course mth 158
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11:25:37 query 3.4.14 (was 3.3.6). Concave down then concave up. Does this graph represent a constant, linear, square, cube, square root, reciprocal, abs value or greatest integer function?
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RESPONSE --> this graph represents a cube function
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11:25:41 ** A linear function, represented most simply by y = x, has no curvature. A quadratic function, represented most simply by y = x^2, has a parabolic graph, which is either concave up or concave down. A cubic function, represented most simply by y = x^3, has a graph which changes concavity at the origin. This is the first power which can change concavity. A square root function, represented most simply by y = sqrt(x), have a graph which consists of half of a parabola opening to the right or left. Its concavity does not change. A reciprocal function, represented most simply by y = 1 / x, has a graph with a vertical asymptote at the origin and horizontal asymptotes at the x axis, and opposite concavity on either side of the asymptote. An absolute value function, represented most simply by y = | x |, has a graph which forms a 'V' shape. The greatest integer function [[ x ]] is a 'step' function whose graph is made up of horizontal 'steps'. The concavity of the reciprocal function changes, as does the concavity of the cubic function. The cubic function is the only one that matches the graph, which lacks the vertical and horizontal asymptotes. **
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RESPONSE --> ok
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22:29:41 query 3.4.20 (was 3.3.12). Does your sketch of f(x) = sqrt(x) increase or decrease, and does it do so at an increasing or decreasing rate?
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RESPONSE --> the sketch increases and at an increasing rate
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22:30:43 ** y = sqrt(x) takes y values sqrt(0), sqrt(2) and sqrt(4) at x = 0, 2 and 4. sqrt(0) = 0, sqrt(2) = 1.414 approx., and sqrt(4) = 2. The change in y from the x = 0 to the x = 2 point is about 1.414; the change from the x = 2 to the x = 4 point is about .586. The y values therefore increase, but by less with each 'jump' in x. So the graph contains points (0,0), (2, 1.414) and (4, 2) and is increasing at a decreasing rate. **
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RESPONSE --> see where I went wrong I was half right.
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22:33:41 What three points did you label on your graph?
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RESPONSE --> (1,1),(2,1.414), and (5,2.236)
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22:49:55 query 3.4.24 (was 3.3.24) f(x) = 2x+5 on (-3,0), -3 at 0, -5x for x>0. Given the intercepts, domain and range of the function
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RESPONSE --> you mean query 3.4. 34 intercepts (0,5) and (-3,0) domain {xlx>or = to -3} range {yl<=to -3 y <= 5}
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22:52:23 ** From x = -3 to x = 0 the graph coincides with that of y = 2 x + 5. The graph of y = 2x + 5 has y intercept 5 and slope 2, and its x intecept occurs when y = 0. The x intercept therefore occurs at 2x + 5 = 0 or x = -5/2. Since this x value is in the interval from -3 to 0 it is part of the graph. At x = -3 we have y = 2 * (-3) + 5 = -1. So the straight line which depicted y = 2x + 5 passes through the points (-3, -1), (-5/2, 0) and (0, 5). The part of this graph for which -3 < x < 0 starts at (-3, -1) and goes right up to (0, 5), but does not include (0, 5). For x > 0 the value of the function is -5x. The graph of -5x passes through the origin and has slope -5. The part of the graph for which x > 0 includes all points of the y = -5x graph which lie to the right of the origin, but does not include the origin. For x = 0 the value of the function is given as -3. So the graph will include the single point (0, -3). **
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RESPONSE --> I was way off but understand now
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۾z}r~jzj}YWͨy assignment #025 ؾ{]Uӂ͛ξ College Algebra 11-19-2005
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22:58:28 query 3.5.12 (was 3.4.6). Upward parabola vertex (0, 2). What equation matches this function?
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RESPONSE --> g(x)=x^2+2 maths this function
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23:03:09 The correct equation is y = x^2 + 2. How can you tell it isn't y = 2 x^2 + 2?. How can you tell it isn't y = (x+2)^2.
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RESPONSE --> because y=(x+2)^2 would be a horizontal shift of y=x^2 by looking at the points we know that it was a vertical shift.and y=2x^2+2 would be a vertical stretch not a vertical shift.
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23:03:35 GOOD STUDENT ANSWERS: it isnt y = 2x^2 + 2 because that would make it a verical stretch and it isnt vertical it is just a parabola. it isnt y = (x+2)^2 because that would make it shift to the left two units and it did not. INSTRUCTOR NOTE: Good answers. Here is more detail: The basic y = x^2 parabola has its vertex at (0, 0) and also passes through (-1, 1) and (1, 1). y = 2 x^2 would stretch the basic y = x^2 parabola vertically by factor 2, which would move every point twice as far from the x axis. This would leave the vertex at (0, 0) unchanged and would move the points (-1, 1) and (1, 1) to (-1, 2) and (1, 2). Then y = 2x^2 + 2 would shift this function vertically 2 units, so that the points (0, 0), (-1, 2) and (1, 2) would shift 2 units upward to (0, 2), (-1, 4) and (1, 4). The resulting graph coincides with the given graph at the vertex but because of the vertical stretch it is too steep to match the given graph. The function y = (x + 2)^2 shifts the basic y = x^2 parabola 2 units to the left so that the vertex would move to (-2, 0) and the points (-1, 1) and (1, 1) to (-3, 1) and (-1, 1). The resulting graph does not coincide with the given graph. y = x^2 + 2 would shift the basic y = x^2 parabola vertically 2 units, so that the points (0, 0), (-1, 1) and (1, 1) would shift 2 units upward to (0, 2), (-1, 3) and (1, 3). These points do coincide with points on the given graph, so it is the y = x^2 + 2 function that we are looking for. **
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RESPONSE --> o.k.
