)\({}=300^\circ\)
Answer: \(x=60^\circ\) or \(x=300^\circ\)
\(\sin^2x\); \(\cos^2x\); \(\frac12\sin2x\) are the first three terms of an arithmetic sequence with a non-zero common difference. Find the general solution for \(x\).
Arithmetic means equal differences: \(\cos^2x-\sin^2x\)\({}=\frac12\sin2x-\cos^2x\)
Use \(\frac12\sin2x=\sin x\cos x\) and collect: \(2\cos^2x-\sin x\cos x-\sin^2x\)\({}=0\)
Factorise like a trinomial: \((2\cos x+\sin x)(\cos x-\sin x)\)\({}=0\)
Reject \(\cos x=\sin x\): then \(d=\cos^2x-\sin^2x=0\), which is not allowed.
So \(\sin x\)\({}=-2\cos x\)\({}\Rightarrow\tan x\)\({}=-2\). Reference angle \(63{,}43^\circ\); tan is negative in quadrant 2: \(180^\circ-63{,}43^\circ\)\({}=116{,}57^\circ\)
Answer: \(x\)\({}=221{,}81^\circ+k\cdot360^\circ\) or \(x\)\({}=318{,}19^\circ+k\cdot360^\circ,\ k\in\mathbb{Z}\)
✗ Using \(+k\cdot360^\circ\) for tan. ✓ Tan repeats every 180°: \(+k\cdot180^\circ\).
✗ Dividing both sides by \(\cos x\). ✓ Factorise instead, or you can lose solutions where \(\cos x=0\).
✗ Working on both sides of an identity at once. ✓ Simplify one side until it matches the other.
✗ Forgetting “\(k\in\mathbb{Z}\)”. ✓ It is part of a general solution and carries a mark.
Reduction\(\sin(180^\circ+x)=\)
180° keeps the ratio. Quadrant 3 is the T quadrant, so sin is negative.
Co-ratios\(\cos(90^\circ+x)=\)
90° switches cos to sin. \(90^\circ+x\) is in quadrant 2, where cos is negative.
General solutionGeneral solution of \(\cos x=0{,}5\)?
Cos: plus or minus the reference angle, every 360°.
This is the biggest block on Paper 2. Revisit the CAST diagram until you can draw it from memory.
\(y=a\sin bx\) is a wave. \(a\) sets how tall it is (the amplitude) and \(b\) sets how squashed it is (the period). Use the sliders to see both.
Period: 360° divided by \(b\) for sin and cos, 180° divided by \(b\) for tan. So \(\cos2x\) repeats every 180°.
Graph facts\(y=a\sin bx+q\): amplitude \(|a|\), period \(\frac{360^\circ}{b}\), range \([q-|a|;q+|a|]\) \(y=\sin(x+p)\): shifts left by \(p\)
Tan\(y=\tan bx\): period \(\frac{180^\circ}{b}\), asymptotes where \(\cos bx=0\)
Sine rule\(\dfrac{a}{\sin A}\)\({}=\dfrac{b}{\sin B}\)\({}=\dfrac{c}{\sin C}\) Use when you know a side and its opposite angle.
Cosine rule\(a^2=b^2+c^2-2bc\cos A\) Use for two sides and the angle between them, or all three sides.
Area rule\(\text{Area}=\frac12ab\sin C\) (two sides and the angle between them)
3D strategyFind the right-angled triangles first (vertical poles, walls). Solve those with SOH CAH TOA, then carry the side into the slanted triangle.
Calculator in DEG mode, every time. A “D” must show at the top of the screen.
\(f(x)=\cos2x\) for \(x\in[-135^\circ;135^\circ]\). Write down the period of \(f\). If \(f\) is translated 45° left to form \(h\), find \(h\) in simplest form and its range.
Period: \(\frac{360^\circ}{2}\)\({}=180^\circ\)
45° left means replace \(x\) with \((x+45^\circ)\): \(h(x)\)\({}=\cos2(x+45^\circ)\)\({}=\cos(2x+90^\circ)\)
Reduce: \(\cos(90^\circ+2x)=-\sin2x\)
Range: a shift left doesn’t change the height, so \(-1\le y\le1\)
Answer: 180°; \(h(x)=-\sin2x\); \(y\in[-1;1]\)
\(f(x)=2\sin(x-45^\circ)\), \(x\in[-180^\circ;180^\circ]\). Write down the amplitude, and find where \(f'(x)\ge0\).
Amplitude: \(|a|=2\)
\(f'(x)\ge0\) means \(f\) is increasing (or flat). Find the turning points.
Minimum where \(x-45^\circ\)\({}=-90^\circ\)\({}\Rightarrow x\)\({}=-45^\circ\). Maximum where \(x-45^\circ\)\({}=90^\circ\)\({}\Rightarrow x\)\({}=135^\circ\).
A puppet stage has a horizontal base BKLC with \(\hat{KBC}=104^\circ\) and \(\hat{BKC}=52{,}6^\circ\). The vertical rectangle ABCD has area 648 cm² and AB : BC = 1 : 2.Show AB = 18 cm, then find AC and KC. If AB = BK, find \(\hat{KAC}\).
