High School

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For \( T(t) = -0.3t^2 + 2.4t + 98.3 \) where \( 0 \leq t \leq 8 \), what are the relative extreme points?

A. \( (4, 108.5) \)
B. \( (1.3, 100.1) \)
C. \( (4, 98.5) \)
D. \( (1.3, 100.5) \)

Answer :

Final answer:

The relative extreme point of the function T(t) = -0.3t² + 2.4t + 98.3 is found using the vertex formula, resulting in the point (4, 108.3). Among the provided options, a) ((4, 108.5)) is the closest, despite a slight typo.

Explanation:

The student is asking for the relative extreme points of the quadratic function T(t) = -0.3t² + 2.4t + 98.3. An extreme point on a quadratic function can be a maximum or a minimum, and it's found by using the vertex formula. Here, 'a' is the coefficient of t², 'b' is the coefficient of t, and 'c' is the constant term.

To find the t-value of the vertex (which corresponds to the extreme point), you use the formula -b/(2a). After finding the t-value, you can substitute it back into the original function to find the y-value (T(t)).

In this case, a = -0.3, b = 2.4. Plugging these into the formula gives us: t = -2.4/(2 imes -0.3) = -2.4/(-0.6) = 4. Substituting t=4 back into the function T(t) gives us T(4) = -0.3(4)² + 2.4(4) + 98.3 = 108.3. Therefore, the correct answer is a) ((4, 108.3)), which is a slight typo; however, it's the closest to the actual calculated value among the provided options.

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