ASSN 3 qa

#$&*

course MTH163

003.

NOTE THAT THE FIRST QUESTIONS IN THIS ASSIGNMENT CONTINUE q_a_ ASSIGNMENT 2. ********************************************* Question: `q001. Note that this assignment has 9 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A=-0.45833 ,b=5.33333, c=-6.875 X= -b +/- sqrt b^2 -4ac 2a X= 1.476379797 , x= 10.16006119 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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. STUDENT COMMENT This whole problem confuses me. I put in these numbers in my calculator but I get a different answer everytime. INSTRUCTOR RESPONSE It is possible your calculator doesn't follow the order of operations. Most do, but some do not. It is also possible that you are making an error with the order of operations. You do have some out-of-place parentheses (e.g., in the expression 4 * ( -0.4583) * ) -6.875)). You should evaluate the various quantities separately, then put them together. For example to calculate [ -5.33 + square rt (5.33^2 -4 * ( -0.4583) * ) -6.875) ] / (2 * (-0.45833) , begin by calculating 4 * ( -0.4583) * -6.875, then calculate 5.33^2, then subtract, then take the square root. Calculate 2 * (-0.45833) . Combine your results to calculate [-5.33333 + `sqrt(5.33333^2 - 4 * (-0.45833 ) * (-6.875)) ] / ( 2 * (-0.45833)). If you tell me what you enter into your calculator at each step, and what you get, I can tell you if you're making an error and, if so, how to avoid it in the future. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Originally I had no clue how to solve this, however I reached out to vhcc and they have an online tutoring program . I had a session with a tutor today and they helped me understand this equation. Self-critique rating:

@&

@& Very good.

You should go back and review the full if-then statement of the quadratic formula as presented in the first assignment (and which you are required to know verbatim). Having worked out this example you'll understand that statement much better. *@

*@

********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ? confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand how to graph the two (x) point I just solved to get, however I am unclear on where the other points are coming from.

@& This question needs to be answered based on your sketch. You should have a graph with a smooth curve, and it should be possible to extend that curve, following its trend, until you pass through a point where y = 0. *@

------------------------------------------------ Self-critique rating:

********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: X=1.48, x=10.16 X=91.48+10.16)/2 =5.82 Ax^2+bx+c -0.45833 (5.82)^2 +5.33333(5.82)-6.875 =8.64 Max occurs at (5.82,8.64) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I first didn’t understand what I was supposed to do, but after looking at the solution I now understand what you want. Self-critique rating:

********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: X=-b/ (2a) X=-5.33333/2(-0.458333) X=5.81822 Y= -0.458333(5.81822)^2 + 5.33333(5.81822)-6.875 Y=8.64014 Vertex (5.81822, 8.64014) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Understood ------------------------------------------------ Self-critique rating:

********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Shift 1 right = 6.8182 Shift 1 left = 4.8182 Plugging both into the equation y= - 0.45833 x^2 + 5.33333 x - 6.875 Y= 8.1818 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? ----this I don’t really understand. Could you dumb down your reasoning for me??? 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Good except the last part about : 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. *** this I don’t really understand. ------------------------------------------------ Self-critique rating:

@& Your x values 1 unit to the right and left of the vertex are

Shift 1 right = 6.8182 Shift 1 left = 4.8182

as you have found.

Both of these x values give you your y value 8.1818.

The vertex lies at (5.8182, 8.6402).

By how much does the y value 8.1818 differ from the y value at the vertex?

Based on this result and on your graph, which you should have sketched, does it make sense that moving 1 unit in either direction from the vertex changes the y value by -.45833?

You are welcome to submit a copy of this note and your response for review using the question form. *@

********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y=-1x^2+10x+100 X= -b/(2a) X=5 Y=-1(5)^2 + 10(5) +100 Y=125 -shift right = (6,124) -shift left = (4,124) 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. STUDENT COMMENT This problem was a little confusing. I wasn’t really sure the point of seeing if the parabola would touch the x axis? INSTRUCTOR RESPONSE On the x axis, the y value is zero. The points where the graph intersects the x axis are called the zeros of the function. The quadratic formula gives you the zeros of the function a x^2 + b x + c. These points are very important in applications, as you will see very soon. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Can you explain alittle more why the (y) goes from 125 to 124 when the problems we were doing in assignment 2 the (y) didn’t change??

