Algebraic Functions & Equations: Types, Graphs & Business Uses

TAKEAWAYS:

• Functions map inputs to outputs: linear ($f(x)=mx+b$), quadratic ($f(x)=a{x}^2+bx+c$), and exponential ($f(x)=a^x$) have distinct graphs and business uses (e.g., depreciation, growth).
• Equations solve for unknowns: linear ($ax+b=0$), quadratic ($a{x}^2+bx+c=0$), and simultaneous equations model equilibrium points in markets (e.g., $Q_d=Q_s$).
• Graphs visualize relationships: demand curves slope downward, supply curves upward; intersections show market-clearing prices.
• Applications include profit maximization (marginal revenue = marginal cost), loan amortization (arithmetic sequences), and production functions (Cobb-Douglas).
• Exam focus: Solve equations, sketch graphs, and interpret real-world scenarios (e.g., "If Daraz’s revenue is $R=50Q-0.1Q^2$, find the quantity that maximizes profit").
• Common pitfalls: Misidentifying function types (e.g., confusing $f(x)=x^2$ with $f(x)=2^x$), incorrect domain/range, and algebraic errors in solving.

1. Algebraic Functions: Types and Graphs

Functions describe relationships between variables. In business, they model costs, revenues, demands, and supplies.

1.1 Linear Functions

Definition: $f(x)=mx+b$, where $m$ is the slope and $b$ is the y-intercept. Graph: Straight line with slope $m$. Business Use:

• Cost functions: $C=FC+VC\cdot Q$ (fixed + variable costs).
• Revenue functions: $R=P\cdot Q$ (price × quantity).
• Break-even analysis: $R=C$ → $P\cdot Q=FC+VC\cdot Q$.

Worked Example 1: Demand Function Daraz’s demand for a product is $Q_d=100-2P$. Sketch the graph and find the quantity demanded when $P=30$. Solution:

1. When $P=30$, $Q_d=100-2(30)=40$.
2. Graph: Line from $(0,100)$ to $(50,0)$, passing through $(30,40)$.

1.2 Quadratic Functions

Definition: $f(x)=a{x}^2+bx+c$ (parabola). Graph: U-shaped (if $a>0$) or inverted (if $a<0$). Business Use:

• Profit functions: $\pi =R-C=-0.1Q^2+50Q-1000$.
• Optimization: Vertex gives max/min (e.g., max profit).

Worked Example 2: Profit Maximization Pathao’s profit function is $\pi =-0.5Q^2+200Q-2000$. Find the quantity that maximizes profit. Solution:

1. Vertex formula: $Q=-\frac{b}{2a}=-\frac{200}{2(-0.5)}=200$.
2. Max profit: $\pi (200)=-0.5(200{)}^2+200(200)-2000=18,000$.

1.3 Rational and Exponential Functions

Worked Example 3: Depreciation (Exponential) Ncell’s equipment depreciates at 10% annually. If initial value is Rs 500,000, find its value after 3 years. Solution:

1. Formula: $V(t)=500,000\cdot (0.9{)}^t$.
2. After 3 years: $V(3)=500,000\cdot (0.9{)}^3=364,500$.

2. Solving Equations

Equations solve for unknowns in business scenarios.

2.1 Linear Equations

Method: Isolate $x$. Example: Solve $3x+5=20$. Solution: $x=\frac{15}{3}=5$.

2.2 Quadratic Equations

Methods:

1. Factoring: $x^2-5x+6=0$ → $(x-2)(x-3)=0$ → $x=2,3$.
2. Quadratic formula: $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.
3. Completing the square: $x^2+6x+5=(x+3{)}^2-4$.

Worked Example 4: Break-Even Quantity eSewa’s cost is $C=100+2Q$ and revenue $R=10Q$. Find break-even quantities. Solution:

1. Set $R=C$: $10Q=100+2Q$ → $8Q=100$ → $Q=12.5$.

2.3 Simultaneous Equations

Method: Substitution or elimination. Example: Solve $Q_d=26-0.4P$ and $Q_s=4+0.6P$ (market equilibrium).

Solution:

1. Equate $Q_d=Q_s$: $26-0.4P=4+0.6P$.
2. Solve for $P=22$, then $Q=17$.

3. Applications in Business

3.1 Demand and Supply Functions

• Demand: $Q_d=a-bP$ (downward slope).
• Supply: $Q_s=c+dP$ (upward slope).
• Equilibrium: $Q_d=Q_s$ → solve for $P$ and $Q$.

Worked Example 5: NTC’s Tariff Adjustment NTC’s demand and supply for electricity are $Q_d=100-2P$ and $Q_s=20+3P$. Find equilibrium price and quantity. Solution:

1. Set $Q_d=Q_s$: $100-2P=20+3P$ → $80=5P$ → $P=16$.
2. $Q=100-2(16)=68$.

3.2 Profit Maximization

Rule: Set Marginal Revenue (MR) = Marginal Cost (MC). Example: If $MR=20-0.4Q$ and $MC=5+0.2Q$, find optimal $Q$. Solution:

1. $20-0.4Q=5+0.2Q$ → $15=0.6Q$ → $Q=25$.

4. Common Mistakes and How to Avoid Them

## In the Real World

1. Khalti’s Transaction Fees

   • Idea: Linear function for fee calculation.
   • How: Fee = $0.5\%\times \text{Amount}$ → $F=0.005A$.
   • Graph: Straight line through origin with slope $0.005$.

2. Daraz’s Order Fulfillment Queue

   • Idea: Rational function for processing time.
   • How: Time per order = $\frac{100}{Q}$ (more orders → faster per-unit processing).
   • Example: If Daraz processes 500 orders/day, time per order = $100/500=0.2$ hours.

3. NEPSE Stock Prices

   • Idea: Exponential growth/decay for stock trends.
   • How: $P_t=P_0\cdot e^{rt}$ (e.g., a stock growing at 5% annually).
   • Example: If a stock starts at Rs 1000 and grows at 5%, after 2 years: $P_2=1000\cdot e^{0.1}\approx 1105$.

## Exam Tip

1. Graphs: Always label axes, plot key points, and shade regions (e.g., profit > 0).
2. Equations: Show all steps—examiners deduct for skipped algebra.
3. Business Context: Relate answers to real scenarios (e.g., "This equilibrium price ensures no surplus/shortage").
4. Units: Include units in final answers (e.g., "Rs 500 per unit").
5. Common Questions:
   • Sketch graphs for demand/supply.
   • Solve for equilibrium in markets.
   • Find max/min using calculus (vertex formula or derivatives).
6. Past Exam Patterns:
   • 30%: Solving equations (linear/quadratic).
   • 40%: Graph interpretation (equilibrium, optimization).
   • 30%: Real-world applications (profit, cost, demand).

Visual Summary:

mindmap
  root((Algebraic Functions & Equations))
    Types
      Linear
      Quadratic
      Rational
      Exponential
    Graphs
      Demand (↓ slope)
      Supply (↑ slope)
      Profit (↑ then ↓)
    Applications
      Break-even
      Optimization
      Market Equilibrium
    Exam Focus
      Solve & Graph
      Interpret Business Scenarios