Open QA 3

#$&*

course Mth 163

003.

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

*********************************************

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.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

To solve the quadratic function for the value of x we use the formula :

X = [-b +- `sqrt(b^2 - 4ac) / 2a

X = [-5.3 +- `sqrt(5.33333^2 - 4(-0.45833)(-6.875)) / 2(-0.45833)]

X = [-5.3 +- `3.97 / -0.91667]

X = [-5.3 +- (-4.3)]

X = -9.6 or -1

confidence rating #$&*:

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

.............................................

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): ?????I believe I have input the equation correctly but I’m unsure. When you take the square root of (b - 4(a)(c)) do we then divide by 2 and proceed to multiply this number by (a), or do we divide by the product of 2 and A?????

------------------------------------------------

Self-critique rating:

@&

The quadratic formula is written above as

[ -b + `sqrt(b^2 - 4 a c ] / (2 a) .

It has been written like that for several years, but your question caused me to go back and start explaining the formula using that representation.

Which is when I realized that the expression isn't properly formed, which I imagine is the source of the confusion.

The expression should have been

[ -b + `sqrt(b^2 - 4 a c) ] / (2 a) .

(previously the parenthesis after the sqrt wasn't closed).

So you first evaluate b^2 - 4 a c inside the (now-closed) parentheses.

Then you take the square root.

You're still within the braces [ ] so you then add the square root to -b. This gives you the numerator of the expression.

Now you evaluate the 2 a, which is in parentheses. This result is the denominator.

Finally you divide numerator by denominator.

Evaluating

[ -b - `sqrt(b^2 - 4 a c) ] / (2 a)

in an analogous manner you get the other solution.

*@

*********************************************

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: I do not recall from exactly which sketch in assignment one we are pulling this data from, therefore I cannot validate my results from the previous problem, I am not sure if there is a relation between the two.

dence Assessment:

.............................................

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): Oh I see, I made an arithmetic error, I obtained values of -1 and -9, which should have been 1.4768 and 10.16006 if I had not divided up the order of operations in my problem on the calculator, and instead transferred over the radicals, my answers would have been closer to yours. I plotted the equation on a graph program and I see how the graph peaks and where the values were obtained from. I can see that my values were very close…

------------------------------------------------

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:

Halfway between the x values is 1.48+10.16 / 2 =

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

Y = 8.642

confidence rating #$&*:

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

.............................................

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):

If the x value 5.82 and y value 8.64 are plotted on the graph of y= - 0.45833x^2 + 5.33333x - 6.875 the point is at the peak of the curve.

------------------------------------------------

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 = -5.33 /(2(-0.458)

X = 5.82

confidence rating #$&*:

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

.............................................

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): I was unknowingly working ahead on that one. It’s a good feeling.

------------------------------------------------

Self-critique rating:

@&

Not the first time you've done so. Very good.

*@

*********************************************

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:

Shifted to the left the original value of 5.82 becomes 4.82 and shifted to the right becomes 6.82

y = - 0.45833(4.82)^2 + 5.33333(4.82) - 6.875

y=8.1835

y = - 0.45833 (6.82)^2 + 5.33333 (6.82) - 6.875

y=8.1803

These y value differ by 0.0032, quite a small amount.

We get near the same number, which in the solution you said we should get the same number, but mine weren’t exactly the same. The difference in your value 8.1818 with original y value of 8.6402 = -0.4584

the same number for our A value

confidence rating #$&*:

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

.............................................

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): The vertex of a quadratic function, if moved to the left or right, the resulting values when plugged into x for the quadratic function will give the same number, which is always the value of A in the quadratic function. The quadratic function is taken from 3 points on the graph.

------------------------------------------------

Self-critique rating:

*********************************************

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:

The vertex of this parabola will be at x = -b / (2a)

X = -10/(2*-1)

X = 5

The y value will therefore be

125

Plotting this on a graph and shifting the vertex left or right 1 gives us 3 points

(4, 124), (5, 125), and (6, 124).

confidence rating #$&*:

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

.............................................

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): A negative A value results in a parabola with its two curves moving in the downward direction.

------------------------------------------------

Self-critique rating:

""

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

""

"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#*&!

@&

Very well done. Check my notes.

*@