Question 503728
The monthly value of a share of ACME Corporation, in dollars, since January 2009 are shown in the table below. The values of the table below model the behavior of a quadratic function when it is graphed.
Month.......... Share Value (in dollars)
0...... 55
3...... 16
5...... 0
7...... -8
9..... -8
11.... 0
13.... 16
15..... 40


1. Using the table above, find the quadratic equation that represents the monthly value of a share of ACME Corporation since January 2009. Use the variable x to represent the number of months after January 2009 and the variable y to represent the monthly value.
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Note that when x = 5 and x = 11, y = 0, therefore these are the x intercepts
(as the hint says); derive two factors and use FOIL to create a quadratic equation
(x-5)*(x-11) = x^2 - 11x - 5x + 55
y = x^2 - 16x + 55; is the equation
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2. Find the y-intercept. Interpret the meaning of the y-intercept in relation to the given real-life scenario.
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The y intercept occurs when x = 0, then y = 55
This would be the initial value in Jan 2009
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3. Find the x-intercepts. Interpret the meaning of the x-intercepts in relation to the given real-life scenario.
From this we have to assume the share value was par (0) in March and June
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4. From the equation you found in problem 1, what is the concavity of the parabola? How do you know? What is the lowest or maximum monthly value of a share of ACME Corporation and when did that occur?
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we know from the fact that the coefficient of x^2 is positive, it is concave and has a minimum
Here is a graph of that equation, you can see for yourself
{{{ graph( 300, 200, -6, 20, -30, 80, x^2-16x+55) }}}
you can see that the min value is in Aug, (x=8), when the value is -9
which is the vertex, 8,-9
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5. Using the quadratic equation you found in problem 1, predict the value of a share of ACME Corporation in October 2010?
Oct 2010, x = 22
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y = x^2 - 16x + 55
y = 22^2 - 16(22) + 55
y = 484 - 352 + 55
y = 187 in Oct 2010
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redraw our graph to show this
{{{ graph( 300, 200, -10, 30, -100, 220, x^2-16x+55) }}}