Assignment 05

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19:01:58 `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?

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RESPONSE --> -3 = -3^2 = 9 -2 = -2^2 = 4 -1 = -1^2 = 1 0 = 0^2 = 0 1 = 1^2 = 1 2 = 2^2 = 4 3 = 3^2 = 9

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19:02:10 You should have obtained y values 9, 4, 1, 0, 1, 4, 9, in that order.

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RESPONSE --> great

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19:06:27 `q002. Evaluate the function y = 2^x for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?

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RESPONSE --> y = 2^-3 = 1/2^3 = 1/8 y = 2^-2 = 1/2^2 = 1/4 y = 2^-1 = 1/2^1 = 1/2 y = 2^0 = 1 y = 2^1 = 2 y = 2^2 = 4 y = 2^3 = 8

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19:07:19 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.

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RESPONSE --> great

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19:33:05 `q003. Evaluate the function y = x^-2 for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?

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RESPONSE --> y= -3^-2 = 1/-3^2 = 1/9 y = -2^-2 = 1/-2^2 = 1/4 y = -1^-2 = 1/-1^2 = 1 y = 0^-2 = 0 y = 1^-2 = 1/1^2 = 1 y = 2^-2 = 1/2^2 = 1/4 y = 3^-2 = 1/3^2 = 1/9

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19:37:06 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.

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RESPONSE --> Division by 0 is undefined. I was looking more at the exponent then I was paying attention to where in the fraction the ""0"" was located. Noted...

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19:41:09 `q004. Evaluate the function y = x^3 for x values -3, -2, -1, 0, 1, 2, and 3. What are your y values?

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RESPONSE --> y = -3^3 = -27 y = -2^3 = -8 y = -1^3 = -1 y = 0^3 = 0 y = 1^3 = 1 y = 2^3 = 8 y = 3^3 = 27

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19:45:19 The y values should be -27, -8, -1, 0, 1, 4, 9.

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RESPONSE --> My calculation of 2^3 = 2*2*2 = 8, where you have the solution as ""4"" and my calculation of 3^3 is 3*3*3 = 27, where your solution is 9...I am concerned that I may have copied the problem incorrectly or miscalculated in some way. Would you please respond?

You're right. That was an editing error on my part.

I've been aware of this for some time and have chosen not to change it, mainly to see who's paying attention. There was never a question of that in your case.

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21:25:59 `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.

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RESPONSE --> The equation of y = x^2 is a quadradic equation which gives me a parabala with the coordinates of (-3,9) (-2,4) (-1,1) (0,0) (1,1) (2,4) (3,9) The equation on y = 2^x is a vertical asymptote (I think it's called vertical) that passes thru the points of (-3,1/8) (-2, -1/4) (-1,1/2) (0,1) (1,2) (2,4) (3,8) The equation y = x^-2 gives me two identical asymptote graphs on the positive side of the y axis, both are identical on either side of the (0,0) coordinate. The one to the left of negative side of the y axis passes thru the points (-3,1/9) (-2,1/4) (-1,1)...since the 0 coordinate is undefined, the graph starts again on the positive side of the y axis passing thru the points (1,1) (2,1/4) (3,1/9) The equation y = x^3 gives me an inverted parabola that passes thru the points (-3,-27) (-2,-8) (-1,-1) (0,0) (1,1) (2,8) (3,27) with the vertex at (0,0)...given the problem is correct (refer to the question on the previous problem)

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21:27:47 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.

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RESPONSE --> My responses were not as specific as the questions...I've noted my failure in being this specific, but I think that the description that was given was sufficient to describe the graph that I have drawn

Your description was good. No failure on your part, but of course you want to pick up some additional ideas from the given solution.

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21:37:01 `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?

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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. The graph of y = x^2 is a parabola that has it's vertex at (0,0)...the graph of y = x^2 =3 is the same graph of the parabola but stretched three units positive vertically on the y axis

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21:37:19 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.

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RESPONSE --> I have the same graph

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21:53:06 `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.

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RESPONSE --> In the equation y = (x-1)^3 the coordinates on the graph are (-3,-64) (-2,-27) (-1,-8) (0,-1) (1,0) (2,1) (3,8) and the corrdinates on the equation y = x^3 are (-3,27) (-2,-8) (-1,-1) (0,0) (1,1) (2,8) (3,27)...The graphs are identical except for a horizontal shift.

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21:53:41 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.

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RESPONSE --> I didn't include in my response that the shift was one unit to the right, but I see that on my graph

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22:26:14 `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.

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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. The equation y = 3 - 2^x have the coordinates of (-3,23/8(2.875)), 9-2,11/4(2.75)) (-1,5/2(2.5)), (0,2) (1,1) (2,-1) (3,-5). The coordinates for the equation y = 2^x are (-3,1/8) (-2,1/4) (-1, 1/2) 0,1) (1,2) (2,4) (3,8) Both graphs start on the ""negative"" side of the y axis with the graph of y = 3 - 2^x at the coordinate (-3,23/8) passing thru the y axis at the point 0,2) on a downward slope pasing thru the x axis at approx 0.1.75) continuing down. The y = 2^x graph starts at the point (-3,1/8) passint thru the y axis at (0,1) on an upward slope crossing the graph of y = 3-2^x at the approx coordinate (.6,1.5) continuing on an upward slope

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22:28:35 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.

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RESPONSE --> self critique: I did my elementary math incorrect. I see that the 3 in the equation y = 2^x shows that the graphs should look alike with the points being 3 times as far apart

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Your work on this assignment is quite good. See my notes and let me know if you have questions.