The function y = -0.25t^2 + 5t + 180 models the estimated attendance y at a city park t days...

1. TRANSLATE the problem information

• Given information:
  • Function: \(\mathrm{y = -0.25t^2 + 5t + 180}\)
  • \(\mathrm{y}\) = estimated attendance at park
  • \(\mathrm{t}\) = days after the end of a holiday weekend
  • Need to interpret the y-intercept

• The y-intercept is the y-value when \(\mathrm{t = 0}\)

2. INFER the calculation approach

• To find the y-intercept, substitute \(\mathrm{t = 0}\) into the function
• This will give us the attendance value at the y-intercept

3. SIMPLIFY by substitution

• Substitute \(\mathrm{t = 0}\) into \(\mathrm{y = -0.25t^2 + 5t + 180}\):

\(\mathrm{y = -0.25(0)^2 + 5(0) + 180}\)

\(\mathrm{y = 0 + 0 + 180}\)

\(\mathrm{y = 180}\)

4. TRANSLATE the mathematical result back to context

• The y-intercept is 180
• Since \(\mathrm{t = 0}\) represents "zero days after the end of the holiday weekend"
• This means "at the end of the holiday weekend"
• Therefore: At the end of the holiday weekend, the estimated attendance was 180

Answer: B

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students misinterpret what \(\mathrm{t = 0}\) represents in the context.

Many students think that since \(\mathrm{t}\) measures "days after the end of the holiday weekend," \(\mathrm{t = 0}\) must mean "one day after." They fail to recognize that "zero days after" actually means "at that moment" or "at the end of." This fundamental misunderstanding of the timeline leads them to select Choice C or D (which refer to "one day after").

The Bottom Line:

The mathematical calculation is straightforward, but the real challenge lies in correctly interpreting what \(\mathrm{t = 0}\) means in the real-world context. Students must carefully translate between mathematical language and everyday language about time.