course mth 158 7/17 4 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 #$&* ********************************************* Question: * 2.2.43 / 19 (was 2.2.15). Parabola vertex origin opens to left. **** Give the intercepts of the graph and tell whether the graph is symmetric to the x axis, to the y axis and to the origin. Explain how you determined the answer to each question. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If the parabola opens to the left, then it is imaginable that the points of interception on the x and y axis is right in the origin… or, is zero and zero. 0,0 X axis symmetry: x2 +(-y)= 0 or 0sqrd – 0 = 0 it is symmetrical with the x axis -x2 + y=0 -0sqrd + 0=0 it is symmetrical with the y axis -x2 + -y=0 0sqrd -0=0 it is symmetrical with the origin. confidence rating #$&* 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 #$&* ********************************************* Question: * 2.2.48 / 24 (was 2.2.20). basic cubic poly arb vert stretch **** Give the intercepts of the graph and tell whether the graph is symmetric to the x axis, to the y axis and to the origin. Explain how you determined the answer to each question. The graph s strictly increasing except perhaps at the origin where it might level off for just an instant, in which case the only intercept is at the origin (0, 0). The graph is symmetric with respect to the origin, since for every x we have f(-x) = - f(x). For example, f(2) = 8 and f(-2) = -8. It looks like f(1) = 1 and f(-1) = -1. Whatever number you choose for x, f(-x) = - f(x). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I’m a little confused but I think I understand the concept of f(x)=x and f(-x) = - f (x) ------------------------------------------------ Self-critique rating #$&* ********************************************* 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. Your solution: X intercept: 4x2 +02= 4 4x2 = 4 sqr root 4x2 = sqr root 4 x=2 (2,0) Y intercept: 0x2 +y2= 4 sqr root y2 = sqr root 4 y=2 (0,2) (2,2) x symmetry: X2 + -y2=4 2sqrd -2sqrd =4 4-2=4 none y symmetry -X2 + -y2=4 -2sqrd -2sqrd =-8 none origin symmetry: -xsqrd + -ysqrd=4 -2sqrd + -2sqrd= -8 none ?????????????? confidence rating #$&* 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok, I see we worked the problem very closely but I’m confused on where the 4x^2 + y^2 = 4 went.
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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. ** "