course Mth 163
02/07 2:34pm
If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
003.
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Question: `q001. Note that this assignment has 6 questions
The function y = a x^2 + b x + c takes the value y = 0 when x = [ -b + `sqrt(b^2 - 4 a c ] / (2 a) or when x = [ -b - `sqrt(b^2 - 4 a c ] / (2 a).
For the function y = - 0.45833 x^2 + 5.33333 x - 6.875, which you obtained as a quadratic model of the points (1, -2), (3, 5) and (7, 8) in the preceding assignment, find the values of x for which y = 0.
Compare to the estimates you made from the graph through (1,-2), (3, 5) and (7, 8) in Assignment 1.
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Your solution:
X=[-5.33333+`sqrt(5.33333^2-4(-0.45833)(-6.875)]/(2*-0.45833)
X=[-5.33333+ `sqrt(28.4444-4(-7.33333)]/.916666
(-0.45833)(-6.875) is about 3, not -7.3333. It looks like you added instead of multiplying.
In the step below, if the -7.3333 was correct, then your calculation 28.3 - 4 * (-7.333) would be about 28 + 29 = 57; wouldn't be negative.
X=[-5.33333+`sqrt(-57.778689)/0.916666
X=-5.33333+7.60122/0.916666
X=2.26789/0.916666
X=2.47
X=[-5.33333-`sqrt(5.33333^2-4(-0.45833)(-6.875)]/2(-0.45833)
X=[-5.33333-`sqrt(28.4444-4(-7.33333)/-0.916666
X=[-5.33333-`sqrt(28.4444-29.33332)/-0.916666 I have gone wrong right here and get back to where I need to be, I discovered that I had used the wrong value for a, when I corrected that I am now really confused.
X=[-5.33333-`sqrt(22.27772)/-0.916666
X=-5.33333-4.719992/-0.916666
X=-10.05332/-0.916666
X=-10.9699
If Y is 0 then x would be2.47 or -10.969
confidence rating #$&* -1
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Given Solution:
For the function y = - 0.45833 x^2 + 5.33333 x - 6.875 we have a = -0.45833, b = 5.33333 and c = -6.875. The quadratic formula therefore tells us that for our function we have y = 0 when
x = [-5.33333 + `sqrt(5.33333^2 - 4 * (-0.45833 ) * (-6.875)) ] / ( 2 * (-0.45833)) = 1.47638 and when
x = [-5.33333 - `sqrt(5.33333^2 - 4 * (-0.45833 ) * (-6.875)) ] / ( 2 * (-0.45833)) = 10.16006.
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Self-critique (if necessary):
I worked this problem several times and came up some different answers this is as close as I could get to the solution given. I think it is probably a very basic order of operation error or bad math, but I can’t find what I did wrong. I tried to put all the steps in my solution so that perhaps you could tell me where I went wrong.
see the note I inserted in your solution; procedure was fine, just a couple of arithmetic errors
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Self-critique rating #$&* Not OK
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Question: `q002. Extend the smooth curve in your sketch to include both points at which y = 0. Estimate the x value at which y takes its maximum value.
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Your solution:
When I extend the curve it does come out to a nice curve and my best guess is about 5.5 using the numbers that you have for your solution rather than my answers. I am very confused about how to explain this without the graph.
confidence rating #$&* -1
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Given Solution:
Your graph should clearly show how the parabola passes through the x axis at the points where x is approximately 1.5 (corresponding to the more accurate value 1.47638 found in the preceding problem) and where takes is a little more than 10 (corresponding to the more accurate value 10.16006 found in the preceding problem).
The graph of the parabola will peak halfway between these x values, at approximately x = 6 (actually closer to x = 5.8), where the y value will be between 8 and 9.
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Self-critique (if necessary): I don’t really understand this at all is there something I get to read or are there other problems like this that I can try to do and see if I can make sense out of this.
Your estimate of 5.5 is good.
What specifically do you not understand about each sentence in the given explanation?
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Self-critique rating #$&* Not OK
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Question: `q003. For the function of the preceding two questions, y will take its maximum value when x is halfway between the two values at which y = 0. Recall that these two values are approximately x = 1.48 and x = 10.16. At what x value will the function take its maximum value? What will be this value? What are the coordinates of the highest point on the graph?
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Your solution:
10.16+1.48/2=11.64/2=5.82
X=5.82 so y?
Y=-0.45833(5.82)^2+5.33333(5.82)-6.875
Y=-0.45833(33.8724)+5.33333(5.82)-6.875
Y=-15.52473+31.03998-6.875
Y=8.6402
confidence rating #$&* 1/2
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Given Solution:
The x value halfway between x = 1.48 and x = 10.16 is the average x = (1.48 + 10.16) / 2 = 5.82.
At x = 5.82 we have y = - 0.45833 x^2 + 5.33333 x - 6.875 = -.45833 * 5.82^2 + 5.33333 * 5.82 - 6.875 = 8.64 approx..
Thus the graph of the function will be a parabola whose maximum occurs at its vertex (5.82, 8.64).
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Self-critique (if necessary):I took a guess at the first part and got lucky with the second part.
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Self-critique rating #$&*Shaky
It doesn't look bad at all from here.
