diff --git a/.agents/skills/ofb/SKILL.md b/.agents/skills/ofb/SKILL.md
index 26150a0..a276e0e 100644
--- a/.agents/skills/ofb/SKILL.md
+++ b/.agents/skills/ofb/SKILL.md
@@ -50,6 +50,7 @@ bun run build # production: JS + Zola + pagefind → public/
| Chapter URLs | `/{subjectId}/{chapter-slug}/` (e.g. `/math/quadratic-equations/`) |
| Map | Mermaid DAG view (`?view=tree`) driven by `graph.edges` |
| Graph | `graph.edges` must be **acyclic**; builder throws on cycles |
+| Chart axes (all widgets) | Every coordinate chart must render **x/y axis labels, positive-direction arrowheads, and visible graduations (ticks + scale labels)**. Apply this consistently across all subjects/widgets. |
| Accessibility | Avoid decorative characters that may render as emoji on mobile. Prefer broadly supported symbols/text so desktop and mobile stay visually consistent. |
## Common tasks → where to edit
@@ -130,6 +131,8 @@ Live chapters live under `content/{subject}/{slug}/`:
**Checkpoints (all subjects):** use `book-question`, `book-question__prompt`, `book-question__solution` (styled in theme `_book.scss`). Never use `book-question__answer`.
+**Charts (all subjects):** coordinate charts in chapter widgets must include axis labels, arrowheads, and graduation/tick scale by default. Do not ship unlabeled axes.
+
**Math checkpoints / dots:** follow [math-chapter-patterns.md](math-chapter-patterns.md); shared dot classes in `content/math/subject.scss`.
Reference: `content/math/quadratic-equations/`.
diff --git a/content/math/complex-numbers/_index.md b/content/math/complex-numbers/_index.md
new file mode 100644
index 0000000..5ee41ca
--- /dev/null
+++ b/content/math/complex-numbers/_index.md
@@ -0,0 +1,228 @@
++++
+title = "Complex numbers"
+description = "Represent complex numbers in Cartesian form and perform operations including conjugates and division."
+template = "chapter.html"
+weight = 14
+[extra]
+subject = "math"
+chapter_id = "complex-numbers"
+strand = "Number & Algebra"
++++
+
+
+
+ Complex numbers extend the real number system so equations like
+ \(x^2+1=0\) can be solved. In high school, you usually write a complex number in
+ Cartesian form:
+ \[
+ z = a + bi,
+ \]
+ where \(a,b\in\mathbb{R}\) and \(i^2=-1\).
+
+
+
The imaginary unit \(i\)
+
+
Definition
+
\[ i^2 = -1 \]
+
+
+ Useful cycle:
+ \[
+ i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1,
+ \]
+ and then the pattern repeats every 4 powers.
+
+
+
Real and imaginary parts
+
+ For \(z=a+bi\):
+
+
+
\(\Re(z)=a\) (real part)
+
\(\Im(z)=b\) (imaginary coefficient)
+
+
Example. If \(z=3-4i\), then \(\Re(z)=3\), \(\Im(z)=-4\).
+ Plot \(z=a+bi\) as the point \((a,b)\): horizontal axis is real part, vertical axis is
+ imaginary part. This lets you interpret \(|z|\) as geometric distance.
+
+
+
Why this matters: solving quadratics
+
+ If \(x^2+4=0\), then \(x^2=-4\), so
+ \[
+ x=\pm 2i.
+ \]
+ Complex numbers ensure every quadratic has two roots (counting multiplicity), and later this
+ extends to all polynomial equations.
+
+
+
+
+
History
+
+ Complex numbers first appeared when mathematicians tried to solve cubic equations in the
+ 16th century. Algebraic formulas sometimes produced square roots of negative numbers, even when
+ final answers were real.
+
+
+ Mathematicians like Cardano, Bombelli, and later
+ Euler developed rules for these numbers. What began as a formal trick became a
+ major mathematical system used in engineering, physics, and signal processing.
+
+
+
Derivation
+
Why the conjugate removes \(i\) in denominators
+
+ For \(c+di\):
+ \[
+ (c+di)(c-di)=c^2-d^2i^2=c^2+d^2,
+ \]
+ which is real and positive (unless \(c=d=0\)). This is why multiplying by conjugates converts
+ complex division into standard real-number division.
+
+
+
Why \(|z|^2=z\overline{z}\)
+
+ For \(z=a+bi\):
+ \[
+ z\overline{z}=(a+bi)(a-bi)=a^2+b^2=|z|^2.
+ \]
+ So modulus is directly linked to conjugation and the Pythagorean distance in the Argand plane.
+
+
+
Checkpoints
+
+ Pause and reason before continuing. Discuss with a classmate or write your reasoning in a
+ notebook.
+
+
+
+
Checkpoint
+
+ Simplify \(i^{23}\) using the 4-step power cycle.
+
+
+
+
+
Checkpoint
+
+ Compute \((4-3i)+(2+5i)\) and \((4-3i)-(2+5i)\).
+
+
+
+
+
Checkpoint
+
+ Expand and simplify \((1+2i)(3-i)\) into \(a+bi\) form.
+
+
+
+
+
Checkpoint
+
+ Find \(\overline{z}\) and \(|z|\) when \(z=-5+12i\).
+
+
+
+
+
Checkpoint
+
+ Write \(\dfrac{2-i}{1+3i}\) in \(a+bi\) form.
+
+
+
+
+
Checkpoint
+
+ Solve \(x^2-6x+13=0\). Why are the roots complex?
+
+
+
+
Applications
+
AC circuits and impedance
+
+ Electrical engineers model alternating-current circuits with complex numbers because amplitude
+ and phase are naturally represented together as one complex quantity.
+
+
+
Waves and signals
+
+ Sound, radio, and image-processing systems use complex numbers to describe oscillations and
+ frequency content efficiently (e.g., Fourier methods).
+
+
+
Geometry and rotations
+
+ Multiplying by certain complex numbers corresponds to rotating and scaling points in the plane,
+ making complex numbers useful in geometry and computer graphics.
+
+
+
References
+
+
Bombelli, R. (1572). L'Algebra (early systematic rules for imaginary numbers).
+
Euler, L. (18th century works) connecting complex numbers with trigonometry and exponentials.
+
Standard high-school extension algebra curricula on Cartesian form, conjugates, and division.
+
+
diff --git a/content/math/complex-numbers/core.html b/content/math/complex-numbers/core.html
new file mode 100644
index 0000000..5e69379
--- /dev/null
+++ b/content/math/complex-numbers/core.html
@@ -0,0 +1,109 @@
+
+
+ Complex numbers extend the real number system so equations like
+ \(x^2+1=0\) can be solved. In high school, you usually write a complex number in
+ Cartesian form:
+ \[
+ z = a + bi,
+ \]
+ where \(a,b\in\mathbb{R}\) and \(i^2=-1\).
