Sara opened a savings account at a bank. The table shows the exponential relationship between the time t, in years,...

1. TRANSLATE the problem information

• Given information:
  • Table shows exponential relationship between time t and amount d
  • At \(\mathrm{t = 0}\): \(\mathrm{d = 670.00}\)
  • At \(\mathrm{t = 1}\): \(\mathrm{d = 674.02}\)
  • At \(\mathrm{t = 2}\): \(\mathrm{d = 678.06}\)
• This tells us we need an exponential equation of the form \(\mathrm{d = a(1 + r)^t}\)

2. INFER the approach

• Since we know this is exponential growth, we need to find:
  • The initial value a (what happens when \(\mathrm{t = 0}\))
  • The growth rate r (how much it grows each year)
• Strategy: Use the \(\mathrm{t = 0}\) data point to find a, then use \(\mathrm{t = 1}\) to find r

3. TRANSLATE the initial condition

• When \(\mathrm{t = 0}\): \(\mathrm{d = a(1 + r)^0 = a(1) = a}\)
• From the table: when \(\mathrm{t = 0}\), \(\mathrm{d = 670}\)
• Therefore: \(\mathrm{a = 670}\)

4. SIMPLIFY to find the growth rate

• Now we have: \(\mathrm{d = 670(1 + r)^t}\)
• Using \(\mathrm{t = 1}\), \(\mathrm{d = 674.02}\): \(\mathrm{674.02 = 670(1 + r)^1}\)
• This simplifies to: \(\mathrm{674.02 = 670(1 + r)}\)
• Divide both sides by 670: \(\mathrm{674.02 ÷ 670 = 1 + r}\) (use calculator)
• \(\mathrm{1.006 = 1 + r}\)
• Therefore: \(\mathrm{r = 0.006}\)

5. TRANSLATE to final equation form

• Substituting back: \(\mathrm{d = 670(1 + 0.006)^t = 670(1.006)^t}\)
• This matches choice B exactly

Answer: B. \(\mathrm{d = 670(1.006)^t}\)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak INFER skill: Students see the amounts increasing and think "this looks linear" instead of recognizing the exponential pattern. They calculate that from year 0 to 1, the increase is \(\mathrm{674.02 - 670 = 4.02}\), and assume this continues linearly.

This leads them to think \(\mathrm{d = 670 + 4.02t}\) and select Choice C (\(\mathrm{d = 670 + 4.02t}\)).

Second Most Common Error:

Poor TRANSLATE reasoning: Students correctly identify this as exponential but confuse the growth rate with the growth factor. They see that \(\mathrm{674.02 ÷ 670 = 1.006}\) and think this IS the growth rate (rather than 1 + growth rate), leading them to write \(\mathrm{d = 670(1 + 0.006t)}\) instead of \(\mathrm{d = 670(1.006)^t}\).

This may lead them to select Choice D (\(\mathrm{d = 670(1 + 0.006t)}\)).

The Bottom Line:

This problem tests whether students can distinguish between linear and exponential growth patterns, and whether they understand that exponential functions have the form \(\mathrm{a(1 + r)^t}\), not \(\mathrm{a(1 + rt)}\).