Mathematics
SAT-45: Writing Exponential Functions (y = ab^x) from Word Problems
Turn a real-world growth or decay story into y = a·b^x by finding the start value a and the factor b.
SAT-45: Writing Exponential Functions (y = abx) from Word Problems
Description: Once you know a situation is exponential (SAT-44), you write it as y = a·bx. This lesson shows how to pull a and b out of the words.
What a and b mean
- a = the starting value (when x = 0).
- b = the growth/decay factor per step.
Build b from the percent rate:
- Growth of r% → b = 1 + r/100 (e.g. +8% → b = 1.08).
- Decay of r% → b = 1 − r/100 (e.g. −8% → b = 0.92).
(Oʻzbekcha: a — boshlangʻich qiymat, b — har bir qadamdagi oʻzgarish koeffitsiyenti.)
Worked Example (growth)
A town has 5,000 people and grows 3% per year. Write the population after x years.
- Starting value a = 5000.
- Growth 3% → b = 1 + 0.03 = 1.03.
- Function: y = 5000·(1.03)x.
(Oʻzbekcha: 3% oʻsish uchun b = 1.03.)
Worked Example (decay)
A car worth $20,000 loses 15% of its value each year.
- a = 20000, decay 15% → b = 1 − 0.15 = 0.85.
- Function: y = 20000·(0.85)x.
Tip: "doubles" → b = 2; "halves" → b = 0.5; "triples" → b = 3. (Oʻzbekcha: "ikki barobar" → b = 2; "yarmiga" → b = 0.5.)
Practice
A colony of 200 bacteria doubles every hour. Write y for hours x.
Show answer
a = 200, doubling → b = 2. So y = 200·2x.
Key words — Kalit soʻzlar
- Exponential function — eksponensial funksiya
- Starting value (initial) — boshlangʻich qiymat
- Growth factor — oʻsish koeffitsiyenti
- Decay factor — kamayish koeffitsiyenti
- Rate (percent) — foiz sur'ati
- Per year / per hour — har yili / har soatda
- Double / Halve — ikki barobar / yarmiga
- Base — asos
- Exponent — daraja koʻrsatkichi
Summary
- Model: y = a·bx; a = starting value, b = factor per step.
- Growth r% → b = 1 + r/100; decay r% → b = 1 − r/100.
- "Doubles" → b = 2, "halves" → b = 0.5, etc.