+
+
+
The imaginary unit \(i\)
+
+
Definition
+
\[ i^2 = -1 \]
+
+
+ Useful cycle:
+ \[
+ i^1=i,\quad i^2=-1,\quad i^3=-i,\quad i^4=1,
+ \]
+ and then the pattern repeats every 4 powers.
+
+
+
Real and imaginary parts
+
+ For \(z=a+bi\):
+
+
+
\(\Re(z)=a\) (real part)
+
\(\Im(z)=b\) (imaginary coefficient)
+
+
Example. If \(z=3-4i\), then \(\Re(z)=3\), \(\Im(z)=-4\).
+ Plot \(z=a+bi\) as the point \((a,b)\): horizontal axis is real part, vertical axis is
+ imaginary part. This lets you interpret \(|z|\) as geometric distance.
+
+
+
Why this matters: solving quadratics
+
+ If \(x^2+4=0\), then \(x^2=-4\), so
+ \[
+ x=\pm 2i.
+ \]
+ Complex numbers ensure every quadratic has two roots (counting multiplicity), and later this
+ extends to all polynomial equations.
+
+
diff --git a/content/math/complex-numbers/supplement.html b/content/math/complex-numbers/supplement.html
new file mode 100644
index 0000000..548d636
--- /dev/null
+++ b/content/math/complex-numbers/supplement.html
@@ -0,0 +1,107 @@
+
+
History
+
+ Complex numbers first appeared when mathematicians tried to solve cubic equations in the
+ 16th century. Algebraic formulas sometimes produced square roots of negative numbers, even when
+ final answers were real.
+
+
+ Mathematicians like Cardano, Bombelli, and later
+ Euler developed rules for these numbers. What began as a formal trick became a
+ major mathematical system used in engineering, physics, and signal processing.
+
+
+
Derivation
+
Why the conjugate removes \(i\) in denominators
+
+ For \(c+di\):
+ \[
+ (c+di)(c-di)=c^2-d^2i^2=c^2+d^2,
+ \]
+ which is real and positive (unless \(c=d=0\)). This is why multiplying by conjugates converts
+ complex division into standard real-number division.
+
+
+
Why \(|z|^2=z\overline{z}\)
+
+ For \(z=a+bi\):
+ \[
+ z\overline{z}=(a+bi)(a-bi)=a^2+b^2=|z|^2.
+ \]
+ So modulus is directly linked to conjugation and the Pythagorean distance in the Argand plane.
+
+
+
Checkpoints
+
+ Pause and reason before continuing. Discuss with a classmate or write your reasoning in a
+ notebook.
+
+
+
+
Checkpoint
+
+ Simplify \(i^{23}\) using the 4-step power cycle.
+
+
+
+
+
Checkpoint
+
+ Compute \((4-3i)+(2+5i)\) and \((4-3i)-(2+5i)\).
+
+
+
+
+
Checkpoint
+
+ Expand and simplify \((1+2i)(3-i)\) into \(a+bi\) form.
+
+
+
+
+
Checkpoint
+
+ Find \(\overline{z}\) and \(|z|\) when \(z=-5+12i\).
+
+
+
+
+
Checkpoint
+
+ Write \(\dfrac{2-i}{1+3i}\) in \(a+bi\) form.
+
+
+
+
+
Checkpoint
+
+ Solve \(x^2-6x+13=0\). Why are the roots complex?
+
+
+
+
Applications
+
AC circuits and impedance
+
+ Electrical engineers model alternating-current circuits with complex numbers because amplitude
+ and phase are naturally represented together as one complex quantity.
+
+
+
Waves and signals
+
+ Sound, radio, and image-processing systems use complex numbers to describe oscillations and
+ frequency content efficiently (e.g., Fourier methods).
+
+
+
Geometry and rotations
+
+ Multiplying by certain complex numbers corresponds to rotating and scaling points in the plane,
+ making complex numbers useful in geometry and computer graphics.
+
+
+
References
+
+
Bombelli, R. (1572). L'Algebra (early systematic rules for imaginary numbers).
+
Euler, L. (18th century works) connecting complex numbers with trigonometry and exponentials.
+
Standard high-school extension algebra curricula on Cartesian form, conjugates, and division.
+ Exponential and logarithmic functions are inverse ideas. They model fast growth, rapid decay,
+ and scale changes that linear models cannot describe well.
+
+
+
Exponential basics
+
+ An exponential function has variable in the exponent:
+ \[
+ y=a^x,\quad a>0,\ a\ne1.
+ \]
+
+
+ If \(a>1\), the function grows as \(x\) increases. If \(0<a<1\), it decays.
+
+ A common model:
+ \[
+ A(t)=A_0(1+r)^t
+ \]
+ for growth (\(r>0\)) or decay (\(r<0\), with \(1+r>0\)).
+
+
+ Continuous model:
+ \[
+ A(t)=A_0e^{kt},
+ \]
+ where \(k>0\) gives growth and \(k<0\) gives decay.
+
+
+
Exam strategy
+
+
State domain restrictions first in log equations.
+
When combining logs, check arguments stay positive.
+
Use exact forms before calculator rounding.
+
For modeling, define variables and units clearly.
+
+
+
+
+
History
+
+ Logarithms were developed in the early 17th century (notably by John Napier) to simplify long
+ multiplication and division into addition and subtraction. Before electronic calculators, this was
+ a major speed boost for astronomy, navigation, and engineering.
+
+
+ Exponential functions became central in science through models of population growth, radioactive
+ decay, and continuously changing quantities.
+
+
+
Derivation and reasoning
+
Why log product law works
+
+ Let \(\log_a M=p\) and \(\log_a N=q\). Then \(M=a^p\), \(N=a^q\), so
+ \[
+ MN=a^{p+q}.
+ \]
+ Taking \(\log_a\) gives:
+ \[
+ \log_a(MN)=p+q=\log_a M+\log_a N.
+ \]
+
+
+
Inverse relationship view
+
+ If \(y=a^x\), swapping \(x,y\) gives \(x=a^y\), so \(y=\log_a x\). That is why exponential and
+ logarithmic graphs are mirror images across \(y=x\).
+
+
+
Checkpoints
+
+
+ Evaluate \(\log_3 81\) and \(\log_{10}0.01\).
+
+
Answer: \(x+1=\log_2 7\), so \(x=\log_2 7-1 \approx 1.807\).
+
+
+
+
+
+ Solve \(\log_5(x-4)=2\) and state domain condition.
+
+
+
Domain: \(x-4>0\Rightarrow x>4\).
+
Answer: \(x-4=25\Rightarrow x=29\), valid.
+
+
+
+
+
+ A population starts at 500 and grows 6% per year. Write the model and find \(A(4)\).
+
+
+
Model: \(A(t)=500(1.06)^t\).
+
Value: \(A(4)=500(1.06)^4\approx 631.24\), about 631.
