Certainly! Let's solve Kayla's problem step-by-step to find the lowest temperature and the time it occurs.
To find the minimum (or maximum) of a quadratic function of the form [tex]\( t(h) = ah^2 + bh + c \)[/tex], we employ the vertex formula for the quadratic function. The vertex formula for the [tex]\( h \)[/tex]-coordinate of the vertex is given by:
[tex]\[
h = -\frac{b}{2a}
\][/tex]
Given the provided function:
[tex]\[
t(h) = 0.5h^2 - 5h + 27.5
\][/tex]
we can identify the coefficients as:
[tex]\( a = 0.5 \)[/tex],
[tex]\( b = -5 \)[/tex],
[tex]\( c = 27.5 \)[/tex].
Now, plug these values into the vertex formula to find [tex]\( h \)[/tex]:
[tex]\[
h = -\frac{-5}{2 \cdot 0.5} = \frac{5}{1} = 5
\][/tex]
Thus, [tex]\( h = 5 \)[/tex]. This means the lowest temperature occurs 5 hours after 10:00 p.m.
Next, we need to compute the temperature at this time by substituting [tex]\( h = 5 \)[/tex] back into the function:
[tex]\[
t(5) = 0.5(5)^2 - 5(5) + 27.5
\][/tex]
Calculate each term separately:
[tex]\[
0.5(25) = 12.5,
\][/tex]
[tex]\[
-5(5) = -25,
\][/tex]
[tex]\[
27.5
\][/tex]
Now, sum these values:
[tex]\[
t(5) = 12.5 - 25 + 27.5 = 15.0
\][/tex]
So, the lowest temperature is [tex]\( 15^{\circ}F \)[/tex].
Next, we determine the actual time when this temperature occurs. Since it happens 5 hours after 10:00 p.m., we simply add 5 hours to 10:00 p.m., which gives us:
[tex]\[
10:00 \text{ p.m.} + 5 \text{ hours} = 3:00 \text{ a.m.}
\][/tex]
Therefore, the lowest temperature recorded was [tex]\( 15^{\circ}F \)[/tex] and it occurred at [tex]\( 3:00 \text{ a.m.} \)[/tex].
So the answer is:
A. [tex]\( 15^{\circ} F \)[/tex] at [tex]\( 3:00 \text{ a.m.} \)[/tex].
