Answer:
The most helpful function in an attempt to determine the time it takes for the dolphin to re-enter is h(x) = 2·(x - 1)·(x - 4)·(x - 6)
Step-by-step explanation:
For 2·x²·(x - 11) + 4·(17·x - 12)
h(5) = 2×5^2×(5 - 11) + 4×(17×5 - 12) = -8
For the function h(x) = 2·x·(x² - 11·x + 34) -48 we have;
h(5) = 2×5×(5^2 - 11×5 + 34) -48 = -8
For the function h(x) = 2·(x - 1)·(x - 4)·(x - 6) we have;
h(5) = 2×(5 - 1)×(5 - 4)×(5 - 6) = -8
For the function h(x) = 2·x³ - 22·x² + 68·x -48 we have;
h(5) = 2×5^3 - 22×5^2 + 68× 5 - 48 = -8
Given that the values of the function are all equal at x = 5, the function that will be most helpful in determining the time it takes for the dolphin to re-enter the sea after leaping out of the water is the function that is already factorized
Thereby where the value of the function h(x) at which the dolphin re-enters the the sea is h(x) = 0, we have the function h(x) = 2·(x - 1)·(x - 4)·(x - 6), readily gives the time values, x, as x = 1 second or 4 second or 6 second, therefore, the most helpful function is h(x) = 2·(x - 1)·(x - 4)·(x - 6).
