course mth 163 end programvŊռ~cm݃
......!!!!!!!!...................................
08:13:12 `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 --> confidence assessment: 2
.................................................
......!!!!!!!!...................................
08:17:08 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 --> x= [-5.3333+'sqrt(5.3333^2 - 4*(-0.458330*(-6.875))] / (2*(-0.45833))= 1.47638 x=[-5.33333-'sqrt(5.3333^2-4*(-0.45833)* (-6.875))]/(2*(-0.45833))=10.16006 self critique assessment: 2
.................................................
......!!!!!!!!...................................
08:17:37 `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 --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
08:17:40 `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 --> confidence assessment:
.................................................
......!!!!!!!!...................................
08:18:39 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.
......!!!!!!!!...................................
RESPONSE --> x=5.5 y=8.5 self critique assessment: 2
.................................................
......!!!!!!!!...................................
08:24:21 `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 --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
08:28:03 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 --> x halfway between x=1.48 & x=10.16 = x= (1.48+ 10.16)/2= 5.82 x=5.82; y= -0.45833x^2+5.33333x-6.875= -.45833*5.82^2+5.33333*5.82-6.875= 8.64 max @ vertex= (5.82 , 8.64) self critique assessment: 3
.................................................
......!!!!!!!!...................................
08:31:37 `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 --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
08:40:55 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 --> x= -b/(2a); 5.33333/2(0.45833) = 5.81822 y = -0.45833(5.81822^2)+5.33333(5.81822)-6.875 y=8.64024 Vertex occurs between (5.81822 , 8.64024) self critique assessment: 3
.................................................
......!!!!!!!!...................................
08:42:10 `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 --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
08:52:06 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 --> x=5.81822 + 1=6.81822 x=5.81822 - 1= 4.81822 y=-0.45833(6.81822^2) + 5.33333(6.81822)-6.875 y=8.8189 self critique assessment: 3
.................................................
......!!!!!!!!...................................
09:00:30 `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 --> x=-b/(2a); 10/((2(-1)); x=5 y=-1(5^2)+10(5)+100 y=125 Vertex between= (5,125) 1 unit left and right; (5,124) will not touch x axis confidence assessment: 3
.................................................
......!!!!!!!!...................................
09:01:05 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 --> forgot to add the minus 1 unit... (4,124) self critique assessment: 3
.................................................
dÔztі愅 assignment #003 003. Precalculus I 08-12-2008
......!!!!!!!!...................................
09:04:59 `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 --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
09:05:36 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 --> x = [-5.33333 + `sqrt(5.33333^2 - 4 * (-0.45833 ) * (-6.875)) ] / ( 2 * (-0.45833)) = 1.47638; x = [-5.33333 - `sqrt(5.33333^2 - 4 * (-0.45833 ) * (-6.875)) ] / ( 2 * (-0.45833)) = 10.16006. self critique assessment: 3
.................................................
......!!!!!!!!...................................
09:05:46 `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 --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
09:06:19 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.
......!!!!!!!!...................................
RESPONSE --> x=6.5 y=8.5 self critique assessment: 3
.................................................
......!!!!!!!!...................................
09:07:04 `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 --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
09:09:24 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 --> 1.48+10.16/2= 5.82 y=-0.45833x^2+5.33333x-6.875 -.45833(5.82^2)+5.33333(5.82)-6.875 y=8.64 Vertex= (5.82 , 8.64) self critique assessment: 3
.................................................
......!!!!!!!!...................................
09:10:09 `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 --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
09:12:18 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 --> x=-b/2a; 5.33333/(2*-.45833) x=5.81822 y=-.45833(5.81822^2)+5.33333(5.81822)-6.875 y=8.64024 self critique assessment: 3
.................................................
......!!!!!!!!...................................
09:15:13 `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 --> y=-1x^2+10x+100 x=-b/(2a); x=5 y=-1(5^2)+10(5)+100 y=125 vertex= (5 , 125) +1 unit= (6, 124) -1 unit= (4 , 124) confidence assessment: 3
.................................................
......!!!!!!!!...................................
09:15:57 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: 3
.................................................
......!!!!!!!!...................................
09:18:02 `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 --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
09:19:03 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 --> will touch x axis self critique assessment: 3
.................................................
E݆òӝ assignment #004 004. Precalculus I 08-12-2008
......!!!!!!!!...................................
14:34:22 `q001. Note that this assignment has 4 questions If f(x) = x^2 + 4, then find the values of the following: f(3), f(7) and f(-5). Plot the corresponding points on a graph of y = f(x) vs. x. Give a good description of your graph.
