Answer:
See below because there are 9 parts (A through I)
Explanation:
Part A: write an equation that models the area of the figure. Let y represent the area, and write your answer in the form y = ax2 + bx + c.
The figure shows a rectangular table with these dimensions:
- Length: - x + 64
- Witdth: x + 4
The area of a rectangle is width × length:
- [tex](x + 4)\times (-x+64)[/tex]
Use distributive property:
- [tex]x\cdot (-x)+x\cdot(64)+4\cdot (-x)+4\cdot (64)=-x^2+64x-4x+256[/tex]
Simplify:
- [tex]-x^2+64x-4x+256=-x^2+60x+256[/tex]
Part B. Graph the equation you wrote in part A. Adjust the zoom of the graphing window so the vertex, x-intercepts, and y-intercept can be seen.
1. Factor the equation:
[tex]-x^2+60x+256=-(x^2-60x-256)[/tex]
- Find two numbers that add - 60 and whose product is -256. Theyb are -64 and + 4
[tex]-(x-64)(x+4)[/tex]
2. Find the roots:
Equal the expression to zero:
[tex]-(x-64)(x+4)=0\\ \\ x-64=0\implies x=64\\ \\ x+4=0\implies x=-4[/tex]
Those are the x-intercepts: (-4,0) and (64,0)
3. Find the symmetry axis:
The simmetry axis is the line x = the middle value between the two roots:
[tex]x=(64-4)/2=60/2=30[/tex]
4. Find the vertex
The vertex has x-coordinate equal to the x axis (30 in this case).
Substitute in the equation of find the y-coordinate:
[tex]y=-(30-64)(30+4)=-(-34)(34)=1,156[/tex]
Hence, the vertex is (30, 1,156)
5. Find the y-intercept
Make x = 0
[tex]y=-(x^2-60x-256)=-(0-256)=256[/tex]
Hence, the y-intercept is (0, 256)
With the x-incercepts, the y-intercept, the axis of symmetry, and the vertex, you can sketch the graph.
You can see now the graph in the attached figure
Part C. Extreme location of the graph
The graph shows that the parabola opens downward. That is due to the fact that the coefficient of the leading term (x²) is negative.
The parabola starts in the second quadrant. starts growing, crosses the x-axis at (-4,0), crosses the y-axis at (0,256), reaches the maximum value at (30, 1156), and then decreases toward the fouth quadrant, crossing the x-axis at (64,0).
Thus the vertex is a maximun, and the coordinates of the maximum are (30, 1156).
Part D. According to the graph, what is the maximum possible area of the game board? Give your answer to the nearest whole number. (Assume that the maximum area is not reduced by the open hole in the game board.)
The maximum possible area of the game is the maximum value of the function y = -x² + 60x + 256.
This value was calculated as y = 1156.
Part E. Use the original expressions for the length and width, and substitute the x-coordinate from the extreme location. What are the length and width of the game board at the extreme location?
The length is:
- length = - 30 + 64 = 34 inches
The width is:
- width = 30 + 4 = 34 inches
Part F. What type of quadrilateral will be formed when the game board covers the maximum possible area?
Since the length and the width are equal, the quadrilateral is a square.
Part G. Suppose the carnival director asks you to create a game board that is 1,120 square inches. Find the dimensions that would meet this request by setting the area equation equal to 1,120, solving for x, and substituting x into the expressions for the length and width.
[tex]y=-x^2+60x+256\\ \\ 1,120=-x^2+60x+256\\ \\ x^2-60x-256+1120=0\\ \\ x^2-60x+864=0[/tex]
Factor:
Find two numbers whose sum is - 60 and the product os 864. They are -24 and - 34:
[tex]x^2-60x+864=(x-24)(x-36)[/tex]
Use the zero product rule:
[tex](x-24)(x-36)=0\\ \\ x-24=0\implies x=24\\ \\ x-36=0\implies x=36[/tex]
Now substitute to find the dimensions:
x = 36
- width = x + 4 = 36 + 4 = 40
Hence, legth = 28, width = 40
x = 24
- length = - x + 64 = -24 + 64 = 40
- width = x + 4 = 24 + 4 = 28
Part H. When you solved the area equation for x, did any extraneous solutions result? Describe how an extraneous solution would arise in this situation.
The two solutions are valid (non extraneous) because both leads to positive real dimensions for which the areas can be 1,120 in².
An extraneous solution could arise if you try to find areas for which x is greater than or equal to 64, because in that case - x + 64 would be zero or negative and dimensions must be positive.
For the same reason, also an extraneous solution would arise if you try to fix areas for which x is less than or equal to - 4.
So, the domain of your function has to be - 4 < x < 64.
Part I. What method of solving quadratics did you use to solve the equation set equal to 1,120? Why did you choose this method?
The method use was factoring.
Discuss the usefulness of other methods of solving quadratics as they pertain to this scenario.
The other importants methods are graphical and the quadratic equation.
For graphical method you graph your parabola and find the values of x that sitisfies the area searched (value of y).
The quadratic equation gives the y-values (areas) without factoring:
[tex]\frac{-b+/-\sqrt{b^2-4(a)(c)} }{2(a)}[/tex]