Area with the ratio: let AB \(=x\), BC \(=2x\). \(x\cdot2x\)\({}=648\)\({}\Rightarrow x^2\)\({}=324\)\({}\Rightarrow x\)\({}=18\) ✓
AC (right angle at B in the vertical rectangle): \(AC\)\({}=\sqrt{18^2+36^2}\)\({}=\sqrt{1620}\)\({}=40{,}25\) cm
KC in △BKC (sine rule): \(\dfrac{KC}{\sin104^\circ}\)\({}=\dfrac{36}{\sin52{,}6^\circ}\)\({}\Rightarrow KC\)\({}=43{,}97\) cm
AK: AB is vertical, so △ABK has a right angle at B. With BK = 18: \(AK=\sqrt{18^2+18^2}=25{,}46\) cm
Cosine rule in △AKC for the angle at A: \(\cos\hat{KAC}\)\({}=\dfrac{AK^2+AC^2-KC^2}{2\cdot AK\cdot AC}\)\({}=\dfrac{648+1620-1933{,}39}{2(25{,}46)(40{,}25)}\)
Answer: AC = 40,25 cm; KC = 43,97 cm; \(\hat{KAC}=80{,}60^\circ\)
AB is a vertical flagpole, 4 units tall. D, B and C are on level ground. \(\hat{BAC}=\hat{BCD}=\theta\) and \(\hat{BDC}=\alpha\).Show that BC \(=4\tan\theta\), and that \(DC\)\({}=4\sin\theta\left(1+\dfrac{\tan\theta}{\tan\alpha}\right)\).
△ABC is right-angled at B (the pole is vertical): \(\tan\theta\)\({}=\frac{BC}{AB}\)\({}=\frac{BC}{4}\)\({}\Rightarrow BC\)\({}=4\tan\theta\) ✓
In △BDC, the third angle: \(\hat{DBC}\)\({}=180^\circ-(\theta+\alpha)\), so \(\sin\hat{DBC}\)\({}=\sin(\theta+\alpha)\)
Sine rule: \(\dfrac{DC}{\sin(\theta+\alpha)}\)\({}=\dfrac{BC}{\sin\alpha}\)\({}\Rightarrow DC\)\({}=\dfrac{4\tan\theta\sin(\theta+\alpha)}{\sin\alpha}\)
Expand the compound angle: \(DC\)\({}=\dfrac{4\tan\theta(\sin\theta\cos\alpha+\cos\theta\sin\alpha)}{\sin\alpha}\)\({}=4\tan\theta\left(\dfrac{\sin\theta}{\tan\alpha}+\cos\theta\right)\)
Take out \(\sin\theta\) using \(\tan\theta\cos\theta\)\({}=\sin\theta\): \(DC\)\({}=4\sin\theta\left(\dfrac{\tan\theta}{\tan\alpha}+1\right)\) ✓
Answer: Both results shown
✗ Calculator in RAD mode. ✓ Check for the “D” before every trig question.
✗ Using the sine rule with the wrong opposite angle. ✓ Each side sits opposite its own angle. Label them on the sketch first.
✗ Shifting \(\cos(x+45^\circ)\) to the right. ✓ \(+45^\circ\) inside the bracket moves the graph left.
PeriodWhat is the period of \(y=\tan2x\)?
Tan’s basic period is 180°. Divide by \(b=2\): 90°.
AmplitudeWhat is the amplitude of \(f(x)=2\sin(x-45^\circ)\)?
The number in front of sin: \(|a|=2\). The 45° only shifts it sideways.
Which rule?You know two sides, 5 and 8, and the 60° angle between them. Which rule finds the third side?
Two sides and the included angle: cosine rule. \(c^2=25+64-80\cos60^\circ=49\), so \(c=7\).
Topics 10 and 11 together cover the full 50 trig marks.
You don’t need to be “good at geometry”. You need the theorem list and a habit: mark every angle you know on the diagram, then ask which theorem links it to the one you want.
The theorems as flash cards: tap to flip
The proofs you can be asked to write out in full: angle at the centre, opposite angles of a cyclic quad, the tan-chord theorem, the proportionality theorem, and similar triangles. Learn them like a song: same diagram, same lines, every time.
Before you calculateCopy the given angles onto the diagram. Look for: radii (isosceles triangles), diameters (90°), tangents (90° and tan-chord), parallel lines (alternate and corresponding angles).
Writing a statementOne statement per line, with its reason in brackets beside it. Name angles clearly: \(\hat{O}_2\) or \(\hat{LOF}\).
Ratio questionsLook for the parallel lines. Each parallel line gives you one proportion. Use a “\(k\)” to turn ratios into lengths.
Area ratiosSimilar triangles: area ratio = (side ratio)². Same height: area ratio = base ratio.
Prove that the angle subtended by a chord at the centre is twice the angle at the circumference: prove \(\hat{TOP}=2\hat{TKP}\).