......!!!!!!!!...................................
RESPONSE --> f(x)=x^2 + 4 f(3)=3^2+4 f(3)= 13 f(7)=7^2+4 f(7)= 53 f(-5)=-5^2+4 f(-5)= -21 confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:35:05 f(x) = x^2 + 4. To find f(3) we replace x by 3 to obtain f(3) = 3^2 + 4 = 9 + 4 = 13. Similarly we have f(7) = 7^2 + 4 = 49 + 4 = 53 and f(-5) = (-5)^2 + 9 = 25 + 4 = 29. Graphing f(x) vs. x we will plot the points (3, 13), (7, 53), (-5, 29). The graph of f(x) vs. x will be a parabola passing through these points, since f(x) is seen to be a quadratic function, with a = 1, b = 0 and c = 4. The x coordinate of the vertex is seen to be -b/(2 a) = -0/(2*1) = 0. The y coordinate of the vertex will therefore be f(0) = 0 ^ 2 + 4 = 0 + 4 = 4. Moving along the graph one unit to the right or left of the vertex (0,4) we arrive at the points (1,5) and (-1,5) on the way to the three points we just graphed.
......!!!!!!!!...................................
RESPONSE --> The shape is a parabola self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:35:55 `q002. If f(x) = x^2 + 4, then give the symbolic expression for each of the following: f(a), f(x+2), f(x+h), f(x+h)-f(x) and [ f(x+h) - f(x) ] / h. Expand and/or simplify these expressions as appropriate.
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:37:30 If f(x) = x^2 + 4, then the expression f(a) is obtained by replacing x with a: f(a) = a^2 + 4. Similarly to find f(x+2) we replace x with x + 2: f(x+2) = (x + 2)^2 + 4, which we might expand to get (x^2 + 4 x + 4) + 4 or x^2 + 4 x + 8. To find f(x+h) we replace x with x + h to obtain f(x+h) = (x + h)^2 + 4 = x^2 + 2 h x + h^2 + 4. To find f(x+h) - f(x) we use the expressions we found for f(x) and f(x+h): f(x+h) - f(x) = [ x^2 + 2 h x + h^2 + 4 ] - [ x^2 + 4 ] = x^2 + 2 h x + 4 + h^2 - x^2 - 4 = 2 h x + h^2. To find [ f(x+h) - f(x) ] / h we can use the expressions we just obtained to see that [ f(x+h) - f(x) ] / h = [ x^2 + 2 h x + h^2 + 4 - ( x^2 + 4) ] / h = (2 h x + h^2) / h = 2 x + h.
......!!!!!!!!...................................
RESPONSE --> f(a)=a^2+4 f(x+2)=(x+2)^2+4 f(x+h)=(x+h)^2+4 self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:39:58 `q003. If f(x) = 5x + 7, then give the symbolic expression for each of the following: f(x1), f(x2), [ f(x2) - f(x1) ] / ( x2 - x1 ). Note that x1 and x2 stand for subscripted variables (x with subscript 1 and x with subscript 2), not for x * 1 and x * 2. x1 and x2 are simply names for two different values of x. If you aren't clear on what this means please ask the instructor.
......!!!!!!!!...................................
RESPONSE --> f(x)=5x+7; f(x1)=5(x1)+7 f(x2)=5(x2)+7 confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:40:09 Replacing x by the specified quantities we obtain the following: f(x1) = 5 * x1 + 7, f(x2) = 5 * x2 + 7, [ f(x2) - f(x1) ] / ( x2 - x1) = [ 5 * x2 + 7 - ( 5 * x1 + 7) ] / ( x2 - x1) = [ 5 x2 + 7 - 5 x1 - 7 ] / (x2 - x1) = (5 x2 - 5 x1) / ( x2 - x1). We can factor 5 out of the numerator to obtain 5 ( x2 - x1 ) / ( x2 - x1 ) = 5.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
14:41:36 `q004. If f(x) = 5x + 7, then for what value of x is f(x) equal to -3?
......!!!!!!!!...................................
RESPONSE --> f(x)=5x+7 if x= -2, then f(x) will equal -3 confidence assessment: 3
.................................................
......!!!!!!!!...................................
14:41:49 If f(x) is equal to -3 then we right f(x) = -3, which we translate into the equation 5x + 7 = -3. We easily solve this equation (subtract 7 from both sides then divide both sides by 5) to obtain x = -2.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
Γxۦ寍 assignment #005 005. Precalculus I 08-12-2008
......!!!!!!!!...................................
15:04:21 `q001. Note that this assignment has 8 questions Evaluate the function y = x^2 for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?