@& When the value of x changes, the value of y typically changes. Otherwise the graph would be a horizontal straight line. *@

@& In this case the vertex lies at x = 5, and the y value at the vertex is 125.

If you move 1 unit to the right, you get x = 6, and you can verify that the corresponding y value is 124. *@

Also, it would be very difficult to make a graph with values of 125 and 124, is there an easier way to know if it will cross the x axis?? ------------------------------------------------ Self-critique rating:

@& It isn't difficult to sketch this graph. Just mark off the y axis in intervals of 10, and the x axis in intervals of 1.

Graphs typically need to be scaled to fit the situation, and the scale of the y axis will typically differ from the scale of the x axis. *@

If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties. ********************************************* Question: `q007. Sketch a parabola through points (-4, 4), (1, -1) and (2, 4). The parabola should be symmetric about some vertical line. Estimate the coordinates at which the parabola passes through the x axis. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First intercept is around -3.2 ish and the other about 1.1 ish confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q008. Solve the quadratic equation x^2 + 2 x - 4 = 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: X= 1.23606798 ,x =-3.236067978

@& You haven't shown the steps of your solution, but I trust that you worked this out correctly. *@

confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q009. Verify that the points (-4, 4), (1, -1), (2, 4) lie on the graph of the parabola y = x^2 + 2 x - 4. What is the value of y at the vertex of this parabola? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: -b/2a -2/2 X=-1 Y=x^2+2x-4 Y=(-1)^2 +2(1) -4 Y=1-6 Y=-5 Vertex (-1,-5)

@& Very good. You have the correct vertex. But you haven't verified that the given points lie on the parabola.

So do so you would plug the x coordinate of each point into the function and evaluate it, verifying that the resulting y values matche the y coordinates of the corresponding points. *@

Confidence rating: 3 ------------------------------------------------ Self-critique rating:

I have been using an online tutor through vhcc and I believe it is helping !!  "

Self-critique (if necessary): ------------------------------------------------ Self-critique rating:

I have been using an online tutor through vhcc and I believe it is helping !!  "

Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!

@& You are asking good questions. Check my responses.

You are also making a very good effort. Keep it up. *@

ASSN 3 qa

#$&*

course MTH163

003.

NOTE THAT THE FIRST QUESTIONS IN THIS ASSIGNMENT CONTINUE q_a_ ASSIGNMENT 2.

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Question: `q001. Note that this assignment has 9 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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

A=-0.45833 ,b=5.33333, c=-6.875

X= -b +/- sqrt b^2 -4ac

2a

X= 1.476379797 , x= 10.16006119

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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.

STUDENT COMMENT

This whole problem confuses me. I put in these numbers in my calculator but I get a different answer everytime.

INSTRUCTOR RESPONSE

It is possible your calculator doesn't follow the order of operations. Most do, but some do not.

It is also possible that you are making an error with the order of operations. You do have some out-of-place parentheses (e.g., in the expression 4 * ( -0.4583) * ) -6.875)).

You should evaluate the various quantities separately, then put them together. For example to calculate [ -5.33 + square rt (5.33^2 -4 * ( -0.4583) * ) -6.875) ] / (2 * (-0.45833) , begin by calculating

4 * ( -0.4583) * -6.875, then calculate 5.33^2, then subtract, then take the square root.

Calculate 2 * (-0.45833) .

Combine your results to calculate

[-5.33333 + `sqrt(5.33333^2 - 4 * (-0.45833 ) * (-6.875)) ] / ( 2 * (-0.45833)).

If you tell me what you enter into your calculator at each step, and what you get, I can tell you if you're making an error and, if so, how to avoid it in the future.

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Self-critique (if necessary):

Originally I had no clue how to solve this, however I reached out to vhcc and they have an online tutoring program . I had a session with a tutor today and they helped me understand this equation.

Self-critique rating:

@&

@&

Very good.

You should go back and review the full if-then statement of the quadratic formula as presented in the first assignment (and which you are required to know verbatim). Having worked out this example you'll understand that statement much better.

*@

*@

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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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

?

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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 understand how to graph the two (x) point I just solved to get, however I am unclear on where the other points are coming from.

@&

This question needs to be answered based on your sketch. You should have a graph with a smooth curve, and it should be possible to extend that curve, following its trend, until you pass through a point where y = 0.

