The given function for the patient's temperature, $T(t)$, is:

$T(t) = -0.02t^2 + 0.42t + 97.9$

This is a quadratic function in the form $T(t) = at^2 + bt + c$, where:

• $a = -0.02$
• $b = 0.42$
• $c = 97.9$

Since $a$ is negative, the parabola opens downward, meaning the function has a maximum value at the vertex.

To find the time $t$ at which the maximum temperature occurs, we use the vertex formula:

$t = \frac{-b}{2a}$

Substituting $a = -0.02$ and $b = 0.42$ into the formula:

$t = \frac{-0.42}{2(-0.02)} = \frac{-0.42}{-0.04} = 10.5$

So, the maximum temperature occurs 10.5 hours after the illness begins.

Next, we substitute $t = 10.5$ into the temperature function to find the maximum temperature:

$T(10.5) = -0.02(10.5)^2 + 0.42(10.5) + 97.9$

$T(10.5) = -0.02(110.25) + 4.41 + 97.9$

$T(10.5) = -2.205 + 4.41 + 97.9 = 100.105$

Thus, the patient's maximum temperature during the illness is approximately 100.11°F.

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Would you like a more detailed explanation or have any further questions about this?

Here are some related questions you might find interesting:

1. How does the maximum temperature change if the coefficients of the quadratic function change?
2. What does the negative coefficient of the quadratic term indicate about the temperature function's behavior?
3. How would the function look if the illness lasted longer than 24 hours?
4. What is the patient's temperature at the beginning and end of the illness (at $t = 0$ and $t = 24$)?
5. How would you determine the patient's temperature at any given time if the formula was different?

Tip: When dealing with quadratic functions, always check the sign of the leading coefficient. If it's negative, the function has a maximum; if positive, it has a minimum.