course Mth 163 ??????????rz?assignment #003003. Precalculus I 07-24-2008
......!!!!!!!!...................................
23:22:44 `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.
......!!!!!!!!...................................
RESPONSE --> if y = 0 x = -.126868 or -.87313 confidence assessment: 1
.................................................
......!!!!!!!!...................................
23:32:04 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.
......!!!!!!!!...................................
RESPONSE --> I found a mathematical error after rechecking my after my answer did not match the student solution. After recalculating my answer I found the same solution for x = 1.47627 or 10.160168 self critique assessment: 2
.................................................
......!!!!!!!!...................................
23:34:48 `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.
......!!!!!!!!...................................
RESPONSE --> 1 confidence assessment: 0
.................................................
......!!!!!!!!...................................
23:51:19 `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?
......!!!!!!!!...................................
RESPONSE --> I got a value of x= 5.81822, but when I try to find the vertex I come up with a negative coordinate for y = -17.0569. I can see the sketch that is described to the left, I also thought if a was a negative number the parabala would be inverted, which would make sense with the vertex coordinates (5.8, -17.06) I solved for. Am I off the mark and completely confused? confidence assessment: 1
.................................................
......!!!!!!!!...................................
23:56:44 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).
......!!!!!!!!...................................
RESPONSE --> I have found my mathematical error. I left out an x value when solving for y. I solved for y and got y = 8.6 self critique assessment: 3
.................................................
......!!!!!!!!...................................
23:58:42 `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?
......!!!!!!!!...................................
RESPONSE --> the vertex coordinates are (5.81822, 8.64024) confidence assessment: 3
.................................................
......!!!!!!!!...................................
23:59:13 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).
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
00:06:19 `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?
......!!!!!!!!...................................
RESPONSE --> The coordinates for 1 unit to the right of the vertex are (6.81822, 8.18191) and the coordinates for 1 unit to the left of the vertex are (4.81822, 8.18191) the y values do not differ confidence assessment: 2
.................................................
......!!!!!!!!...................................
00:06:53 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.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 2
.................................................
......!!!!!!!!...................................
00:21:15 `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?
......!!!!!!!!...................................
RESPONSE --> vertex= (5, 125) 1 unit R = (6, 124) 1 unit L = (4, 124) mathematically it should not cross the x axis confidence assessment: 1
.................................................
......!!!!!!!!...................................
00:27:32 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.
......!!!!!!!!...................................
RESPONSE --> It would cross the x axis at x values of -6.18035 and 16.18035 self critique assessment: 3
.................................................