To determine how long it takes for the rock to hit the water, we need to find the value of \( x \) when the height \( f(x) = 0 \), since the height of the rock is 0 feet when it hits the water.
The function given is:
[tex]\[ f(x) = -16(x^2 - 3x - 18) \][/tex]
First, we set the function equal to zero to solve for \( x \):
[tex]\[ -16(x^2 - 3x - 18) = 0 \][/tex]
To simplify, divide both sides by -16:
[tex]\[ x^2 - 3x - 18 = 0 \][/tex]
Now, we need to solve the quadratic equation:
[tex]\[ x^2 - 3x - 18 = 0 \][/tex]
We can factor the quadratic equation:
[tex]\[ (x - 6)(x + 3) = 0 \][/tex]
Setting each factor equal to zero gives us:
[tex]\[ x - 6 = 0 \quad \text{or} \quad x + 3 = 0 \][/tex]
Solving these equations, we get the solutions:
[tex]\[ x = 6 \quad \text{or} \quad x = -3 \][/tex]
Since \( x \) represents the number of seconds since the rock was thrown, it cannot be negative. Therefore, we take the positive solution:
[tex]\[ x = 6 \][/tex]
Thus, it takes [tex]\( \boxed{6} \)[/tex] seconds for the rock to hit the water.