+
+
+
+
+
+ A substance follows \(A(t)=120e^{-0.18t}\). Is this growth or decay? Why?
+
+
+
Answer: Decay, because exponent coefficient is negative (\(k=-0.18\)).
+
+
+
+
Applications
+
Compound interest
+
+ Savings and loans often follow exponential rules. Logarithms are used to solve for unknown time
+ or interest rate.
+
+
+
Half-life and decay
+
+ Radioactive substances and medicine concentration in the body are modeled with exponential decay.
+
+
+
Logarithmic scales
+
+ pH, decibels, and Richter magnitude use logarithmic scales to represent very large ratio ranges
+ in manageable numbers.
+
+
+
References
+
+
Standard high-school curricula on exponential and logarithmic functions (IGCSE, IB, A-Level, AP).
+
Precalculus and algebra references on log laws, inverse functions, and modeling.
+
+
diff --git a/content/math/exponential-logarithmic/core.html b/content/math/exponential-logarithmic/core.html
new file mode 100644
index 0000000..1b5609b
--- /dev/null
+++ b/content/math/exponential-logarithmic/core.html
@@ -0,0 +1,138 @@
+
+
+ Exponential and logarithmic functions are inverse ideas. They model fast growth, rapid decay,
+ and scale changes that linear models cannot describe well.
+
+
+
Exponential basics
+
+ An exponential function has variable in the exponent:
+ \[
+ y=a^x,\quad a>0,\ a\ne1.
+ \]
+
+
+ If \(a>1\), the function grows as \(x\) increases. If \(0<a<1\), it decays.
+
+ A common model:
+ \[
+ A(t)=A_0(1+r)^t
+ \]
+ for growth (\(r>0\)) or decay (\(r<0\), with \(1+r>0\)).
+
+
+ Continuous model:
+ \[
+ A(t)=A_0e^{kt},
+ \]
+ where \(k>0\) gives growth and \(k<0\) gives decay.
+
+
+
Exam strategy
+
+
State domain restrictions first in log equations.
+
When combining logs, check arguments stay positive.
+
Use exact forms before calculator rounding.
+
For modeling, define variables and units clearly.
+
+
diff --git a/content/math/exponential-logarithmic/supplement.html b/content/math/exponential-logarithmic/supplement.html
new file mode 100644
index 0000000..16de7a1
--- /dev/null
+++ b/content/math/exponential-logarithmic/supplement.html
@@ -0,0 +1,112 @@
+
+
History
+
+ Logarithms were developed in the early 17th century (notably by John Napier) to simplify long
+ multiplication and division into addition and subtraction. Before electronic calculators, this was
+ a major speed boost for astronomy, navigation, and engineering.
+
+
+ Exponential functions became central in science through models of population growth, radioactive
+ decay, and continuously changing quantities.
+
+
+
Derivation and reasoning
+
Why log product law works
+
+ Let \(\log_a M=p\) and \(\log_a N=q\). Then \(M=a^p\), \(N=a^q\), so
+ \[
+ MN=a^{p+q}.
+ \]
+ Taking \(\log_a\) gives:
+ \[
+ \log_a(MN)=p+q=\log_a M+\log_a N.
+ \]
+
+
+
Inverse relationship view
+
+ If \(y=a^x\), swapping \(x,y\) gives \(x=a^y\), so \(y=\log_a x\). That is why exponential and
+ logarithmic graphs are mirror images across \(y=x\).
+
+
+
Checkpoints
+
+
+ Evaluate \(\log_3 81\) and \(\log_{10}0.01\).
+
+
Answer: \(x+1=\log_2 7\), so \(x=\log_2 7-1 \approx 1.807\).
+
+
+
+
+
+ Solve \(\log_5(x-4)=2\) and state domain condition.
+
+
+
Domain: \(x-4>0\Rightarrow x>4\).
+
Answer: \(x-4=25\Rightarrow x=29\), valid.
+
+
+
+
+
+ A population starts at 500 and grows 6% per year. Write the model and find \(A(4)\).
+
+
+
Model: \(A(t)=500(1.06)^t\).
+
Value: \(A(4)=500(1.06)^4\approx 631.24\), about 631.
+
+
+
+
+
+ A substance follows \(A(t)=120e^{-0.18t}\). Is this growth or decay? Why?
+
+
+
Answer: Decay, because exponent coefficient is negative (\(k=-0.18\)).
+
+
+
+
Applications
+
Compound interest
+
+ Savings and loans often follow exponential rules. Logarithms are used to solve for unknown time
+ or interest rate.
+
+
+
Half-life and decay
+
+ Radioactive substances and medicine concentration in the body are modeled with exponential decay.
+
+
+
Logarithmic scales
+
+ pH, decibels, and Richter magnitude use logarithmic scales to represent very large ratio ranges
+ in manageable numbers.
+
+
+
References
+
+
Standard high-school curricula on exponential and logarithmic functions (IGCSE, IB, A-Level, AP).
+
Precalculus and algebra references on log laws, inverse functions, and modeling.
+
+
diff --git a/content/math/factor-remainder-theorems/_index.md b/content/math/factor-remainder-theorems/_index.md
new file mode 100644
index 0000000..095fe6b
--- /dev/null
+++ b/content/math/factor-remainder-theorems/_index.md
@@ -0,0 +1,226 @@
++++
+title = "Factor & remainder theorems"
+description = "Use the factor and remainder theorems to factorise higher-degree polynomials and solve related equations."
+template = "chapter.html"
+weight = 13
+[extra]
+subject = "math"
+chapter_id = "factor-remainder-theorems"
+strand = "Number & Algebra"
++++
+
+
+
+ This chapter introduces two connected ideas for polynomial division:
+ Remainder Theorem and Factor Theorem.
+ They help you test factors quickly, find roots, and break a cubic (or higher) polynomial into
+ simpler factors.
+
+
+
Quick recap
+
+ A polynomial in \(x\) is an expression like
+ \(P(x)=2x^3-3x^2+5x-7\). If you divide \(P(x)\) by \((x-a)\), you get:
+
+
+ \[
+ P(x) = (x-a)Q(x) + R,
+ \]
+
+
+ where \(Q(x)\) is the quotient and \(R\) is a constant remainder.
+
+
+
Remainder Theorem
+
+
Remainder Theorem
+
+ \[
+ \text{When } P(x)\text{ is divided by }(x-a),\ \text{remainder}=P(a).
+ \]
+
+
+
+
Example 1. Find the remainder when \(P(x)=x^3-4x+1\) is divided by \((x-2)\).
+
+ By the theorem, remainder \(=P(2)=2^3-4(2)+1=8-8+1=1\).
+
+
+
Example 2. Find the remainder when \(2x^3+x^2-5\) is divided by \((x+1)\).
+
+ \((x+1)=(x-(-1))\), so use \(a=-1\):
+ \(P(-1)=2(-1)^3+(-1)^2-5=-2+1-5=-6\).
