Question

5
select the correct answer
kayla set up an outdoor digital thermometer to record the temperature overnight as part of her science fair project. she began recording the temperature, in degrees fahrenheit, at 10:00 p.m. kayla modeled the overnight temperature with function ( t ), where ( h ) represents the number of hours since 10:00 p.m.
( t(h) = 0.5h^2 - 5h + 27.5 )
graph of the function ( t(h) ) with ( h ) (hours since 10:00 p.m.) on the x - axis and ( t(h) ) (temperature ((^circ\text{f}))) on the y - axis, showing a parabola opening upwards
what is the lowest temperature and at what time did it occur?
a. ( 5^circ\text{f} ) at 5:00 a.m
b. ( 15^circ\text{f} ) at 5:00 a.m
c. ( 5^circ\text{f} ) at 3:00 a.m
d. ( 15^circ\text{f} ) at 3:00 a.m

Explanation:

Step1: Identify the function type

The function \( t(h) = 0.5h^2 - 5h + 27.5 \) is a quadratic function. For a quadratic function in the form \( ax^2 + bx + c \) (where \( a>0 \)), the vertex represents the minimum point. The \( h \)-coordinate of the vertex is given by \( h = -\frac{b}{2a} \).

Step2: Calculate the time (h) of the minimum

Here, \( a = 0.5 \) and \( b = -5 \). Plugging into the formula:
\( h = -\frac{-5}{2 \times 0.5} = \frac{5}{1} = 5 \) hours.

Step3: Find the corresponding time

Since \( h \) is the number of hours since 10:00 p.m., adding 5 hours to 10:00 p.m. gives 3:00 a.m.? Wait, no—wait, 10:00 p.m. + 5 hours is 3:00 a.m.? Wait, 10 p.m. + 5 hours: 10 + 5 = 15, 15 - 12 = 3 a.m. Wait, no, 10 p.m. + 5 hours: 10 + 5 = 15, which is 3 a.m.? Wait, no, 10 p.m. + 5 hours: 10 + 5 = 15, 15 - 12 = 3 a.m.? Wait, no, 10 p.m. + 5 hours: 10 + 5 = 15, which is 3 a.m.? Wait, no, 10 p.m. + 5 hours: 10 + 5 = 15, 15 - 12 = 3 a.m. Wait, but let's check the calculation again. Wait, 10 p.m. + 5 hours: 10 + 5 = 15, which is 3 a.m. (since 12 a.m. is 12, 1 a.m. is 13, 2 a.m. is 14, 3 a.m. is 15). Wait, no, 10 p.m. is 22:00, adding 5 hours: 22 + 5 = 27, 27 - 24 = 3, so 3:00 a.m. Wait, but let's check the function value at \( h = 5 \).

Step4: Calculate the minimum temperature

Plug \( h = 5 \) into \( t(h) \):
\( t(5) = 0.5(5)^2 - 5(5) + 27.5 \)
\( = 0.5(25) - 25 + 27.5 \)
\( = 12.5 - 25 + 27.5 \)
\( = (12.5 + 27.5) - 25 \)
\( = 40 - 25 = 15 \) °F.

Wait, but earlier I thought 10 p.m. + 5 hours is 3 a.m.? Wait, no, 10 p.m. + 5 hours: 10 + 5 = 15, which is 3 a.m. (since 12 a.m. is 12, 1 a.m. is 13, 2 a.m. is 14, 3 a.m. is 15). Wait, but let's check the options. Option B is 15°F at 5:00 a.m., option D is 15°F at 3:00 a.m. Wait, my calculation for time: \( h = 5 \) hours since 10:00 p.m. So 10:00 p.m. + 5 hours is 3:00 a.m.? Wait, no, 10 p.m. + 5 hours: 10 + 5 = 15, which is 3 a.m. (because 12 a.m. is 12, 1 a.m. is 13, 2 a.m. is 14, 3 a.m. is 15). Wait, but 10 p.m. + 5 hours: 10 + 5 = 15, which is 3 a.m. So the time is 3 a.m., temperature is 15°F. So option D.

Wait, but let's recheck the vertex formula. The quadratic is \( t(h) = 0.5h^2 -5h +27.5 \). The vertex \( h \)-coordinate is \( -b/(2a) = -(-5)/(20.5) = 5/1 = 5 \). So \( h = 5 \) hours after 10 p.m. 10 p.m. + 5 hours is 3 a.m. (since 10 + 5 = 15, 15 - 12 = 3 a.m.). Then \( t(5) = 0.525 -25 +27.5 = 12.5 -25 +27.5 = 15 \). So the minimum temperature is 15°F at 3:00 a.m., which is option D.

Answer:

D. 15°F at 3:00 a.m