#$&* course Mth 158 12:37 from 6/28/2012 If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.Your solution, attempt at solution:
.............................................
Given Solution: * * There are three points: The point symmetric to (-1, -1) with respect to the x axis is (-1 , 1). The point symmetric to (-1, -1) with respect to the y axis is y axis (1, -1) The point symmetric to (-1, -1) with respect to the origin is ( 1,1). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 2.2.43 / 19 (was 2.2.15). Parabola vertex origin opens to left. **** Give the intercepts of the graph and tell whether the graph is symmetric to the x axis, to the y axis and to the origin. Explain how you determined the answer to each question. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It is intercepting the x and y axis at the origin (0,0). If you draw a line of symmetry, it will cross the x axis. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ------------------------------------------------ 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The intercept is the origin and the only possible point of symmetry is at the origin. 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): ------------------------------------------------ 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: I break this into two problems. Problem 1: 4x^2 = 4. I would divide by 4 and take the sqrt of 1. Your x intercept is (1,0). Problem 2: y^2 = 4. The sqrt of 4 is 2. Your y intercept is (0,2). (Mr. Smith, Please don’t think I am being rude, but you copied the problem down wrong in your solution. You changed =4 to =1.) For the intercepts, you change the signs for x intercept, you would also have (-1,0) and for y intercept, you would have (0,-2). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 2.2.68 / 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I substitute -x into the equation. This will not make the equations equal so there isn’t symmetry at the y axis. You substitute -y into the equation and the equations are not the same so there isn’t symmetry at the x axis. You will substitute -x and -y into the solution. This will end up with the same equation so the symmetry is at the origin. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!