+
+
+
Factor Theorem
+
+
Factor Theorem
+
+ \[
+ (x-a)\text{ is a factor of }P(x)\iff P(a)=0.
+ \]
+
+
+
+
+ This is a direct consequence of the Remainder Theorem: factor means remainder \(0\).
+
+
+
Example 3. Is \((x-3)\) a factor of \(P(x)=x^3-4x^2-3x+18\)?
+
+ \(P(3)=27-36-9+18=0\), so yes, \((x-3)\) is a factor.
+
+
+
Finding factors and roots (high-school workflow)
+
+
List possible rational roots using factors of constant term (and leading coefficient).
+
Test candidates with \(P(a)\).
+
When \(P(a)=0\), divide by \((x-a)\) to reduce degree.
+
Factor the smaller polynomial and solve for all roots.
+
+
+
Example 4. Solve \(x^3-6x^2+11x-6=0\).
+
+ Try integer candidates \(\pm1,\pm2,\pm3,\pm6\). We get \(P(1)=0\), so \((x-1)\) is a factor.
+ Dividing gives \(x^2-5x+6\), which factorises to \((x-2)(x-3)\).
+ Hence:
+
+
+ \[
+ x^3-6x^2+11x-6=(x-1)(x-2)(x-3),
+ \]
+
+
so roots are \(x=1,2,3\).
+
+
Synthetic division (quick table method)
+
+ Synthetic division is a compact way to divide by \((x-a)\). It is fast for checking many
+ candidates in exam settings.
+
+
Example 5. Divide \(x^3-6x^2+11x-6\) by \((x-1)\):
Always rewrite \((x+b)\) as \((x-(-b))\) before substituting.
+
If remainder is not zero, that candidate is not a factor.
+
After finding one factor, reduce degree before continuing.
+
Check final roots by substitution when time permits.
+
+
+
+
+
History
+
+ Polynomial equations became central in algebra from the Renaissance onward, when mathematicians
+ studied systematic methods for solving cubic and quartic equations. As symbolic algebra matured,
+ methods for division and factor testing became standard tools.
+
+
+ What we now call the factor and remainder theorems are part of the broader
+ polynomial division framework developed in 18th–19th century algebra texts.
+ They remain important in modern high-school courses because they connect substitution,
+ division, and root-finding in one clear structure.
+
+
+
Derivation
+
From the polynomial division algorithm
+
+ For any polynomial \(P(x)\), dividing by \((x-a)\) gives:
+ \[
+ P(x)=(x-a)Q(x)+R.
+ \]
+ Now substitute \(x=a\):
+ \[
+ P(a)=(a-a)Q(a)+R=R.
+ \]
+ So the remainder equals \(P(a)\), proving the Remainder Theorem.
+
+
+ If \((x-a)\) is a factor, remainder \(R=0\), so \(P(a)=0\). Conversely, if \(P(a)=0\), then
+ remainder is zero, so \((x-a)\) divides \(P(x)\). This proves the Factor Theorem.
+
+
+
Checkpoints
+
+ Pause and reason before continuing. Discuss with a classmate or write your reasoning in a
+ notebook.
+
+
+
+
Checkpoint
+
+ Find the remainder when \(P(x)=3x^3-2x+7\) is divided by \((x-2)\).
+
+
+
+
+
Checkpoint
+
+ Is \((x+2)\) a factor of \(x^3+4x^2+5x+2\)? Show the one-line test.
+
+
+
+
+
Checkpoint
+
+ Given \(P(x)=2x^3+kx^2-8x-4\), if \((x-2)\) is a factor, find \(k\).
+
+
+
+
+
Checkpoint
+
+ A polynomial leaves remainder \(5\) when divided by \((x-1)\). What is \(P(1)\)?
+
+
+
+
+
Checkpoint
+
+ If \(P(3)=0\) and \(P(-1)=0\) for a cubic \(P(x)\), what two linear factors must appear?
+
+
+
+
+
Checkpoint
+
+ Why is synthetic division especially useful after you already know one root?
+
+
+
+
Applications
+
Graph intercepts and curve sketching
+
+ Solving \(P(x)=0\) gives x-intercepts of polynomial graphs. Factor Theorem helps find those
+ intercepts quickly, then sketch end behavior and turning points more accurately.
+
+
+
Constraint values in models
+
+ In applied models, a condition like “output is zero at \(x=a\)” means \(P(a)=0\), so
+ \((x-a)\) must be a factor. This appears in physics calibration, optimization constraints, and
+ coding of polynomial approximations.
+
+
+
Gateway to numerical methods
+
+ Understanding factor tests and polynomial reduction supports later methods such as Newton’s
+ method and polynomial interpolation, where roots and remainders are central ideas.
+
+
+
References
+
+
Standard high-school algebra curricula (IB, A-Level, DSE extension) on polynomial theorems.
+
Classical algebra texts covering polynomial division and root-factor relationships.
+
+
diff --git a/content/math/factor-remainder-theorems/core.html b/content/math/factor-remainder-theorems/core.html
new file mode 100644
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--- /dev/null
+++ b/content/math/factor-remainder-theorems/core.html
@@ -0,0 +1,108 @@
+
+
+ This chapter introduces two connected ideas for polynomial division:
+ Remainder Theorem and Factor Theorem.
+ They help you test factors quickly, find roots, and break a cubic (or higher) polynomial into
+ simpler factors.
+
+
+
Quick recap
+
+ A polynomial in \(x\) is an expression like
+ \(P(x)=2x^3-3x^2+5x-7\). If you divide \(P(x)\) by \((x-a)\), you get:
+
+
+ \[
+ P(x) = (x-a)Q(x) + R,
+ \]
+
+
+ where \(Q(x)\) is the quotient and \(R\) is a constant remainder.
+
+
+
Remainder Theorem
+
+
Remainder Theorem
+
+ \[
+ \text{When } P(x)\text{ is divided by }(x-a),\ \text{remainder}=P(a).
+ \]
+
+
+
+
Example 1. Find the remainder when \(P(x)=x^3-4x+1\) is divided by \((x-2)\).
+
+ By the theorem, remainder \(=P(2)=2^3-4(2)+1=8-8+1=1\).
+
+
+
Example 2. Find the remainder when \(2x^3+x^2-5\) is divided by \((x+1)\).
+
+ \((x+1)=(x-(-1))\), so use \(a=-1\):
+ \(P(-1)=2(-1)^3+(-1)^2-5=-2+1-5=-6\).
+
+
+
Factor Theorem
+
+
Factor Theorem
+
+ \[
+ (x-a)\text{ is a factor of }P(x)\iff P(a)=0.
+ \]
+
+
+
+
+ This is a direct consequence of the Remainder Theorem: factor means remainder \(0\).
+
+
+
Example 3. Is \((x-3)\) a factor of \(P(x)=x^3-4x^2-3x+18\)?
