#$&* course MTH 158 017. `* 17
.............................................
Given Solution: * * There are three points: The point symmetric to (-1, -1) with respect to the x axis is (-1 , 1). The point symmetric to (-1, -1) with respect to the y axis is y axis (1, -1) The point symmetric to (-1, -1) with respect to the origin is ( 1,1). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: * 2.2.43 / 19 (was 2.2.15). Parabola vertex origin opens to left. **** Give the intercepts of the graph and tell whether the graph is symmetric to the x axis, to the y axis and to the origin. Explain how you determined the answer to each question. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A) X intercept = 0,0 ; y intercept = 0,0 B) The graph is symmetric in respect to the x - axis confidence rating #$&*: 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): 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). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph is symmetric to the origin. I couldn’t find the intercepts for this problem. Confidence Assesment: 2
.............................................
Given solution: 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 need to review this problem and try to find the intercepts the way that you did. ------------------------------------------------ Self-critique Rating: 2 ********************************************* 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 took the square root of both sides and then put the result in y = mx + b form. This gave me y = -2x + 2 After this point I got confused as far as the tests go. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): Again I did the problem totally wrong, yet the way the problem was actually worked was very easy to me and I do understand the tests after finding the right section in the book. ------------------------------------------------ Self-critique Rating:1 ********************************************* 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: y = (x^2-4)/(2x) 0 = (x^2-4)/(2x) multiply both sides by 2x 0 = x^2 -4 4 = x^2 take the square root of both sides 2 = x Y intercept = (0, 2) y = (x^2-4)/(2x) y = (0^2-4)/(2(0)) y = undefined confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): I failed to do this problem in the correct way. I tried to repeat the process from the previous problem. I need to closely review my work before taking the test. ------------------------------------------------ Self-critique Rating: 1 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!