The path of water from a hose on a fire tugboat can be approximated by the equation
y = −0.006x2 + 1.2x + 10,
where y is the height, in feet, of the water above the ocean when the water is x feet from the tugboat. When the water from the hose is 6 feet above the ocean, at what distance from the tugboat is it? Round answer to nearest hundredth.
This is the same kind of question I just answered for you in the post below
In this case, set
−0.006x2 + 1.2x + 10 = 6
arrange it into the standard form of a quadratic and use the formula.
To find the distance from the tugboat at which the water from the hose is 6 feet above the ocean, we need to solve the equation for x when y = 6.
Given equation: y = -0.006x^2 + 1.2x + 10
Substituting y = 6 into the equation, we have:
6 = -0.006x^2 + 1.2x + 10
Rearrange the equation to set it equal to zero:
0 = -0.006x^2 + 1.2x + 4
We now have a quadratic equation. To solve it, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = -0.006, b = 1.2, and c = 4. Plug these values into the quadratic formula:
x = (-1.2 ± √(1.2^2 - 4(-0.006)(4))) / (2(-0.006))
Let's solve this equation:
x = (-1.2 ± √(1.44 + 0.096))/(-0.012)
x = (-1.2 ± √1.536)/(-0.012)
x = (-1.2 ± 1.24)/(-0.012)
Now we have two possible x-values:
x₁ = (-1.2 + 1.24)/(-0.012)
x₁ = 0.04/(-0.012)
x₁ ≈ -3.33
x₂ = (-1.2 - 1.24)/(-0.012)
x₂ = -2.44/(-0.012)
x₂ ≈ 203.33
Since distance cannot be negative in this context, we discard x₁ and round x₂ to the nearest hundredth:
x ≈ 203.33
Therefore, when the water from the hose is 6 feet above the ocean, it is approximately 203.33 feet from the tugboat.