+
+ \(P(3)=27-36-9+18=0\), so yes, \((x-3)\) is a factor.
+
+
+
Finding factors and roots (high-school workflow)
+
+
List possible rational roots using factors of constant term (and leading coefficient).
+
Test candidates with \(P(a)\).
+
When \(P(a)=0\), divide by \((x-a)\) to reduce degree.
+
Factor the smaller polynomial and solve for all roots.
+
+
+
Example 4. Solve \(x^3-6x^2+11x-6=0\).
+
+ Try integer candidates \(\pm1,\pm2,\pm3,\pm6\). We get \(P(1)=0\), so \((x-1)\) is a factor.
+ Dividing gives \(x^2-5x+6\), which factorises to \((x-2)(x-3)\).
+ Hence:
+
+
+ \[
+ x^3-6x^2+11x-6=(x-1)(x-2)(x-3),
+ \]
+
+
so roots are \(x=1,2,3\).
+
+
Synthetic division (quick table method)
+
+ Synthetic division is a compact way to divide by \((x-a)\). It is fast for checking many
+ candidates in exam settings.
+
+
Example 5. Divide \(x^3-6x^2+11x-6\) by \((x-1)\):
Always rewrite \((x+b)\) as \((x-(-b))\) before substituting.
+
If remainder is not zero, that candidate is not a factor.
+
After finding one factor, reduce degree before continuing.
+
Check final roots by substitution when time permits.
+
+
diff --git a/content/math/factor-remainder-theorems/supplement.html b/content/math/factor-remainder-theorems/supplement.html
new file mode 100644
index 0000000..0c6a27d
--- /dev/null
+++ b/content/math/factor-remainder-theorems/supplement.html
@@ -0,0 +1,106 @@
+
+
History
+
+ Polynomial equations became central in algebra from the Renaissance onward, when mathematicians
+ studied systematic methods for solving cubic and quartic equations. As symbolic algebra matured,
+ methods for division and factor testing became standard tools.
+
+
+ What we now call the factor and remainder theorems are part of the broader
+ polynomial division framework developed in 18th–19th century algebra texts.
+ They remain important in modern high-school courses because they connect substitution,
+ division, and root-finding in one clear structure.
+
+
+
Derivation
+
From the polynomial division algorithm
+
+ For any polynomial \(P(x)\), dividing by \((x-a)\) gives:
+ \[
+ P(x)=(x-a)Q(x)+R.
+ \]
+ Now substitute \(x=a\):
+ \[
+ P(a)=(a-a)Q(a)+R=R.
+ \]
+ So the remainder equals \(P(a)\), proving the Remainder Theorem.
+
+
+ If \((x-a)\) is a factor, remainder \(R=0\), so \(P(a)=0\). Conversely, if \(P(a)=0\), then
+ remainder is zero, so \((x-a)\) divides \(P(x)\). This proves the Factor Theorem.
+
+
+
Checkpoints
+
+ Pause and reason before continuing. Discuss with a classmate or write your reasoning in a
+ notebook.
+
+
+
+
Checkpoint
+
+ Find the remainder when \(P(x)=3x^3-2x+7\) is divided by \((x-2)\).
+
+
+
+
+
Checkpoint
+
+ Is \((x+2)\) a factor of \(x^3+4x^2+5x+2\)? Show the one-line test.
+
+
+
+
+
Checkpoint
+
+ Given \(P(x)=2x^3+kx^2-8x-4\), if \((x-2)\) is a factor, find \(k\).
+
+
+
+
+
Checkpoint
+
+ A polynomial leaves remainder \(5\) when divided by \((x-1)\). What is \(P(1)\)?
+
+
+
+
+
Checkpoint
+
+ If \(P(3)=0\) and \(P(-1)=0\) for a cubic \(P(x)\), what two linear factors must appear?
+
+
+
+
+
Checkpoint
+
+ Why is synthetic division especially useful after you already know one root?
+
+
+
+
Applications
+
Graph intercepts and curve sketching
+
+ Solving \(P(x)=0\) gives x-intercepts of polynomial graphs. Factor Theorem helps find those
+ intercepts quickly, then sketch end behavior and turning points more accurately.
+
+
+
Constraint values in models
+
+ In applied models, a condition like “output is zero at \(x=a\)” means \(P(a)=0\), so
+ \((x-a)\) must be a factor. This appears in physics calibration, optimization constraints, and
+ coding of polynomial approximations.
+
+
+
Gateway to numerical methods
+
+ Understanding factor tests and polynomial reduction supports later methods such as Newton’s
+ method and polynomial interpolation, where roots and remainders are central ideas.
+
+
+
References
+
+
Standard high-school algebra curricula (IB, A-Level, DSE extension) on polynomial theorems.
+
Classical algebra texts covering polynomial division and root-factor relationships.
+
+
diff --git a/content/math/functions-graphs/_index.md b/content/math/functions-graphs/_index.md
new file mode 100644
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--- /dev/null
+++ b/content/math/functions-graphs/_index.md
@@ -0,0 +1,257 @@
++++
+title = "Functions & graphs"
+description = "Model input-output rules, work with domain and range, and sketch key function graphs and transformations."
+template = "chapter.html"
+weight = 15
+[extra]
+subject = "math"
+chapter_id = "functions-graphs"
+strand = "Number & Algebra"
++++
+
+
+
+ A function is a rule that assigns exactly one output to each allowed input.
+ In high school, functions help you connect equations, tables, and graphs.
+
+
+
What is a function?
+
+ If \(f\) is a function, then \(f(x)\) means “the output when input is \(x\).”
+ Example:
+
+
+ \[
+ f(x)=2x+3.
+ \]
+
+
+ Then \(f(1)=5\), \(f(4)=11\), and \(f(-2)=-1\).
+
Check whether a requested equation is function form or relation form.
+
+
+
+
+
History
+
+ The modern idea of a function developed over centuries. Early algebra connected formulas to
+ unknown quantities, while analytic geometry (Descartes) linked equations to curves on coordinate
+ axes.
+
+
+ Later, Euler and others formalized function notation such as \(f(x)\). Today, function thinking
+ is central in mathematics, science, economics, and computing because it models how one quantity
+ depends on another.
+
+
+
Derivation and reasoning
+
From two points to line equation
+
+ If a line passes through \((x_1,y_1)\) and \((x_2,y_2)\), slope is:
+ \[
+ m=\frac{y_2-y_1}{x_2-x_1}.
+ \]
+ Then use \(y=mx+c\) and substitute one point to find \(c\).
+
+
+
Why vertex form helps for quadratics
+
+ Writing \(y=a(x-h)^2+k\) shows the turning point immediately: vertex \((h,k)\). This is why
+ completing the square is useful for sketching and transformations.
+
+
+
Checkpoints
+
+
+ For \(f(x)=3x-2\), find \(f(5)\) and \(f(-1)\).
+
+
+
Answer: \(f(5)=13\), \(f(-1)=-5\).