*@

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Self-critique rating:

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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:

X=1.48, x=10.16

X=91.48+10.16)/2 =5.82

Ax^2+bx+c

-0.45833 (5.82)^2 +5.33333(5.82)-6.875

=8.64

Max occurs at (5.82,8.64)

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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 first didn’t understand what I was supposed to do, but after looking at the solution I now understand what you want.

Self-critique rating:

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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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

X=-b/ (2a)

X=-5.33333/2(-0.458333)

X=5.81822

Y= -0.458333(5.81822)^2 + 5.33333(5.81822)-6.875

Y=8.64014

Vertex (5.81822, 8.64014)

confidence rating #$&*: 3

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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):

Understood

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Self-critique rating:

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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?

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Shift 1 right = 6.8182

Shift 1 left = 4.8182

Plugging both into the equation y= - 0.45833 x^2 + 5.33333 x - 6.875

Y= 8.1818

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?

----this I don’t really understand. Could you dumb down your reasoning for me???

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):

Good except the last part about :

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.

*** this I don’t really understand.

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Self-critique rating:

@&

Your x values 1 unit to the right and left of the vertex are

Shift 1 right = 6.8182

Shift 1 left = 4.8182

as you have found.

Both of these x values give you your y value 8.1818.

The vertex lies at (5.8182, 8.6402).

By how much does the y value 8.1818 differ from the y value at the vertex?

Based on this result and on your graph, which you should have sketched, does it make sense that moving 1 unit in either direction from the vertex changes the y value by -.45833?

You are welcome to submit a copy of this note and your response for review using the question form.

*@

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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:

Y=-1x^2+10x+100

X= -b/(2a)

X=5

Y=-1(5)^2 + 10(5) +100

Y=125

-shift right = (6,124)

-shift left = (4,124)

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.

STUDENT COMMENT

This problem was a little confusing. I wasn’t really sure the point of seeing if the parabola would

touch the x axis?

INSTRUCTOR RESPONSE

On the x axis, the y value is zero. The points where the graph intersects the x axis are called the zeros of the function.

The quadratic formula gives you the zeros of the function a x^2 + b x + c.

These points are very important in applications, as you will see very soon.

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Self-critique (if necessary):

Can you explain alittle more why the (y) goes from 125 to 124 when the problems we were doing in assignment 2 the (y) didn’t change??

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When the value of x changes, the value of y typically changes. Otherwise the graph would be a horizontal straight line.

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In this case the vertex lies at x = 5, and the y value at the vertex is 125.

If you move 1 unit to the right, you get x = 6, and you can verify that the corresponding y value is 124.

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Also, it would be very difficult to make a graph with values of 125 and 124, is there an easier way to know if it will cross the x axis??

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Self-critique rating:

@&

It isn't difficult to sketch this graph. Just mark off the y axis in intervals of 10, and the x axis in intervals of 1.

Graphs typically need to be scaled to fit the situation, and the scale of the y axis will typically differ from the scale of the x axis.

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If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily. If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.

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Question: `q007. Sketch a parabola through points (-4, 4), (1, -1) and (2, 4). The parabola should be symmetric about some vertical line.

Estimate the coordinates at which the parabola passes through the x axis.

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Your solution:

First intercept is around -3.2 ish and the other about 1.1 ish

confidence rating #$&*:

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Question: `q008. Solve the quadratic equation x^2 + 2 x - 4 = 0.

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Your solution:

X= 1.23606798 ,x =-3.236067978

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You haven't shown the steps of your solution, but I trust that you worked this out correctly.

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confidence rating #$&*:

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Question: `q009. Verify that the points (-4, 4), (1, -1), (2, 4) lie on the graph of the parabola y = x^2 + 2 x - 4.

What is the value of y at the vertex of this parabola?

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Your solution:

-b/2a

-2/2

X=-1

Y=x^2+2x-4

Y=(-1)^2 +2(1) -4

Y=1-6

Y=-5

Vertex (-1,-5)

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Very good. You have the correct vertex. But you haven't verified that the given points lie on the parabola.

So do so you would plug the x coordinate of each point into the function and evaluate it, verifying that the resulting y values matche the y coordinates of the corresponding points.

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Confidence rating: 3

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Self-critique rating:

I have been using an online tutor through vhcc and I believe it is helping !! 

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Self-critique (if necessary):

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Self-critique rating:

I have been using an online tutor through vhcc and I believe it is helping !! 

"

Self-critique (if necessary):

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Self-critique rating:

#*&!

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You are asking good questions. Check my responses.

You are also making a very good effort. Keep it up.

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