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Question: `q004. The function y = a x^2 + b x + c has a graph which is a parabola. This parabola will have either a highest point or a lowest point, depending upon whether it opens upward or downward. In either case this highest or lowest point is called the vertex of the parabola. The vertex of a parabola will occur when x = -b / (2a).
At what x value, accurate to five significant figures, will the function y = - 0.458333 x^2 + 5.33333 x - 6.875 take its maximum value? Accurate to five significant figures, what is the corresponding y value?
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Your solution:
Y=ax^2+bx+c
X=-b/2(-0.45833)
X=-5.33333/-0.91666
X=-6.24999
x would be positive; you divided a negative by a negative, so the result is positive
then you plugged the negative result in below, which of course will give you a completely different answer than if you have used x = +6.25.
Y=-0.458333(-6.24999)^2+5.33333(-6.24999)+-6.875
Y=39.06237+-0.91666+(-7.33333)
Y=30.31238
X=-6.249999
Y=30.312
confidence rating #$&* -1
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Given Solution:
In the preceding problem we approximated the x value at which the function is maximized by averaging 1.48 and 10.16, the x values at which the function is zero. Here we will use x = -b / (2 a) to obtain
x value at which function is maximized: x = -b / (2a) = - 5.33333 / (2 * -0.45833) = 5.81818.
To find corresponding y value we substitute x = 5.81818 into y = - 0.458333 x^2 + 5.33333 x - 6.875 to obtain y = 8.64024.
Thus the vertex of the parabola lies at (5.81818, 8.64024).
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Self-critique (if necessary):I got nothing even close to the solution given, I have shown the steps I took , please advise!!!!
I see only an arithmetic error, easily enough correct. Procedure is fine. See my note.
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Self-critique rating #$&* Not OK
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Question: `q005. As we just saw the vertex of the parabola defined by the function y = - 0.45833 x^2 + 5.33333 x - 6.875 lies at (5.8182, 8.6402).
What is the value of x at a point on the parabola which lies 1 unit to the right of the vertex, and what is the value of x at a point on the parabola which lies one unit to the left of the vertex?
What is the value of y corresponding to each of these x values?
By how much does each of these y values differ from the y value at the vertex, and how could you have determined this number by the equation of the function?
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Your solution: I am so confused I have no idea what to do next, I have looked at the solution and I just don’t get it.
confidence rating #$&*
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Given Solution:
The vertex lies at x = 5.8182, so the x values of points lying one unit to the right and left of the vertex must be x = 6.8182 and x = 4.8182. At these x values we find by substituting into the function y = - 0.458333 x^2 + 5.33333 x - 6.875 that at both points y = 8.1818.
Each of these y values differs from the maximum y value, which occurs at the vertex, by -0.4584. This number is familiar. Within roundoff error is identical to to the coefficient of x^2 in the original formula y = - 0.458333 x^2 + 5.33333 x - 6.875.
This will always be the case. If we move one unit to the right or left of the vertex of the parabola defined by a quadratic function y = a x^2 + b x + c, the y value always differ from the y value at the vertex by the coefficient a of x^2. Remember this.
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Self-critique (if necessary): I really need help to understand what I am doing.
You understand what you're doing; you're just making some arithmetic errors so your answers aren't matching the given solutions.
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Self-critique rating #$&*
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Question: `q006. In the preceding problem we saw an instance of the following rule:
The function y = a x^2 + b x + c has a graph which is a parabola. This parabola has a vertex. If we move 1 unit in the x direction from the vertex, moving either 1 unit to the right or to the left, then move vertically a units, we end up at another point on the graph of the parabola.
In assignment 2 we obtained the solution a = -1, b = 10, c = 100 for a system of three simultaneous linear equations. If these linear equations had been obtained from 3 points on a graph, we would then have the quadratic model y = -1 x^2 + 10 x + 100 for those points.
What would be the coordinates of the vertex of this parabola? What would be the coordinates of the points on the parabola which lie 1 unit to the right and one unit to the left of the vertex?
Sketch a graph with these three points, and sketch a parabola through these points. Will this parabola ever touch the x axis?
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Your solution:
confidence rating #$&*
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Given Solution:
The vertex of the parabola given by y = -1 x^2 + 10 x + 100 will lie at x = -b / (2 a) = -10 / (2 * -1) = 5.
At the vertex the y value will therefore be
y = -1 x^2 + 10 x + 100 = -1 * 5^2 + 10 * 5 + 100 = 125.
It follows that if we move 1 unit in the x direction to the right or left, the y value will change by a = -1. The y value will therefore change from 125 to 124, and we will have the 3 'fundamental points' (4, 124), (5, 125), and (6, 124).
Your graph should show a parabola peaking at (5, 125) and descending slightly as we move 1 unit to the right or left of this vertex. The parabola will then descend more and more rapidly, eventually crossing the x-axis both to the left and to the right of the vertex.
The points to the right and left show clearly that the parabola descends from its vertex. This is because in this case a = -1, a negative value, which effectively pulls the parabola down on either side of the vertex. Had the value of a been positive, the points one unit to the right and left would lie above the vertex and the parabola would asscend from its vertex.
Self-critique (if necessary)
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Self-critique rating #$&*
See my notes. I think you're doing very well, but arithmetic errors (which are fortunately simple to correct) are throwing your answers off.
&#Please let me know if you have questions. &#