+
+
+
+
+
+ State domain and range of \(g(x)=\sqrt{x-4}\).
+
+
+
Answer: Domain \(x\ge 4\), range \(y\ge 0\).
+
+
+
+
+
+ Is \(x=y^2\) a function of \(x\)? Use the vertical line test idea.
+
+
+
Answer: No.
+
+ For a positive \(x\), there are two \(y\)-values (\(+\sqrt{x}\) and \(-\sqrt{x}\)), so one
+ \(x\) maps to two outputs.
+
+
+
+
+
+
+ For \(y=(x+1)^2-3\), state the transformation from \(y=x^2\).
+
+
+
Answer: Shift left 1 and down 3.
+
+
+
+
+
+ Find intercepts of \(y=x^2-4x+3\).
+
+
+
Answer: \(y\)-intercept: \((0,3)\).
+
Answer: \(x\)-intercepts: solve \(x^2-4x+3=0\Rightarrow(x-1)(x-3)=0\), so \((1,0)\), \((3,0)\).
+
+
+
+
+
+ Sketch idea: \(f(x)=\frac{1}{x-2}+1\). What are its asymptotes?
+
+
+ Linear functions model plans with fixed charges plus per-unit costs, such as phone plans or
+ transport fares.
+
+
+
Motion and science
+
+ Position-time or velocity-time graphs are function graphs. Intercepts, gradients, and curvature
+ describe physical behavior.
+
+
+
Data modeling
+
+ Exponential, quadratic, and rational models are used to approximate trends, optimize designs,
+ and make predictions from measured data.
+
+
+
References
+
+
Standard high-school algebra curricula on functions, graphs, and transformations.
+
Analytic geometry and precalculus references covering domain/range and graph families.
+
+
diff --git a/content/math/functions-graphs/core.html b/content/math/functions-graphs/core.html
new file mode 100644
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--- /dev/null
+++ b/content/math/functions-graphs/core.html
@@ -0,0 +1,131 @@
+
+
+ A function is a rule that assigns exactly one output to each allowed input.
+ In high school, functions help you connect equations, tables, and graphs.
+
+
+
What is a function?
+
+ If \(f\) is a function, then \(f(x)\) means “the output when input is \(x\).”
+ Example:
+
+
+ \[
+ f(x)=2x+3.
+ \]
+
+
+ Then \(f(1)=5\), \(f(4)=11\), and \(f(-2)=-1\).
+
Check whether a requested equation is function form or relation form.
+
+
diff --git a/content/math/functions-graphs/supplement.html b/content/math/functions-graphs/supplement.html
new file mode 100644
index 0000000..e02a755
--- /dev/null
+++ b/content/math/functions-graphs/supplement.html
@@ -0,0 +1,114 @@
+
+
History
+
+ The modern idea of a function developed over centuries. Early algebra connected formulas to
+ unknown quantities, while analytic geometry (Descartes) linked equations to curves on coordinate
+ axes.
+
+
+ Later, Euler and others formalized function notation such as \(f(x)\). Today, function thinking
+ is central in mathematics, science, economics, and computing because it models how one quantity
+ depends on another.
+
+
+
Derivation and reasoning
+
From two points to line equation
+
+ If a line passes through \((x_1,y_1)\) and \((x_2,y_2)\), slope is:
+ \[
+ m=\frac{y_2-y_1}{x_2-x_1}.
+ \]
+ Then use \(y=mx+c\) and substitute one point to find \(c\).
+
+
+
Why vertex form helps for quadratics
+
+ Writing \(y=a(x-h)^2+k\) shows the turning point immediately: vertex \((h,k)\). This is why
+ completing the square is useful for sketching and transformations.
+
+
+
Checkpoints
+
+
+ For \(f(x)=3x-2\), find \(f(5)\) and \(f(-1)\).
+
+
+
Answer: \(f(5)=13\), \(f(-1)=-5\).
+
+
+
+
+
+ State domain and range of \(g(x)=\sqrt{x-4}\).
+
+
+
Answer: Domain \(x\ge 4\), range \(y\ge 0\).
+
+
+
+
+
+ Is \(x=y^2\) a function of \(x\)? Use the vertical line test idea.
+
+
+
Answer: No.
+
+ For a positive \(x\), there are two \(y\)-values (\(+\sqrt{x}\) and \(-\sqrt{x}\)), so one
+ \(x\) maps to two outputs.
+
+
+
+
+
+
+ For \(y=(x+1)^2-3\), state the transformation from \(y=x^2\).
+
+
+
Answer: Shift left 1 and down 3.
+
+
+
+
+
+ Find intercepts of \(y=x^2-4x+3\).
+
+
+
Answer: \(y\)-intercept: \((0,3)\).
+
Answer: \(x\)-intercepts: solve \(x^2-4x+3=0\Rightarrow(x-1)(x-3)=0\), so \((1,0)\), \((3,0)\).
+
+
+
+
+
+ Sketch idea: \(f(x)=\frac{1}{x-2}+1\). What are its asymptotes?
+
+
+ Linear programming finds the best value (maximum or minimum) of a linear objective under linear
+ constraints. In high school, we solve two-variable problems graphically.
+
+
+
Core idea
+
+ Optimize an objective like
+ \[
+ P=ax+by
+ \]
+ subject to inequalities such as \(x+y\le 12\), \(x\ge0\), \(y\ge0\).
+
+
+
Modeling from context
+
+ Typical setup:
+
+
+
Define variables (\(x,y\)).
+
Write constraints from limits/resources.
+
Write objective function (profit, cost, time, etc.).
+
Graph constraints and find feasible region.
+
+
+
Graphing constraints
+
+ For each inequality:
+
+
+
Draw boundary line (use solid line for \(\le,\ge\); dashed for \(<,>\)).
+
Test a point (often origin) to choose correct side.
+
Combine all valid half-planes.
+
+
+ Include non-negativity constraints \(x\ge0,y\ge0\) unless problem states otherwise.
+
+
+
Feasible region
+
+ The feasible region is where all constraints hold at once. In school-level cases, it is usually a
+ polygon (possibly unbounded).
+
+
+
Corner-point principle
+
+ If an optimum exists for a bounded feasible region, it occurs at a vertex (corner point).
+
+
+ So compute all feasible vertices, then evaluate objective at each.
+
+ Vertices are \((0,0)\), \((8,0)\), \((0,4)\), and intersection of
+ \(x+y=8\) with \(x+3y=12\), which is \((6,2)\).
+
+
+ Evaluate:
+
+
+
\((0,0)\Rightarrow P=0\)
+
\((8,0)\Rightarrow P=24\)
+
\((0,4)\Rightarrow P=8\)
+
\((6,2)\Rightarrow P=22\)
+
+
+ Maximum is \(24\) at \((8,0)\).
+
+
+
Special cases
+
+
No feasible region: constraints are inconsistent.