......!!!!!!!!...................................
RESPONSE --> y=-3^2; y= -9 y=-2^2; y= -4 y=-1^2; y= -1 y=0^2; y= 0 y=1^2; y= 1 y=2^2; y= 4 y=3^2; y= 9 confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:04:48 You should have obtained y values 9, 4, 1, 0, 1, 4, 9, in that order.
......!!!!!!!!...................................
RESPONSE --> forgot, no negatives!! self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:05:38 `q002. Evaluate the function y = 2^x for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?
......!!!!!!!!...................................
RESPONSE --> y= {9,4,1,0,1,4,9} confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:07:20 By velocity exponents, b^-x = 1 / b^x. So for example 2^-2 = 1 / 2^2 = 1/4. Your y values will be 1/8, 1/4, 1/2, 1, 2, 4 and 8. Note that we have used the fact that for any b, b^0 = 1. It is a common error to say that 2^0 is 0. Note that this error would interfere with the pattern or progression of the y values.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:09:59 `q003. Evaluate the function y = x^-2 for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?
......!!!!!!!!...................................
RESPONSE --> y= {-.11,-.25,-1,0,1,.25,.111} confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:11:31 By the laws of exponents, x^-p = 1 / x^p. So x^-2 = 1 / x^2, and your x values should be 1/9, 1/4, and 1. Since 1 / 0^2 = 1 / 0 and division by zero is not defined, the x = 0 value is undefined. The last three values will be 1, 1/4, and 1/9.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:12:46 `q004. Evaluate the function y = x^3 for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?
......!!!!!!!!...................................
RESPONSE --> y= {-27,-8,-1,0,1,8,27} confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:12:58 The y values should be -27, -8, -1, 0, 1, 4, 9.
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:13:22 `q005. Sketch graphs for y = x^2, y = 2^x, y = x^-2 and y = x^3, using the values you obtained in the preceding four problems. Describe the graph of each function.
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:13:51 The graph of y = x^2 is a parabola with its vertex at the origin. It is worth noting that the graph is symmetric with respect to the y-axis. That is, the graph to the left of the y-axis is a mirror image of the graph to the right of the y-axis. The graph of y = 2^x begins at x = -3 with value 1/8, which is relatively close to zero. The graph therefore starts to the left, close to the x-axis. With each succeeding unit of x, with x moving to the right, the y value doubles. This causes the graph to rise more and more quickly as we move from left to right. The graph intercepts the y-axis at y = 1. The graph of y = x^-2 rises more and more rapidly as we approach the y-axis from the left. It might not be clear from the values obtained here that this progression continues, with the y values increasing beyond bound, but this is the case. This behavior is mirrored on the other side of the y-axis, so that the graph rises as we approach the y-axis from either side. In fact the graph rises without bound as we approach the y-axis from either side. The y-axis is therefore a vertical asymptote for this graph. The graph of y = x ^ 3 has negative y values whenever x is negative and positive y values whenever x is positive. As we approach x = 0 from the left, through negative x values, the y values increase toward zero, but the rate of increase slows so that the graph actually levels off for an instant at the point (0,0) before beginning to increase again. To the right of x = 0 the graph increases faster and faster. Be sure to note whether your graph had all these characteristics, and whether your description included these characteristics. Note also any characteristics included in your description that were not included here.
......!!!!!!!!...................................
RESPONSE --> vertex at origin, shape of parabola. the graph is symmetric. self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:16:54 `q006. Make a table for y = x^2 + 3, using x values -3, -2, -1, 0, 1, 2, 3. How do the y values on the table compare to the y values on the table for y = x^2? How does the graph of y = x^2 + 3 compare to the graph of y = x^2?
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:18:46 A list of the y values will include, in order, y = 12, 7, 4, 3, 4, 7, 12. A list for y = x^2 would include, in order, y = 9, 4, 1, 0, 1, 4, 9. The values for y = x^2 + 3 are each 3 units greater than those for the function y = x^2. The graph of y = x^2 + 3 therefore lies 3 units higher at each point than the graph of y = x^2.
......!!!!!!!!...................................
RESPONSE --> y= {12,7,4,3,4,7,12} y=x^2 {9,4,1,0,1,4,9} values of y=x^2+3 is 3 units greater than the y=x^2, therefore the first lies 3 units higher at ea. point than the latter. self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:19:01 `q007. Make a table for y = (x -1)^3, using x values -3, -2, -1, 0, 1, 2, 3. How did the values on the table compare to the values on the table for y = x^3? Describe the relationship between the graph of y = (x -1)^3 and y = x^3.
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:21:14 The values you obtained should have been -64, -27, -8, -1, 0, 1, 8. The values for y = x^3 are -27, -8, -1, 0, 1, 8, 27. The values of y = (x-1)^3 are shifted 1 position to the right relative to the values of y = x^3. The graph of y = (x-1)^3 is similarly shifted 1 unit to the right of the graph of y = x^3.
......!!!!!!!!...................................
RESPONSE --> {-64,-27,-8,-1,0,1,8} y=x^3 {-27,-8,-1,0,1,8,27} the graph y=(x-1)^3 is shifted to the right 1 unit from y=x^3. self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:21:26 `q008. Make a table for y = 3 * 2^x, using x values -3, -2, -1, 0, 1, 2, 3. How do the values on the table compare to the values on the table for y = 2^x? Describe the relationship between the graph of y = 3 * 2^x and y = 2^x.
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
15:22:57 You should have obtained y values 3/8, 3/4, 3/2, 3, 6, 12 and 24. Comparing these with the values 1/8, 1/4, 1/2, 1, 2, 4, 8 of the function y = 2^x we see that the values are each 3 times as great. The graph of y = 3 * 2^x has an overall shape similar to that of y = 2^x, but each point lies 3 times as far from the x-axis. It is also worth noting that at every point the graph of y = 3 * 2^x is three times as the past that of y = 2^x.
......!!!!!!!!...................................
RESPONSE --> y= {3/8,3/4,3/2,3,6,12,24} y=x^2 {1/8,1/4,1/2,1,2,4,8} values on top are 3x as high. self critique assessment: 3
.................................................
cl˫͒˖ assignment #006 006. Precalculus I 08-12-2008
......!!!!!!!!...................................
16:18:43 `q001. Note that this assignment has 10 questions Recall that the graph of y = x^2 + 3 was identical to the graph of y = x^2, except that it was raised 3 units. This function is of the form y = x^2 + c. In the case of this specific function, c = 3. What function would this form give us for c = -1? How would the graph of this function compare with the graph of y = x^2?
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:19:36 If c = -1 the form y = x^2 + c gives us y = x^2 - 1. Every value on a table of this function would be 1 less than the corresponding value on a table of y = x^2, and the graph of y = x^2 - 1 will lie 1 unit lower at each point then the graph of y = x^2.
......!!!!!!!!...................................
RESPONSE --> y=x^2-1 every value is 1 less than the y=x^2 graph self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:20:41 `q002. Suppose that we wish to graph the functions y = x^2 + c for c = -3, then for c = -2, then for c = -1, then for c = 0, then for c = 1, then for c = 2, then for c = 3. If all these functions were plotted on the same set of coordinate axes, what would the final graph look like? It is suggested that you actually sketch your graph and describe your sketch.
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:22:41 The graph of the c= -3 function y = x^2 - 3 will lie 3 units lower than the graph of y = x^2. The graph of the c= -2 function y = x^2 - 2 will lie 22222 units lower than the graph of y = x^2. The progression should be obvious. The graph of the c= 3 function y = x^2 + 3 will lie 3 units higher than the graph of y = x^2. The final graph will therefore show a series of 7 functions, with the lowest three units below the parabolic graph of y = x^2 and the highest three units above the graph of this function. Each graph will lie one unit higher than its predecessor.
......!!!!!!!!...................................
RESPONSE --> the grasph shifts as c changes. the negative functions (3of them) will be below the parabola y=x^2, the others will be above. self critique assessment:
.................................................
......!!!!!!!!...................................
16:24:03 `q003. The function y = (x -1)^3 is of the form y = (x -k)^3 with k = 1. What function would this form give us for k = 3? How would the graph of this function compare with that of y = x^3?
......!!!!!!!!...................................
RESPONSE --> y=(x-3)^3 confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:24:39 Recall how the graph of y = (x-1)^3 lies one unit to the right of the graph of y = x^3. The k = 3 function y = (x -3)^3 will lie 3 units to the right of the graph of y = x^3.
......!!!!!!!!...................................
RESPONSE --> k=3 will be 3 units to the right of y=x^3 self critique assessment:
.................................................
......!!!!!!!!...................................
16:26:41 `q004. Suppose we wish to graph the functions y = (x -k)^3 for k values 2, then 3, then 4. If we graph all these functions on the same set of coordinate axes, what will the graph look like? It is suggested that you actually sketch your graph and describe your sketch.
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:28:11 The k = 2 graph will lie 2 units to the right of the graph of y = x^3, and the k = 4 graph will lie 4 units to the right. The three graphs will all have the same shape as the y = x^3 graph, but will lie 2, 3 and 4 units to the right.
