a fire hose sprays water in an arch that csn be defined by the function. h(x)=-.25x2 +5x where is the horizontal distance from the hose and h(x) is the height of the water.
What is the maximum height of the water and how far does the water reach?
as with all quadratics ax^2+bx+c, the vertex is at x = -b/2a
In this case, that means x = 5/.5 = 10
Since the graph opens downward, the vertex is the maximum value.
So find h(10)
since h(x) = -.25x(x-20), h=0 at x=0 and x=20
To find the maximum height of the water, we need to determine the vertex of the parabolic function h(x) = -0.25x^2 + 5x. The x-coordinate of the vertex can be found using the formula x = -b / (2a), where a and b are the coefficients in the function.
In this case, a = -0.25 and b = 5. Using the formula, we have:
x = -5 / (2 * -0.25)
x = -5 / (-0.5)
x = 10
So, the maximum height of the water occurs when x = 10. Let's substitute this value back into the function to find the corresponding height:
h(10) = -0.25(10)^2 + 5(10)
h(10) = -0.25(100) + 50
h(10) = -25 + 50
h(10) = 25
Therefore, the maximum height of the water is 25 units.
To determine how far the water reaches, we need to find the x-coordinate(s) where h(x) = 0. In other words, we need to solve the equation -0.25x^2 + 5x = 0.
Factoring out x, we have:
x(-0.25x + 5) = 0
So, either x = 0 or -0.25x + 5 = 0.
For x = 0, the water starts at the hose, so this is not the distance we are looking for.
For -0.25x + 5 = 0, we can solve for x as follows:
-0.25x + 5 = 0
-0.25x = -5
x = -5 / -0.25
x = 20
Therefore, the water reaches a horizontal distance of 20 units from the hose.
To find the maximum height of the water, we need to determine the vertex of the function. The vertex of a parabola defined by the function h(x) = ax^2 + bx + c can be found using the formula: x = -b / (2a)
In this case, the function is h(x) = -0.25x^2 + 5x, where a = -0.25 and b = 5. Plugging these values into the formula, we get:
x = -5 / (2 * -0.25)
x = -5 / -0.5
x = 10
So, the maximum height of the water occurs at x = 10. To find the actual height, we can substitute this value back into the function:
h(10) = -0.25(10)^2 + 5(10)
h(10) = -0.25(100) + 50
h(10) = -25 + 50
h(10) = 25
Therefore, the maximum height of the water is 25 units.
To find how far the water reaches, we need to find the x-intercepts of the function. The x-intercepts are the points where the height of the water is 0. To do this, we set h(x) = 0 and solve for x:
-0.25x^2 + 5x = 0
Factoring out common terms:
x(-0.25x + 5) = 0
Setting each factor equal to 0 gives us two possible solutions:
x = 0 and -0.25x + 5 = 0
Solving the second equation for x gives:
-0.25x = -5
x = -5 / -0.25
x = 20
So, the water reaches a horizontal distance of 20 units from the hose.