Construction: draw KO and produce it to R. Label \(\hat{K}_1=\hat{TKO}\) and \(\hat{K}_2=\hat{OKP}\).
\(OT=OK\) (radii), so \(\hat{T}=\hat{K}_1\) (∠s opp equal sides)
\(\hat{O}_1\)\({}=\hat{T}+\hat{K}_1\)\({}=2\hat{K}_1\) (ext ∠ of Δ), where \(\hat{O}_1=\hat{TOR}\)
Similarly, in △KOP: \(\hat{O}_2=2\hat{K}_2\), where \(\hat{O}_2=\hat{ROP}\)
PQRS is a cyclic quadrilateral. T is on the circle with QT ⊥ SR at V. \(\hat{Q}_2\)\({}=\hat{RQT}\)\({}=35^\circ\) and \(\hat{R}=\hat{S}_1\) (\(\hat{S}_1=\hat{PSR}\)). Find \(\hat{QTS}\), prove PQ ∥ SR, and prove PT is a diameter.
In △QVR: \(\hat{V}=90^\circ\) (given), so \(\hat{QRV}\)\({}=180^\circ-90^\circ-35^\circ\)\({}=55^\circ\) (sum of ∠s in Δ)
\(\hat{QTS}\)\({}=\hat{QRS}\)\({}=55^\circ\) (∠s in same seg, chord QS)
PQ ∥ SR: \(\hat{QPS}+\hat{R}=180^\circ\) (opp ∠s of cyclic quad)
But \(\hat{R}=\hat{S}_1\) (given), so \(\hat{QPS}+\hat{S}_1\)\({}=180^\circ\). These are co-interior angles, so PQ ∥ SR (co-int ∠s supplementary).
PT diameter: QT ⊥ SR and PQ ∥ SR, so \(\hat{PQT}=90^\circ\) (co-int ∠s, PQ ∥ SR)
A chord that subtends 90° at the circumference is a diameter, so PT is a diameter (converse ∠ in semi-circle).
Answer: \(\hat{QTS}=55^\circ\); both results proved
In △ABC, BA is produced to E. F and D lie on BC with AD ∥ EF. AC and EF meet at G. \(\frac{CF}{FB}=\frac25\) and \(\frac{CG}{GA}=\frac32\). Find \(\frac{FD}{CF}\), \(\frac{BA}{EA}\) and \(\frac{\text{Area }\triangle GCF}{\text{Area GFDA}}\).
In △CAD, GF ∥ AD: \(\frac{CF}{FD}\)\({}=\frac{CG}{GA}\)\({}=\frac32\) (line ∥ one side of Δ), so \(\frac{FD}{CF}=\frac23\)
Turn ratios into lengths. \(CF:FD=3:2\) and \(CF:FB=2:5\). Make CF the same in both: CF \(=6k\), FD \(=4k\), FB \(=15k\).
So \(DB=FB-FD=15k-4k=11k\)
In △BEF, AD ∥ EF: \(\frac{BA}{AE}\)\({}=\frac{BD}{DF}\)\({}=\frac{11k}{4k}\)\({}=\frac{11}{4}\) (line ∥ one side of Δ)
△CGF ||| △CAD (∠∠∠, GF ∥ AD) with side ratio \(\frac{CF}{CD}\)\({}=\frac{6k}{10k}\)\({}=\frac35\)
Area ratio = side ratio squared: \(\triangle CGF\)\({}=\frac{9}{25}\triangle CAD\), so GFDA \(=\frac{16}{25}\triangle CAD\)
Geometrythe 12 theorem cards in topic 12, with their reasons
Stop 5 · Exam-day game plan
Three hours, 150 marks, one calm plan.
Most marks are lost to panic and poor timing, not to hard questions. Follow the eight steps, and use the 14-day sprint to get there.
The night before
Calculator with a fresh battery, set to DEG. Two pens, a pencil, ruler and eraser. Sleep beats one more hour of revision.
Read the whole paper first
Use the reading time to tick the questions you recognise. Starting with a win settles your nerves.
Budget 1,2 minutes per mark
A 10-mark question gets about 12 minutes. Write the finishing time for each question in the margin.
Take the friendly marks early
Paper 1 Questions 1–2 and Paper 2 Questions 1–2 are usually the most accessible. You may answer in any order, but number every answer exactly as in the paper.
Show every line
“Answers only will NOT necessarily be awarded full marks.” Write the formula, substitute, then answer. Method marks survive a wrong final number.
Stuck for two minutes? Move on
Leave a gap of half a page and come back later. Fresh eyes often see the trick.
Round only at the end
Two decimal places unless told otherwise. Add units: R, cm, °, units².
Last ten minutes: check
Signs, substitution checks for equations, and a reason next to every geometry statement.
The 14-day sprint
About three 15-minute sprints a day. Tick a day off when it’s done. Your ticks are saved on this device.
Practise with the real papers
Every worked example on this page comes from these Department of Basic Education papers. Download them with their memoranda from the DBE website and write at least one of each under exam conditions.
BIEMANN
