asst 17

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

......!!!!!!!!...................................

Note that you can't use a calculator graph to document your solutions to these problems. You have to use the analytical methods as in the given solutions. Documentation is required on tests, and while you may certainly use the calculator to see symmetry, intercepts etc., you have to support your solutions with the algebraic details of why the graph looks the way it does.

......!!!!!!!!...................................

RESPONSE --> ok

.................................................

......!!!!!!!!...................................

13:55:00 query 2.2.10 (was 2.2.6). Point symmetric to (-1, -1) wrt x axis, y axis, origin. What point is symmetric to the given point with respect to each: x axis, y axis, the origin?

......!!!!!!!!...................................

RESPONSE --> the points symmetric are (-1,1) ; (1,1) ; (1,-1)

.................................................

......!!!!!!!!...................................

13:55:10 ** There are three points: The point symmetric to (-1, -1) with respect to the x axis is (-1 , 1). The point symmetric to (-1, -1) with respect to the y axis is y axis (1, -1) The point symmetric to (-1, -1) with respect to the origin is ( 1,1). **

......!!!!!!!!...................................

RESPONSE --> correct.

.................................................

......!!!!!!!!...................................

13:56:52 **** query 2.1.19 (was 2.2.15). Parabola vertex origin opens to left. **** Give the intercepts of the graph and tell whether the graph is symmetric to the x axis, to the y axis and to the origin. Explain how you determined the answer to each question.

......!!!!!!!!...................................

RESPONSE --> a) intercepts of the graphs are (0,0) b) this graph can be symmetric with respect to the x-axis only.

.................................................

......!!!!!!!!...................................

13:57:04 ** The graph intercepts both axes at the same point, (0,0) The graph is symmetric to the x-axis, with every point above the x axis mirrored by its 'reflection' below the x axis. **

......!!!!!!!!...................................

RESPONSE --> correct.

.................................................

......!!!!!!!!...................................

13:58:42 **** query 2.2.24 (was 2.2.20). basic cubic poly arb vert stretch **** Give the intercepts of the graph and tell whether the graph is symmetric to the x axis, to the y axis and to the origin. Explain how you determined the answer to each question.

......!!!!!!!!...................................

RESPONSE --> The intercepts are (0,0) I think. The graph is symmetric to the x-axis, y-axis and origin.

.................................................

......!!!!!!!!...................................

14:00:27 **** query 2.2.40 (was 2.2.36). 4x^2 + y^2 = 4 **** List the intercepts and explain how you made each test for symmetry, and the results of your tests.

......!!!!!!!!...................................

RESPONSE --> The intersection does cross at -.5,0. This graph was a little hard to read from the book. The graph does pass through only the origin though. I think I am getting used to telling these graphs now.

.................................................

......!!!!!!!!...................................

14:01:38 ** Starting with 4x^2 +y^2 = 1 we find the x intercept by letting y = 0. We get 4x^2 + 0 = 1 so 4x^2 = 1 and x^2=1/4 . Therefore x=1/2 or -1/2 and the x intercepts are (1/2,0) and ( -1/2,0). Starting with 4x^2 +y^2 = 1 we find the y intercept by letting x = 0. We get 0 +y^2 = 1 so y^2 = 1 and y= 1 or -1, giving us y intercepts (0,1) and (0,-1). To test for symmetry about the y axis we substitute -x for x. If there's no change in the equation then the graph will be symmetric to about the y axis. Substituting we get 4 (-x)^2 + y^2 = 1. SInce (-x)^2 = x^2 the result is 4 x^2 + y^2 = 1. This is identical to the original equation so we do have symmetry about the y axis. To test for symmetry about the x axis we substitute -y for y. If there's no change in the equation then the graph will be symmetric to about the x axis. Substituting we get 4 (x)^2 + (-y)^2 = 1. SInce (-y)^2 = y^2 the result is 4 x^2 + y^2 = 1. This is identical to the original equation so we do have symmetry about the x axis. To test for symmetry about the origin we substitute -x for x and -y for y. If there's no change in the equation then the graph will be symmetric to about the origin. Substituting we get 4 (-x)^2 + (-y)^2 = 1. SInce (-x)^2 = x^2 and (-y)^2 - y^2 the result is 4 x^2 + y^2 = 1. This is identical to the original equation so we do have symmetry about the origin. **

......!!!!!!!!...................................

RESPONSE --> When I entered the next question button it showed the answer to the next problem and didn't give me a chance to work it out. Sorry.

.................................................

......!!!!!!!!...................................

14:13:26 **** query 2.2.46 (was 2.2.42). y = (x^2-4)/(2x) **** List the intercepts and explain how you made each test for symmetry, and the results of your tests.

......!!!!!!!!...................................

RESPONSE --> y-intercept: x = 0, then y = (0^2 - 4) / 2(0) = -4/0 = 0 (0,0) x-intercept: y = 0, then 0 = x^2 - 4/2x = 2x = 0 = x = 0 (0,0) test for symmetry: x-axis: replace y by -y: -y = x^2 - 4 / 2x not equivalent to original problem. y-axis: replace x by -x: y = (-x)^2 - 4 / 2(-x) = x^2 - 4 / -2x not equivalent to original problem. origin: replace x by -x and y by -y: -y = x^2 - 4 / -2x is equivalent to new problem. Graph is symmetric.

.................................................

......!!!!!!!!...................................

14:13:44 ** We do not have symmetry about the x or the y axis, but we do have symmetry about the origin: To test for symmetry about the y axis we substitute -x for x. If there's no change in the equation then the graph will be symmetric to about the y axis. }Substituting we get y = ( (-x)^2 - 4) / (2 * (-x) ). SInce (-x)^2 = x^2 the result is y = -(x^2 - 4) / (2 x). This is not identical to the original equation so we do not have symmetry about the y axis. To test for symmetry about the x axis we substitute -y for y. If there's no change in the equation then the graph will be symmetric to about the x axis. Substituting we get -y = (x^2-4)/(2x) , or y = -(x^2-4)/(2x). This is not identical to the original equation so we do not have symmetry about the x axis. To test for symmetry about the origin we substitute -x for x and -y for y. If there's no change in the equation then the graph will be symmetric to about the origin. Substituting we get -y = ((-x)^2-4)/(2(-x)) SInce (-x)^2 = x^2 the result is -y = -(x^2-4)/(2x), or multiplying both sides by -1, our result is y = (x^2-4)/(2x). This is identical to the original equation so we do have symmetry about the origin. **

......!!!!!!!!...................................

RESPONSE --> correct.

.................................................

"

You did well on these questions. Let me know if you have questions.