+
Unbounded region: objective may have no maximum/minimum.
+
Multiple optima: objective line parallel to a feasible edge.
+
+
+
Exam strategy
+
+
Label all boundary lines and intercepts clearly.
+
Show shading logic for each inequality.
+
List vertex coordinates before evaluating objective.
+
Give final answer with units and variable interpretation.
+
+
+
+
+
History
+
+ Linear programming developed in the 20th century for resource allocation and planning. George
+ Dantzig's simplex method made large optimization problems practical in economics and operations
+ research.
+
+
+
Derivation and reasoning
+
Why corner points matter
+
+ Objective lines \(ax+by=c\) are parallel for different \(c\). Sliding this line across a polygonal
+ feasible region reaches the last contact at a vertex (or an entire edge in tie cases).
+
Answer: \(C=6,13,27\). Minimum \(6\) at \((0,2)\).
+
+
+
+
+ In a model, why are \(x\ge0,\ y\ge0\) often included?
+
+
+
Answer: Quantities like items, time, or material cannot be negative.
+
+
+
+
+ How can you detect possible multiple optimal solutions graphically?
+
+
+
Answer: Objective line is parallel to a boundary edge where both edge endpoints give the same optimum value.
+
+
+
+
Applications
+
+
Production planning: maximize profit under labor and material limits.
+
Transport/logistics: minimize cost with capacity constraints.
+
Scheduling: optimize time allocation under availability rules.
+
+
+
References
+
+
Standard high-school extension units on graphical linear programming.
+
Introductory operations research texts on feasible regions and corner-point optimization.
+
+
diff --git a/content/math/linear-programming/core.html b/content/math/linear-programming/core.html
new file mode 100644
index 0000000..c1d90f3
--- /dev/null
+++ b/content/math/linear-programming/core.html
@@ -0,0 +1,92 @@
+
+
+ Linear programming finds the best value (maximum or minimum) of a linear objective under linear
+ constraints. In high school, we solve two-variable problems graphically.
+
+
+
Core idea
+
+ Optimize an objective like
+ \[
+ P=ax+by
+ \]
+ subject to inequalities such as \(x+y\le 12\), \(x\ge0\), \(y\ge0\).
+
+
+
Modeling from context
+
+ Typical setup:
+
+
+
Define variables (\(x,y\)).
+
Write constraints from limits/resources.
+
Write objective function (profit, cost, time, etc.).
+
Graph constraints and find feasible region.
+
+
+
Graphing constraints
+
+ For each inequality:
+
+
+
Draw boundary line (use solid line for \(\le,\ge\); dashed for \(<,>\)).
+
Test a point (often origin) to choose correct side.
+
Combine all valid half-planes.
+
+
+ Include non-negativity constraints \(x\ge0,y\ge0\) unless problem states otherwise.
+
+
+
Feasible region
+
+ The feasible region is where all constraints hold at once. In school-level cases, it is usually a
+ polygon (possibly unbounded).
+
+
+
Corner-point principle
+
+ If an optimum exists for a bounded feasible region, it occurs at a vertex (corner point).
+
+
+ So compute all feasible vertices, then evaluate objective at each.
+
+ Vertices are \((0,0)\), \((8,0)\), \((0,4)\), and intersection of
+ \(x+y=8\) with \(x+3y=12\), which is \((6,2)\).
+
+
+ Evaluate:
+
+
+
\((0,0)\Rightarrow P=0\)
+
\((8,0)\Rightarrow P=24\)
+
\((0,4)\Rightarrow P=8\)
+
\((6,2)\Rightarrow P=22\)
+
+
+ Maximum is \(24\) at \((8,0)\).
+
+
+
Special cases
+
+
No feasible region: constraints are inconsistent.
+
Unbounded region: objective may have no maximum/minimum.
+
Multiple optima: objective line parallel to a feasible edge.
+
+
+
Exam strategy
+
+
Label all boundary lines and intercepts clearly.
+
Show shading logic for each inequality.
+
List vertex coordinates before evaluating objective.
+
Give final answer with units and variable interpretation.
+
+
diff --git a/content/math/linear-programming/supplement.html b/content/math/linear-programming/supplement.html
new file mode 100644
index 0000000..7a08866
--- /dev/null
+++ b/content/math/linear-programming/supplement.html
@@ -0,0 +1,70 @@
+
+
History
+
+ Linear programming developed in the 20th century for resource allocation and planning. George
+ Dantzig's simplex method made large optimization problems practical in economics and operations
+ research.
+
+
+
Derivation and reasoning
+
Why corner points matter
+
+ Objective lines \(ax+by=c\) are parallel for different \(c\). Sliding this line across a polygonal
+ feasible region reaches the last contact at a vertex (or an entire edge in tie cases).
+
+ Ratio between consecutive terms is constant \(r\).
+
+
\[
+ u_n=ar^{n-1}
+ \]
+
Example: \(3,6,12,24,\dots\) has \(a=3,r=2\).
+
+
Geometric series
+
+ For \(r\ne1\), sum of first \(n\) terms:
+
+
\[
+ S_n=a\frac{1-r^n}{1-r}
+ \]
+
+ Equivalent form: \(S_n=a\frac{r^n-1}{r-1}\).
+
+
+
Infinite geometric series
+
+ If \(|r|<1\), the sum to infinity exists:
+
+
\[
+ S_\infty=\frac{a}{1-r}.
+ \]
+
+ If \(|r|\ge1\), no finite sum to infinity.
+
+
+
Sigma notation
+
+ \[
+ \sum_{k=1}^{n} f(k)
+ \]
+ means add values of \(f(k)\) from \(k=1\) to \(k=n\).
+
+
Example: \(\sum_{k=1}^{4}(2k+1)=3+5+7+9=24.\)
+
+
Typical problem-solving workflow
+
+
Identify arithmetic or geometric behavior.
+
Find key parameters (\(a,d\) or \(a,r\)).
+
Use \(u_n\) for term questions, \(S_n\) for sum questions.
+
Check reasonableness (growth/decay, sign, size).
+
+
+
Exam strategy
+
+
Do not mix \(u_n\) and \(S_n\) formulas.
+
For geometric sums, check \(r\ne1\) before applying formula.
+
For infinity sums, always test \(|r|<1\).
+
Define variables clearly before solving word problems.
+
+
+
+
+
History
+
+ Sequences and sums appear in ancient mathematics, including arithmetic progressions used for
+ counting and trade. Later, geometric series became central in finance and calculus.
+
+
+
Derivation and reasoning
+
Why the arithmetic sum formula works
+
+ Write \(S_n=a+(a+d)+\dots+l\) and reverse it. Adding line-by-line gives \(2S_n=n(a+l)\), so
+ \(S_n=\frac{n}{2}(a+l)\).
+
+
Why the geometric sum formula works
+
+ Let \(S_n=a+ar+\dots+ar^{n-1}\). Multiply by \(r\), subtract, and simplify:
+ \[
+ (1-r)S_n=a(1-r^n).
