course 163 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: `a** The 'black' graph takes values 8, 3, 0, -1, 0, 3, 8 at x = -3, -2, -1, 0, 1, 2, 3. The 'blue' graph takes approximate values 1.7, .8, .2, -.1, -.4, -.6, -.8 at the same x values. The 'blue' graph takes value zero at approximately x = -.4. The sum of the two graphs will coincide with the 'blue' graph where the 'black' graph is zero, which occurs at x = -1 and x = 1. The sum will coincide with the 'black' graph where the 'blue' graph is zero, which occurs at about x = -.4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok, I don’t understand and were are you getting this from?
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Given Solution: `a** The sum of the graphs is higher than the 'black' graph where the 'blue' graph is positive, lower where the 'blue' graph is negative. The 'blue' graph is positive on the interval from x = -3 to x = -.4, approx.. This interval can be written [-3, -.4), or -3 <= x < -.4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Blue graph is positive on the interval from x = -3 to x =-.4, interval can be written [-3,-.4), or -3<=x<-.4 ???? how did you get your intervals and am I on the right graph ???? Self-critique rating: ********************************************* Question: `qWhere it is the sum graph higher than the 'blue' graph, and where is it lower? Answer by giving specific intervals. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The sum of the blue graph is higher than the blue graph when the black graph is on the positive side and it is lower than the blue graph when the black graph is on the negative side. confidence rating: mostly ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The sum of the graphs is higher than the 'blue' graph where the 'black' graph is positive, lower where the 'black' graph is negative. The 'black' graph is positive on the interval from x = -1 to x = 1, not including the endpoints of the interval. This interval can be written (-1, 1) or -1 < x < 1. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): the black graph is positive on the interval from x=-1 to x=1, not including endpoints of the interval, can be written (-1,1) or -1< x < 1. Self-critique rating: ********************************************* Question: `qWhere does thus sum graph coincide with the 'black' graph, and why? Give your estimate of the specific coordinates of the point or points where this occurs. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: on the second graph of or sheet the sum doesn’t coincide with the black graph at all because coincide means to occupy the same space at the same time and they don’t. confidence rating: not sure ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The sum coincides with the 'black' graph where the 'blue' graph is zero, which occurs at about x = -.4. The coordinates would be about (-.4, -.7), on the 'black' graph. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok, but I think we are looking at 2 different graphs. Self-critique rating: ********************************************* Question: `qWhere does thus sum graph coincide with the 'blue' graph, and why? Give your estimate of the specific coordinates of the point or points where this occurs. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the sum coincides with the blue graph when the black graph is (0,0) confidence rating: unsure ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The sum coincides with the 'blue' graph where the 'black' graph is zero, which occurs at x = -1 and x = 1. The coordinates would be about (-1, .2) and (1, -.4), on the 'blue' graph. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ???? Mr. Smith it would help me if I knew what Graph you are looking at ????
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Given Solution: `a** The 'black' graph is periodic, passing through 0 at approximately x = -3.1, 0, 3.1, 6.3. This graph has peaks with y = 1.5, approx., at x = 1.6 and 7.8, approx., and valleys with y = -1.5 at x = -1.6 and x = 4.7 approx. The 'blue' graph appears to be parabolic, passing thru the y axis at x = -1 and reaching a minimum value around y = -1.1 somewhere near x = 1. This graph passes thru the x axis at x = 5.5, approx., and first exceeds y = 1 around x = 7.5. The quotient will be further from the x axis than the 'black' graph wherever the 'blue' graph is within 1 unit of the origin, since division by a number whose magnitude is less than 1 gives a result whose magnitude is greater than the number being divided. This will occur to the left of x = 1, and between about x = 2 and x = 7.5. Between about x = 0 and x = 1 the 'blue' graph is more than 1 unit from the x axis and the quotient graph will be closer to the x axis than the 'black' graph. The same is true for x > 7.5, approx.. The 'black' graph is zero at or near x = -3.1, 0, 3.1, 6.3. At both of these points the 'blue' graph is nonzero so the quotient will be zero. The 'blue' graph is negative for x < 5.5, approx.. Since division by a negative number gives us the opposite sign as the number being divided, on this interval the quotient graph will be on the opposite side of the x axis from the 'black' graph. The 'blue' graph is positive for x > 5.5, approx.. Since division by a positive number gives us the same sign as the number being divided, on this interval the quotient graph will be on the same side of the x axis as the 'black' graph. The quotient graph will therefore start at the left with positive y values, about 3 times as far from the x axis as the 'black' graph (this since the value of the 'blue' graph is about -1/3, and division by -1/3 reverses the sign and gives us a result with 3 times the magnitude of the divisor). The quotient graph will have y value about 2.5 at x = -1.6, where the 'black' graph 'peaks', but the quotient graph will 'peak' slightly to the left of this point due to the increasing magnitude of the 'blue' graph. The quotient graph will then reach y = 0 / (-1) = 0 at x = 0 and, since the 'black' graph then becomes positive while the 'blue' graph remains negative, the quotient graph will become negative. Between x = 0 and x = 2 the magnitude of the 'blue' graph is a little greater than 1, so the quotient graph will be a little closer to the x axis than the 'black' graph (while remaining on the other side of the x axis). At x = 3.1 approx. the 'black graph is again zero, so the quotient graph will meet the x axis at this point. Past x = 3.1 the quotient graph will become