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.
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