Question 120787: Hello, I'm completely lost. Can anyone help me with this. Please provide sources if you can. Thanks
Using the Library, web resources, and/or other materials, find a real-life application of a quadratic function. State the application, give the equation of the quadratic function, and state what the x and y in the application represent. Choose at least two values of x to input into your function and find the corresponding y for each. State, in words, what each x and y means in terms of your real-life application. Please see the following example. Do not use any version of this example in your own post. You may use other variables besides x and y, such as t and S depicted in the following example. Be sure to reference all sources using APA style.
Typing hint: To type x-squared, use x^2. Do not use special graphs or symbols because they will not appear when pasted to the Discussion Board.
When thrown into the air from the top of a 50 ft building, a ball’s height, S, at time t can be found by S(t) = -16t^2 + 32t + 50. When t = 1, s = -16(1)^2 + 32(1) + 50 = 66. This implies that after 1 second, the height of the ball is 66 feet. When t = 2, s = -16(2)^2 + 32(2) + 50 = 50. This implies that after 2 seconds, the height of the ball is 50 feet.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Try using Google with search string quaratic function examples.
Here's a sample:
Example 6.
A rancher has 600 meters of fence to enclose a rectangular corral with another fence dividing it in the middle as in the diagram below.
As indicated in the diagram, the four horizontal sections of fence will each be x meters long and the three vertical sections will each be y meters long.
The rancher's goal is to use all of the fence and enclose the largest possible area.
The two rectangles each have area xy, so we have
total area: A = 2xy.
There is not much we can do with the quantity A while it is expressed as a product of two variables. However, the fact that we have only 1200 meters of fence available leads to an equation that x and y must satisfy.
3y + 4x = 1200.
3y = 1200 - 4x.
y = 400 - 4x/3.
We now have y expressed as a function of x, and we can substitute this expression for y in the formula for total area A.
A = 2xy = 2x (400 -4x/3).
We need to find the value of x that makes A as large as possible. A is a quadratic function of x, and the graph opens downward, so the highest point on the graph of A is the vertex. Since A is factored, the easiest way to find the vertex is to find the x-intercepts and average.
2x (400 -4x/3) = 0.
2x = 0 or 400 -4x/3 = 0.
x = 0 or 400 = 4x/3.
x = 0 or 1200 = 4x.
x = 0 or 300 = x.
Therefore, the line of symmetry of the graph of A is x = 150, the average of 0 and 300.
Now that we know the value of x corresponding to the largest area, we can find the value of y by going back to the equation relating x and y.
y = 400 - 4x/3 = 400 -4(150)/3 = 200.
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