course Mth 158 I was unable to find assmnt 16 to punch it in. ??????J~?O??????assignment #017
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15:06:01 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 --> ok
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15:07:51 query 2.2.34 / 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 --> x-axis (-1,1) y-axis ( 1, -1) origin ( 1,1)
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15:08:22 ** 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 --> ok
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15:13:20 **** query 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.
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RESPONSE --> (0,0) symmetric to the x-axis The graph touches at (0,0). It is symmetric to the x axis because it is a mirror image if you were to fold it along the x axis.
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15:13:31 ** 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 --> ok
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15:19:18 **** query 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.
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RESPONSE --> The intercepts appear to be (0, -1/2) and (0, 1/2) It is symmetric to the x axis and the origin. It is symmetric to the origin because for every point on the graph, there is a negative of that point. It is symmetric to the x axis because if you make the x points negative, it doesn't change the equation.
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15:19:54 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).
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RESPONSE --> ok
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15:25:12 **** query 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.
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RESPONSE --> The intercepts are (-1,0); (1,0); (0,-2); and (0,2). The equation remains the same when replaceing x with a negative number, y with a negative number and x & y with negatives.
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15:27:28 ** 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 --> My problem was 4x^2 + y^2 =4.
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15:31:22 **** query 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.
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RESPONSE --> The intercepts are (2,0) and -2, 0). There are no y intercepts because you can't divide by 0. The test for x and y symmetry showed negative numbers changed the original equation. The test for symmetry pertaining to the origin showed that by substituting both a negative x and y, the problem remained essentially the same.
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15:32:06 ** 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 --> ok
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?w????R??????x??ve?assignment #018 018. `query 18 College Algebra 09-23-2008
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18:09:10 query 2.3.34 / 30 (was 2.3.24). Slope 4/3, point (-3,2) Give the three points you found on this line and explain how you obtained them.
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RESPONSE --> With a slope of 4/3, you could start with your x of -3 and go over 3 and up 4. Points could include -3+ 3 =0 and 2 + 4 = 6 or (0,6). 0 + 3 = 3 and 6 + 4 = 10 or (3,10) 3 + 3 = 6 and 10 + 4 = 14 or (6,14).
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18:09:40 STUDENT SOLUTION: (-3,2) slope 4/3. Move 3 units in the x direction, 4 in the y direction to get ((-3+3), (2+4)), which simplifies to (0,6) (-3,2) slope 4/3 = -4/-3 so move -3 units in the x direction and -4 in the y direction to get ((-3-3), (2-4)) which simplifies to (-6,-2) From (0,6) with slope 4/3 we move 4 units in the y direction and 3 in the x direction to get ((0+3), (6+4)), which simplifies to (3,10). The three points I obtained are (-6,-2), (0,6), (3,10).
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RESPONSE --> ok
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18:15:56 query 2.3.40 / 36 (was 2.3.30). Line thru (-1,1) and (2,2) **** Give the equation of the line and explain how you found the equation.
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RESPONSE --> First I found the slope of the line by m = (y1-y2)/(x1-x2) = 1/3. Then I used the formula y-y1 = m(x-x1) y-1 = m(x+1) y= 1/3x + 4/3
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18:16:05 STUDENT SOLUTION: The slope is m = (y2 - y1) / (x2 - x1) = (2-1)/(2- -1) = 1/3. Point-slope form gives us y - y1 = m (x - x1); using m = 1/3 and (x1, y1) = (-1, 1) we get y-1=1/3(x+1), which can be solved for y to obtain y = 1/3 x + 4/3.
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RESPONSE --> ok
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18:19:40 **** query 2.3.54 / 46 (was 2.3.40). x-int -4, y-int 4 **** What is the equation of the line through the given points and how did you find the equation?
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RESPONSE --> The points would be (-4,0) and (0,4) The slope would be m= (4-0)(/0+4) = 1 The equation would be y-0 = 1(x + 4) y= x + 4
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18:20:11 STUDENT SOLUTION: The two points are (0, 4) and (4, 0). The slope is therefore m=rise / run = (4-0)/(0+4) = 1. The slope-intercept form is therefore y = m x + b = 1 x + 4, simplifying to y=x+4.
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RESPONSE --> ok
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18:22:07 **** query 2.3.76 / 56 (was 2.4.48). y = 2x + 1/2. **** What are the slope and the y-intercept of your line and how did you find them?
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RESPONSE --> The slope is 2 and the y-intercept is 1/2. The equation is in slope-intercept form.
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18:22:23 ** the y intercept occurs where x = 0, which happens when y = 2 (0) + 1/2 or y = 1/2. So the y-intercept is (0, 1/2). The slope is m = 2.**
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RESPONSE --> ok
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18:26:10 **** query 2.3.62 / 22 (was 2.4.18) Parallel to x - 2 y = -5 containing (0,0) **** Give your equation for the requested line and explain how you obtained it.
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RESPONSE --> First I put the equation in slope-intercept form: y = 1/2x + 5/2 Then I substituted (0,0) and the slope of 1/2 in the point-slope formula. y = 1/2x
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18:26:17 ** The equation x - 2y = -5 can be solved for y to give us y = 1/2 x + 5/2. A line parallel to this will therefore have slope 1/2. Point-slope form gives us y - 0 = 1/2 * (x - 0) or just y = 1/2 x. **
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RESPONSE --> ok
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18:29:54 **** query 2.3.68 / 28 (was 2.4.24) Perpendicular to x - 2 y = -5 containing (0,4) **** Give your equation for the requested line and explain how you obtained it.
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RESPONSE --> First I converted the equation to slope-intercept form y = 1/2x + 5/2 Then I used the negative reciprocal of the slope in the point-slope form y= -2x +4
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18:30:07 ** The equation x - 2y = -5 can be solved for y to give us y = 1/2 x + 5/2. A line perpendicular to this will therefore have slope -2/1 = -2. Point-slope form gives us y - 4 = -2 * (x - 0) or y = -2 x + 4. **
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RESPONSE --> ok
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