positive, since the signs of both graphs are negative. As we approach x = 5.5, where the value of the 'blue' graph is zero, the quotient will increase more and more rapidly in magnitude (this since the result of dividing a negative number by a negative number near zero is a large positive number, larger the closer the divisor is to zero). The result will be a vertical asymptote at x = 5.5, with the y value approaching +infinity as x approaches 5.5 from the left. Just past x = 5.5 the 'blue' values become positive. Dividing a negative number by a positive number near zero results in a very large negative value, so that on this side of x = 5.5 the asymptote will rise up from -infinity. The quotient graph passes through the x axis near x = 6.3, where the 'black' graph is again zero. To the right of this point both graphs have positive values and the quotient graph will be positive. Around x = 7.5, where the 'blue' value is 1, the graph will coincide with the 'black' graph, giving us a point near (7.5, 1.3). Past this point the 'blue' value is greater than 1 so that the quotient graph will become nearer the x axis than the 'black' graph, increasingly so as x (and hence the 'blue' value) increases. This will result in a 'peak' of the quotient graph somewhere around x = 7.5, a bit to the left of the peak of the 'black' graph. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Blue graph is parabolicand passes through the x axis at x=-1 and reaches minimum value at x = 7.5 Quotent graph y values = 2.5 at x=-1.6 where the black peaks and thequotent peaks a little to the left.quotent will reach y = 0/(-1) = 0 at x = 0, because black will be the oppposite the quotent will become negative. Quotent passes through x axis near x = 6.2, black = 0, right past this point the blavk and blue graphs have a positive value and the quotent is positive At x = 7.5, blue value is 1 , the graph will coincide with black with the points (7.5, 1.3) Past this point blue value is greater than 1, quotent comes nearer to x axis than black. Graph increases so x increases resulting in peak of quotent at x = 7.5 slightly to the left of peak of black. Self-critique rating: ********************************************* Question: `qQuery problems 7-8 Sketch the graph of y = x^2 - 2 x^4 by first sketching the graphs of y = x^2 and y = -2 x^4. How does the result compare to the graph of y = x^2 - x^4, and how do you explain the difference? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Using x =-2 through 2 the graph of y = x^2 has a zero of (0,0) and points to the left (-2,4) and (-1,1) and to the right (1,1) and (2,4) well call this blue the graph y = -2x^4 is a parabola sharing the same zero as y =x^2 of (0,0) and points to the left of (-2, 32) and (-1,4) and to the right (1,4) and (2,32) well call this black the graph of y= x^2 – 2x^4 we have the same zeros as the other two of (0,0) and points to the left (-2,28) and(-1,1) and to the right (1,-1) and(2,-28) we will call this red when the blue was negative and it minused a neg black it gave a positive red number when the blue was positive and it minused a negative black it gave a negative re number. The red graph increases on the left in the negative and increases on the right in the negative confidence rating: unsure ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** At x = 0, 1/2, 1 and 2 we have x^2 values 0, 1/4, 1 and 4, while -x^4 takes values 0, -1/16, -1 and -16, and -2 x^4 takes values 0, -1/8, -2 and -32. All graphs clearly pass through the origin. The graphs of y = x^2 - x^4 and y = x^4 - 2 x^4 are both increasingly negative at far right and far left. Graphical addition will show that y = x^2 - x^4 takes value 0 and hence passes thru the x axis when the graphs have equal but opposite y values, which occurs at x = 1 and x = -1. To the left of x = -1 and to the right of x = 1 the negative values of -x^4 overwhelm the positive values of x^2 and the sum graph will be increasingly negative, with values dominated by -x^4. Near x = 0 the graph of y = -x^4 is 'flatter' than that of y = x^2 and the x^2 values win out, making the sum graph positive. y = x^2 - 2 x^4 will take value 0 where the graphs are equal and opposite in value; this occurs somewhere between x = .8 and x = .9, and also between x = -.9 and x = -.8, which places the zeros closer to the y axis than those of the graph of y = x^2 - x^4. The graph of y = -2 x^4 is still flatter near x = 0 than the graph of y = x^2, but not as flat as the graph of y = -x^4, so while the sum graph will be positive between the zeros the values won't be as great. Outside the zeros the sum graph will be increasingly negative, with values dominated by -2x^4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok, its hard for me to self cretique if we have different data Self-critique rating: ********************************************* Question: `qHow does the shape of the graph change when you add x to get y = -2 x^4 + x^2 + x, and how do you explain this change? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: All graphs pass through (0,0) The y = x^2 – 2 x^4 increases more and more as the x values increase. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** At x = 0 there is no change in the y value, so the graph still passes through (0, 0). As x increases through positive numbers we will have to increase the y values of y = x^2 -2 x^4 by greater and greater amounts. So it will take a little longer for the negative values of -2 x^4 to 'overwhelm' the positive values of x^2 + x than to overcome the positive values of x^2 and the x intercept will shift a bit to the right. As we move away from x = 0 through negative values of x we will find that the positive effect of y = x^2 is immediately overcome by the negative values of y = x, so there is no x intercept to the left of x = 0. The graph in fact stays fairly close to the graph of y = x near (0, 0), gradually moving away from that graph as the values of x^2 and -2 x^4 become more and more significant. ** Query Add comments on any surprises or insights you experienced as a result of this assignment. none &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): It taks longer for the negative value of -2x^4 to overwhelm the positive value of x^2 +x than to overcome positive value of x^2, x interceptshifts to the right As it moves away from x = 0 through negative values of x, we find that the positive effect of y =x^2 is immediately overcome by negative value of y =x, no x intercept to the leftof x = 0. Self-critique rating: