course Mth 158 ????Dw????????assignment #017017. `query 17
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08:51:27 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.
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RESPONSE --> okay
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08:53:44 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?
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RESPONSE --> (a)(-1,1) (b)(1,-1) (c)(1,1) confidence assessment: 2
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08:54:14 ** 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). **
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RESPONSE --> Okay! somewhat the same answer but not enough explaination. self critique assessment: 2
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08:55:51 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.
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RESPONSE --> (a)(0,0) intercepts both the axis at the same point (b) symmetric with respect to the x-axis, mirrors the x-axis above and below like a reflection confidence assessment: 2
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08:56:24 ** 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. **
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RESPONSE --> yeah! self critique assessment: 2
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09:01:54 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.
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RESPONSE --> so, I graph on my calculator and trace the points. maybe (-0.5,0)(0.5,0) or (-1,0)(1,0) symmetric to the orgin confidence assessment: 1
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09:02:12 STUDENT SOLUTION: origin of the graph is (-.5,0) and (.5,0) graph is symmetric to the origin INSTRUCTOR COMMENT: Check and see whether the graph passes thru the origin (0, 0), which according to my note it should (but my note could be wrong). }If so, and if it is strictly increasing except perhaps at the origin where it might level off for just an instant, in which case the only intercept is at the origin (0, 0). I believe the graph is symmetric with respect to the origin, and if so (0, 0) must be an intercept.
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RESPONSE --> okay! self critique assessment: 2
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09:09:23 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.
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RESPONSE --> 4x^2+y^2=4 y-intercept, let x=0 4(0)^2+y^2=4 0+y^2=4 sqrt(y^2)=sqrt(4) y=-+2 (0,-2),(0,2) x-intercept, let y=0 4x^2+(0)^2=4 4x^2=4 divide both sides by 4 x^2=1 sqrt(x^2)=sqrt(1) x=-+1 (-1,0),(1,0) test y=-y x=-x has symmetry at x,y and origin 4(-x)^2+y^2=4 yes 4x^2+(-y)^2=4 yes 4(-x)^2+(-y)^2=4 yes confidence assessment: 2
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09:11:02 ** 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. **
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RESPONSE --> ok so I got some of the same response but a little different answer for the x-intercept. But the Problem in the book says 4x^2+y^2=4 and I went by that, sorry! self critique assessment: 2
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09:18:45 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.
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RESPONSE --> ok so denominator can not =0! y-intercept, let x=0 y=[(0)^2-4]/[2(0)^4] or do you want to go by the problem in the book or what you give!??? y=(0-4)/0 y=0 x-intercept, let y=0 0=(x^2-4)/(2x^4) multiply both sides by 2x^4 0=(x+2)(x-2) ok here is where i get a little confused! (-2,0)(2,0) symmetric to the respect of the x-axis and orgin confidence assessment: 2
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09:19:28 ** 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. **
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RESPONSE --> so, the symmetry was off by the x-axis. self critique assessment: 2
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