The function h(t) = -16t^2 + 98t has a defined domain and range.
Domain: The domain of a quadratic function is all real numbers unless there are any restrictions stated in the problem. In this case, there are no restrictions mentioned, so the domain is all real numbers.
Range: To find the range of a quadratic function, we need to determine the vertex or the minimum/maximum value of the function. In this case, we have a quadratic function in the form h(t) = -16t^2 + 98t, which opens downward since the coefficient of t^2 is negative (-16). The vertex of this quadratic function can be found using the formula t = -b/2a, where "a" is the coefficient of t^2 (-16) and "b" is the coefficient of t (98).
t = -b/2a
t = -98/(2(-16))
t = -98/(-32)
t = 3.0625
The vertex of the quadratic function is (3.0625, h(3.0625)). Since the parabola opens downward, the vertex is the maximum value of the function.
To find the maximum value, we substitute t = 3.0625 into the function:
h(3.0625) = -16(3.0625)^2 + 98(3.0625)
h(3.0625) = -16(9.375) + 299.5
h(3.0625) = -150 + 299.5
h(3.0625) = 149.5
Therefore, the vertex is (3.0625, 149.5), and the range of the function h(t) = -16t^2 + 98t is all real numbers less than or equal to 149.5.
the function h(t)=−16t2+98t
domain
rainge
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