#$&* course Mth 158 7/23/13 11:44 am If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: * * There are three points: The point symmetric to (-1, -1) with respect to the x axis is (-1 , 1). The point symmetric to (-1, -1) with respect to the y axis is y axis (1, -1) The point symmetric to (-1, -1) with respect to the origin is ( 1,1). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * 2.2.43 / 19 (was 2.2.15). Parabola vertex origin opens to left. **** Give the intercepts of the graph and tell whether the graph is symmetric to the x axis, to the y axis and to the origin. Explain how you determined the answer to each question. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (a) The only intercept is at the origin (0, 0) (b) It is symmetric with respect to the x axis confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: * 2.2.48 / 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. The graph s 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). The graph is symmetric with respect to the origin, since for every x we have f(-x) = - f(x). For example, f(2) = 8 and f(-2) = -8. It looks like f(1) = 1 and f(-1) = -1. Whatever number you choose for x, f(-x) = - f(x). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I do not think this question is number 48 in my book, and I am not sure where to find it? ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 2.2.62 / 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To test for the x intercepts, set y = 0 4x^2 + 0 = 4 4x^2 = 4 x^2 = 1 x = +/-1 x intercepts = (1, 0) and (-1, 0) To test for the y intercepts, set x = 0 4(0)^2 + y^2 = 4 y^2 = 4 y = +/-2 y intercepts = (0,2) and (0, -2) To test for symmetry on the x axis set y = -y 4x^2 + (-y)^2 = 4 4x^2 + y^2 = 4 Matches original value Symmetric on x axis To test for symmetry on the y axis set x = -x 4 (-x)^2 + y^2 = 4 4x^2 + y^2 = 4 Matches original value Symmetric on y axis To test for symmetry at the origin, set x = -x and y = -y 4 (-x)^2 + (-y)^2 = 4 4x^2 + y^2 =4 Matches original value Symmetric origin confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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. ** STUDENT COMMENT Ok, I see we worked the problem very closely but I’m confused on where the 4x^2 + y^2 = 4 went. INSTRUCTOR RESPONSE: If you let y = 0 the equation becomes 4 x^2 + 0^2 = 1. As you see we then solve for x to obtain the x intercepts. If you let x = 0 the equation becomes 4 * 0^2 + y^2 = 1. As you see we then solve for y to obtain the y intercepts. Symmetry about the y axis means that you find the same y values at -x as you do at x. Symmetry about the x axis means that you find the same x values at -y as you do at y. Symmetry at the origin says that if (x, y) is a point on the graph, so is (-x, -y). The given solution applies these tests. STUDENT QUESTION I was able to get the x and y axis numbers but I still can’t figure out how to substitute them in to the equations to find the symmetry and the solution confused me more. Got any suggestions??? INSTRUCTOR RESPONSE See if you can answer the following: If you replace x in the equation 4 x^2 + y^2 = 1 with -x, what is your equation? What do you get when you simplify the resulting equation? Is the resulting equation the same as the original equation, or different? If you replace y in the equation 4 x^2 + y^2 = 1 with -y, what is your equation? What do you get when you simplify the resulting equation? Is the resulting equation the same as the original equation, or different? If you replace x in the equation 4 x^2 + y^2 = 1 with -x, and replace y with -y, what is your equation? What do you get when you simplify the resulting equation? Is the resulting equation the same as the original equation, or different? &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I do not understand how the equation changed from being equal to 4 to being equal to 1??? I understand that you have to substitute the x and y values, but I do not see how that immediately changed the 4? ------------------------------------------------ Self-critique Rating: 2
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Given Solution: * * 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. ** PARTIAL STUDENT SOLUTION ((-x)^2 - 4) / (2 * (-x) ) INSTRUCTOR RESPONSE You have substituted -x for x, which is the first step in checking for symmetry about the y axis. Now simplify this expression. Is the simplified expression equal to the original expression? If so you have symmetry about the y axis. To find the x intercept, note that a point is on the x axis if and only if its y coordinate is zero. So you substitute y = 0 and solve for x. To find the y intercept, note that a point is on the y axis if and only if its x coordinate is zero. So you substitute x = 0 and solve for y. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!