#$&* course Mth163 6/13/12 1:56p Question: `q001. Note that this assignment has 6 questions
.............................................
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'm not sure what you asking me to compare between the last problem and the former problem??? ------------------------------------------------ Self-critique rating: ********************************************* 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: Estimating by my graph y = 0 when x = 1.7 and when y is at it's maximum point x=7 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ------------------------------------------------ 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: To figure out the halfway point between the two x values we can add 1.487 + 10.145 and divide by 2 to get 5.816. This wil be the x value of the vertex. So if we plug 5.816 as x into the equation y= -.45833 (5.816^2) + (5.33333)(5.816) - 6.875 we get y = 8.625 so the exact point of the parabole is (5.816, 8.625) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: The x value halfway between x = 1.48 and x = 10.16 is the average x = (1.48 + 10.16) / 2 = 5.82. At x = 5.82 we have y = - 0.45833 x^2 + 5.33333 x - 6.875 = -.45833 * 5.82^2 + 5.33333 * 5.82 - 6.875 = 8.64 approx.. Thus the graph of the function will be a parabola whose maximum occurs at its vertex (5.82, 8.64). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. The function y = a x^2 + b x + c has a graph which is a parabola. This parabola will have either a highest point or a lowest point, depending upon whether it opens upward or downward. In either case this highest or lowest point is called the vertex of the parabola. The vertex of a parabola will occur when x = -b / (2a). At what x value, accurate to five significant figures, will the function y = - 0.458333 x^2 + 5.33333 x - 6.875 take its maximum value? Accurate to five significant figures, what is the corresponding y value? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x= -5.33333 / 2(-.458333) is equal to x = 5.8 will be the maximum value y= -.458333 (5.8^2) + 5.33333 (5.8) - 6.878 y= -15.4 + 30.9 - 6.878 y= 8.622 (5.8, 8.622) will be the parabola confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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):ok ------------------------------------------------ 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: At 1 unit to the right of the vertx the x point will equal 6.8182, at 1 unit left the x will equal 4.8182 y will equal 8.177 when x =6.8182, and y will equal 8.185 when x= 4.8182 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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):ok ------------------------------------------------ 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: Vertex equals x= -10/ 2(-1) so x= 5 when x= 5 then y= -1 (5^2) + 10(5) + 100 so y= 125 vertex equals (5,125) If y differs by the value of the coefficient of a which is x^2 when moving right or left then 5^2= 25 the points would be (-20,125) when moving 1 unit to the left and (30,125) when moving 1 unit to the right for the new vertexs 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): ???I was confused about the coefficient of ""a"". I see how we came to the correct answer using the quadratic formula, but I was trying to use the method of x^2 will equal the distance moved vertically. I became confused and moved the x axis. But still 5^2= 25, and we aren't moving on either axis by 25 units. Maybe you can explain how x^2 is used better to me??? ------------------------------------------------ Self-critique rating:3 "