Bot GPT 4-o mini To define the domain of the volleyball's height \( h \) as described by the equation \( h = -16t^2 + 33t + 6 \), we first need to find the time at which the ball reaches its maximum height. This can be accomplished by identifying the vertex of the quadratic equation provided.
The general form of a quadratic equation is \( ax^2 + bx + c \), and the time \( t \) at which the maximum height occurs for our equation can be found using the formula:
\[ t = -\frac{b}{2a} \]
In our equation:
- \( a = -16 \)
- \( b = 33 \)
Plugging in these values, we calculate:
\[ t = -\frac{33}{2(-16)} = \frac{33}{32} \approx 1.03125 \text{ seconds} \]
Next, we need to define the domain of the function concerning when the volleyball is in the air, which starts from the moment it is served until it reaches the ground again (height \( h = 0 \)).
To find when the ball hits the ground, we need to set \( h = 0 \) and solve for \( t \):
\[ 0 = -16t^2 + 33t + 6 \]
This is a quadratic equation that can be solved using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = -16 \), \( b = 33 \), and \( c = 6 \):
Calculating the discriminant:
\[ b^2 - 4ac = 33^2 - 4(-16)(6) = 1089 + 384 = 1473 \]
Now substituting into the quadratic formula:
\[ t = \frac{-33 \pm \sqrt{1473}}{2(-16)} \]
Calculating \( \sqrt{1473} \):
\[ \sqrt{1473} \approx 38.39 \]
This gives us two potential times:
\[ t = \frac{-33 + 38.39}{-32} \quad \text{and} \quad t = \frac{-33 - 38.39}{-32} \]
Calculating each solution:
- For the positive root:
\[ t = \frac{5.39}{-32} \approx -0.168 \quad \text{(not valid since time cannot be negative)} \]
- For the negative root:
\[ t = \frac{-71.39}{-32} \approx 2.23 \text{ seconds} \]
Thus, the volleyball remains in the air from the moment it is served until it hits the ground, which corresponds to the interval from \( t = 0 \) seconds to \( t \approx 2.23 \) seconds.
Therefore, the domain of the function—indicating the time during which the volleyball is in the air—is:
\[ \boxed{[0, 2.23]} \]
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