+ \]
+
+ Ratio between consecutive terms is constant \(r\).
+
+
\[
+ u_n=ar^{n-1}
+ \]
+
Example: \(3,6,12,24,\dots\) has \(a=3,r=2\).
+
+
Geometric series
+
+ For \(r\ne1\), sum of first \(n\) terms:
+
+
\[
+ S_n=a\frac{1-r^n}{1-r}
+ \]
+
+ Equivalent form: \(S_n=a\frac{r^n-1}{r-1}\).
+
+
+
Infinite geometric series
+
+ If \(|r|<1\), the sum to infinity exists:
+
+
\[
+ S_\infty=\frac{a}{1-r}.
+ \]
+
+ If \(|r|\ge1\), no finite sum to infinity.
+
+
+
Sigma notation
+
+ \[
+ \sum_{k=1}^{n} f(k)
+ \]
+ means add values of \(f(k)\) from \(k=1\) to \(k=n\).
+
+
Example: \(\sum_{k=1}^{4}(2k+1)=3+5+7+9=24.\)
+
+
Typical problem-solving workflow
+
+
Identify arithmetic or geometric behavior.
+
Find key parameters (\(a,d\) or \(a,r\)).
+
Use \(u_n\) for term questions, \(S_n\) for sum questions.
+
Check reasonableness (growth/decay, sign, size).
+
+
+
Exam strategy
+
+
Do not mix \(u_n\) and \(S_n\) formulas.
+
For geometric sums, check \(r\ne1\) before applying formula.
+
For infinity sums, always test \(|r|<1\).
+
Define variables clearly before solving word problems.
+
+
diff --git a/content/math/sequences-series/supplement.html b/content/math/sequences-series/supplement.html
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+
+
History
+
+ Sequences and sums appear in ancient mathematics, including arithmetic progressions used for
+ counting and trade. Later, geometric series became central in finance and calculus.
+
+
+
Derivation and reasoning
+
Why the arithmetic sum formula works
+
+ Write \(S_n=a+(a+d)+\dots+l\) and reverse it. Adding line-by-line gives \(2S_n=n(a+l)\), so
+ \(S_n=\frac{n}{2}(a+l)\).
+
+
Why the geometric sum formula works
+
+ Let \(S_n=a+ar+\dots+ar^{n-1}\). Multiply by \(r\), subtract, and simplify:
+ \[
+ (1-r)S_n=a(1-r^n).
+ \]
+
+ If \(y\) varies jointly as \(x\) and \(z\):
+ \[
+ y\propto xz \Rightarrow y=kxz.
+ \]
+
+
Example: \(y=24\) when \(x=2,z=3\Rightarrow k=4\), so \(y=4xz\).
+
+
Partial variation
+
+ If \(y\) varies partly as \(x\), model has constant and variable part:
+ \[
+ y=mx+c,\quad c\ne0.
+ \]
+
+
+ This appears in pricing: fixed fee + per-unit charge.
+
+
+
Solving workflow
+
+
Translate words into proportional equation with \(k\).
+
Use given data to find \(k\).
+
Substitute required values to answer question.
+
Check units and reasonableness.
+
+
+
Common word-problem forms
+
+
“\(A\) is directly proportional to \(B\)”
+
“\(A\) varies inversely as \(B\)”
+
“\(A\) varies jointly as \(B\) and \(C\)”
+
“\(A\) varies directly as square/cube of \(B\)”
+
+
+
Exam strategy
+
+
Always introduce \(k\) first before substituting numbers.
+
Keep symbolic form until \(k\) is found.
+
For inverse models, domain often excludes zero.
+
State final answer with units when applicable.
+
+
+
+
+
History
+
+ Variation models grew from early physics and astronomy, where scientists needed formulas for how
+ one quantity depends on another. The language of proportionality became a bridge between algebra
+ and real measurements.
+
+
+
Derivation and reasoning
+
Why we use a constant \(k\)
+
+ “Directly proportional” means ratio \(\frac{y}{x}\) stays constant, so \(\frac{y}{x}=k\) and
+ \(y=kx\). For inverse variation, product \(xy\) stays constant, so \(xy=k\).
+
+
+
Checkpoints
+
+ If \(y\propto x\) and \(y=15\) when \(x=3\), find \(y\) when \(x=10\).
+
+
Answer: \(k=5\Rightarrow y=5x\Rightarrow y=50\).
+
+
+
+ If \(y\propto \frac1x\) and \(y=8\) when \(x=5\), find \(y\) when \(x=20\).
+
+ If \(y\) varies jointly as \(x\) and \(z\):
+ \[
+ y\propto xz \Rightarrow y=kxz.
+ \]
+
+
Example: \(y=24\) when \(x=2,z=3\Rightarrow k=4\), so \(y=4xz\).
+
+
Partial variation
+
+ If \(y\) varies partly as \(x\), model has constant and variable part:
+ \[
+ y=mx+c,\quad c\ne0.
+ \]
+
+
+ This appears in pricing: fixed fee + per-unit charge.
+
+
+
Solving workflow
+
+
Translate words into proportional equation with \(k\).
+
Use given data to find \(k\).
+
Substitute required values to answer question.
+
Check units and reasonableness.
+
+
+
Common word-problem forms
+
+
“\(A\) is directly proportional to \(B\)”
+
“\(A\) varies inversely as \(B\)”
+
“\(A\) varies jointly as \(B\) and \(C\)”
+
“\(A\) varies directly as square/cube of \(B\)”
+
+
+
Exam strategy
+
+
Always introduce \(k\) first before substituting numbers.
+
Keep symbolic form until \(k\) is found.
+
For inverse models, domain often excludes zero.
+
State final answer with units when applicable.
+
+
diff --git a/content/math/variations/supplement.html b/content/math/variations/supplement.html
new file mode 100644
index 0000000..b76e8ed
--- /dev/null
+++ b/content/math/variations/supplement.html
@@ -0,0 +1,61 @@
+
+
History
+
+ Variation models grew from early physics and astronomy, where scientists needed formulas for how
+ one quantity depends on another. The language of proportionality became a bridge between algebra
+ and real measurements.
+
+
+
Derivation and reasoning
+
Why we use a constant \(k\)
+
+ “Directly proportional” means ratio \(\frac{y}{x}\) stays constant, so \(\frac{y}{x}=k\) and
+ \(y=kx\). For inverse variation, product \(xy\) stays constant, so \(xy=k\).
+
+
+
Checkpoints
+
+ If \(y\propto x\) and \(y=15\) when \(x=3\), find \(y\) when \(x=10\).
+
+
Answer: \(k=5\Rightarrow y=5x\Rightarrow y=50\).
+
+
+
+ If \(y\propto \frac1x\) and \(y=8\) when \(x=5\), find \(y\) when \(x=20\).
+