To determine the reasonable domain and range of the function \( h = -16t^2 + 80t + 4 \), let's analyze the problem.
The function describes the height of a baseball over time, which implies:
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Domain: The time \( t \) cannot be negative, as you cannot have negative time in this context. Hence, the domain starts from \( t = 0 \) and goes to when the baseball hits the ground. We need to find when \( h(t) = 0 \) to determine the upper limit of \( t \).
To find when the baseball hits the ground, we can solve: \[ -16t^2 + 80t + 4 = 0 \] Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = -16 \), \( b = 80 \), and \( c = 4 \): \[ t = \frac{-80 \pm \sqrt{80^2 - 4 \cdot (-16) \cdot 4}}{2 \cdot (-16)} \] \[ = \frac{-80 \pm \sqrt{6400 + 256}}{-32} \] \[ = \frac{-80 \pm \sqrt{6656}}{-32} \] \[ = \frac{-80 \pm 81.64}{-32} \] This gives two potential values for \( t \): \[ t \approx \frac{1.64}{-32} \quad \text{and} \quad t \approx \frac{-161.64}{-32} \] The positive value \( t \) will be around \( 5.05 \) seconds after calculation.
Therefore, the reasonable domain is: \[ [0, 5.05] \]
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Range: To find the range, we need to determine the maximum height of the baseball. The height of the ball is a quadratic function that opens downwards (since the coefficient of \( t^2 \) is negative). We can find the vertex (maximum height) using the formula \( t = -\frac{b}{2a} \): \[ t = -\frac{80}{2 \times -16} = \frac{80}{32} = 2.5 \]
Now substituting \( t = 2.5 \) back into the height function to get the maximum height: \[ h(2.5) = -16(2.5)^2 + 80(2.5) + 4 \] \[ = -16(6.25) + 200 + 4 \] \[ = -100 + 200 + 4 = 104 \quad \text{feet} \]
The height \( h \) cannot be negative, so the reasonable range is: \[ [0, 104] \]
Combining both findings, the reasonable domain and range for the function are:
- Domain: [0, 5.05]
- Range: [0, 104]
Thus, the correct response from the options given is: Domain: [0, 5.05]; Range: [0, 104].