......!!!!!!!!...................................
RESPONSE --> all graphs have same shape as y=x^3, but the k=2,3,4 will each be 2,3,4 units to the right. self critique assessment: 2
.................................................
......!!!!!!!!...................................
16:32:05 `q005. The function y = 3 * 2^x is of the form y = A * 2^x for A = 3. What function would this form give us for A = 2? How would the graph of this function compare with that of y = 2^x?
......!!!!!!!!...................................
RESPONSE --> y=3(2^x); moved up vertical axis 3 times higher than previous. y=2(2^x); moved up vertical axis 3 times higher than the one before confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:32:17 As we saw earlier, the graph of y = 3 * 2^x lies 3 times as far from the x-axis as a graph of y = 2^x and is at every point three times as steep. We would therefore expect the A = 2 function y = 2 * 2^x to lie 2 times is far from the x-axis as the graph of y = 2^x.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:33:38 `q006. Suppose we wish to graph the functions y = A * 2^x for values A ranging from 2 to 5. If we graph all such functions on the same set of coordinate axes, what will the final graph look like? It is suggested that you actually sketch your graph and describe your sketch.
......!!!!!!!!...................................
RESPONSE --> each moves 2 times, then 3, 4, and 5 times as far away from axis as the previous one. confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:34:30 These graphs will range from 2 times as far to 5 times as far from the x-axis as the graph of y = 2^x, and will be from 2 to 5 times as steep. The y intercepts of these graphs will be (0,2), (0, 3), (0, 4), (0,5).
......!!!!!!!!...................................
RESPONSE --> left out intercepts: (0,2) (0,3) (0,4) (0,5) self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:36:31 `q007. What is the slope of a straight line connecting the points (3, 8) and (9, 12)?
......!!!!!!!!...................................
RESPONSE --> confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:38:38 The rise between the points is from y = 8 to y = 12. This is a rise of 12-8 = 4. The run between these points is from x = 3 to x = 9, a run of 9 - 3 = 6. The slope between these points is therefore rise/run = 4/6 = 2/3, with decimal equivalent .6666....
......!!!!!!!!...................................
RESPONSE --> rise= 12-8 y=8, and y=12; rise=4 run=9-3 x=3, and x=9; run=6 slope=.6666(repeating) self critique assessment:
.................................................
......!!!!!!!!...................................
16:42:01 `q008. What are the coordinates of the t = 5 and t = 9 points on the graph of y = 2 t^2 + 3? What is the slope of the straight line between these points?
......!!!!!!!!...................................
RESPONSE --> y=2(5^2)+3 y=53 y=2(9^2)+3 y=165 rise- 165-53= 112 run-9-5=4 confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:42:49 The t = 5 value is y = 2 * 5^2 + 3 = 2 * 25 + 3 = 50 + 3 = 53. The t = 9 value is similarly calculated. We obtain y = 165. The rise between these points is therefore 165-53 = 112. The run is from t = 5 to t = 9, a run of 9 - 5 = 4. This slope of a straight line connecting these points is therefore rise/run = 112/4 = 28.
......!!!!!!!!...................................
RESPONSE --> I always forget something... The slope is of course; 112/4= 28 self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:45:11 The t = 5 second and t = 9 second depths are easily calculated to be y = 53 cm and y = 165 cm. The depth therefore changes from 53 cm to 165 cm, a change of 165 cm - 53 cm = 112 cm, in the 4 seconds between the to clock times. The average rate of depth changes therefore 112 cm/(4 seconds) = 28 cm/second. We note that this calculation is identical to the calculation of the slope of the straight line connecting the t = 5 and t = 9 points of the graph y = 2 t^2 + 3.
......!!!!!!!!...................................
RESPONSE --> you would figure this problem out the same as the one before. The slope of the line would equal the rate of change so the answer is again 28 self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:47:35 `q010. If we have a graph of depth y vs. clock time t, then how to we know that the slope of a straight line connecting two points of the graph represents the average rate at which depth changes over the corresponding time interval?
......!!!!!!!!...................................
RESPONSE --> rise/run can be considered the same as depth/time. they both represent a change in the data which is represented by; like the example of how much money you make in an hour, you to divide the hours worked by the amount you made. confidence assessment: 3
.................................................
......!!!!!!!!...................................
16:47:45 The rise of the graph represents the change in the depth y and the run represents the change clock time t. The slope, which is rise/run, therefore represents change in depth/change in clock time, which is the average rate at which the depth changes.
......!!!!!!!!...................................
RESPONSE --> self critique assessment: 3
............................................"