Question

Determining the Effect of a New Initial Value
Bus Stop #2:

initial value: 9 miles

rate of change: 0.5 miles each minute

Friend’s House:

A 2-column table with 3 rows. Column 1 is labeled Time (minutes) with entries 3, 6, 9. Column 2 is labeled Distance (miles) with entries 4, 6, 8.

Seth can also get dropped off at Bus Stop #2, which is closer to his school.

Which will get him to school faster? (School is 17 miles away.)

Answer 1

Interpret each option as a distance-from-start function (miles) and solve for when distance = 17 miles.

Bus Stop #2:
D_B(t) = 9 + 0.5t.
Set 9 + 0.5t = 17 → 0.5t = 8 → t = 16 minutes.

Friend’s House:
From the table, distance increases by 2 miles every 3 minutes, so slope = 2/3 miles/min.
Using the point (3,4): D_F(t) = b + (2/3)t, and 4 = b + (2/3)*3 → b = 2.
So D_F(t) = 2 + (2/3)t. Set 2 + (2/3)t = 17 → (2/3)t = 15 → t = 22.5 minutes.

Conclusion: Bus Stop #2 gets him to school faster (16 minutes vs. 22.5 minutes).