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23:09:03 query 3.5.16 (was 3.4.10). Downward parabola.
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RESPONSE --> It would bey= -x^2
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23:10:17 The correct equation is y = -2 x^2. How can you tell it isn't y = 2 x^2?. How can you tell it isn't y = - x^2?
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RESPONSE --> I don't understand it looks to me like it point of origin is (0,0) not - 2 .
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23:10:57 ** The basic y = x^2 parabola has its vertex at (0, 0) and also passes through (-1, 1) and (1, 1). y = -x^2 would stretch the basic y = x^2 parabola vertically by factor -1, which would move every point to another point which is the same distance from the x axis but on the other side. This would leave the vertex at (0, 0) unchanged and would move the points (-1, 1) and (1, 1) to (-1, -1) and (1, -1). y = -2 x^2 would stretch the basic y = x^2 parabola vertically by factor -2, which would move every point to another point which is twice the distance from the x axis but on the other side. This would leave the vertex at (0, 0) unchanged and would move the points (-1, 1) and (1, 1) to (-1, -2) and (1, -2). This graph coincides with the given graph. y = 2 x^2 would stretch the basic y = x^2 parabola vertically by factor 2, which would move every point twice the distance from the x axis and on the same side. This would leave the vertex at (0, 0) unchanged and would move the points (-1, 1) and (1, 1) to (-1, 2) and (1, 2). This graph is on the wrong side of the x axis from the given graph. **
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RESPONSE --> o.k.
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23:13:42 query 3.5.18 (was 3.4.12). V with vertex at origin. What equation matches this function?
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RESPONSE --> y=2lxl
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23:19:35 The correct equation is y = 2 | x | . How can you tell it isn't y = | x | ?. How can you tell it isn't y = | x | + 2?
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RESPONSE --> it isn't y= l x l because if you substitute points in the equation they do not match graph. it is not y=l x l + 2 because that would shift the graph to the left 2 units.
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23:19:59 ** The basic y = | x | graph is in the shape of a V with a 'vertex' at (0, 0) and also passes through (-1, 1) and (1, 1). The given graph has the point of the V at the origin but passes above (-1, 1) and (1, 1). y = | x | + 2 would shift the basic y = | x | graph vertically 2 units, so that the points (0, 0), (-1, 1) and (1, 1) would shift 2 units upward to (0, 2), (-1, 3) and (1, 3). The point of the V would lie at (0, 2). None of these points coincide with points on the given graph y = 2 | x | would stretch the basic y = | x | graph vertically by factor 2, which would move every point twice the distance from the x axis and on the same side. This would leave the point of the 'V' at (0, 0) unchanged and would move the points (-1, 1) and (1, 1) to (-1, 2) and (1, 2). This graph coincides with the given graph. **
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RESPONSE --> o.k.
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23:33:02 query 3.5.30 (was 3.4.24). Transformations on y = sqrt(x). What basic function did you start with and, in order, what transformations were required to obtain the graph of the given function?
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RESPONSE --> it is a square root function we then had to shift up 2 units y=sqrtx+2 we then need to reflect the y axis so we get y=sqrt-x+2 we then need to shift left 3 units we get y= sqrt3-x +2
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23:33:31 What is the function after you shift the graph up 2 units?
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RESPONSE --> y=sqrtx +2
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23:34:02 ERRONEOUS STUDENT RESPONSE: y = x^2 + 2 INSTRUCTOR CORRECTION: y = sqrt(x) is not the same as y = x^2. y = sqrt(x) is the square root of x, not the square of x. Shifting the graph of y = sqrt(x) + 2 up so units we would obtain the graph y = sqrt(x) + 2. **
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RESPONSE --> o.k.
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23:34:43 What is the function after you then reflect the graph about the y axis?
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RESPONSE --> y=sqrt(-x) +2
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23:34:57 ** To reflect a graph about the y axis we replace x with -x. It is the y = sqrt(x) + 2 function that is being reflected so the function becomes y = sqrt(-x) + 2. **
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RESPONSE --> o.k.
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23:35:36 What is the function after you then fhist the graph left 3 units?
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RESPONSE --> y=sqrt(3-x) +2
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23:36:57 ** To shift a graph 3 units to the left we replace x with x + 3. It is the y = sqrt(-x) + 2 function that is being reflected so the function becomes y = sqrt( -(x+3) ) + 2. **
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RESPONSE --> o.k i wasn't sure how to write it using this program I but that aside i had the right response.
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23:52:57 query 3.5.42 (was 3.4.36). f(x) = (x+2)^3 - 3. What basic function did you start with and, in order, what transformations were required to obtain the graph of the given function? Describe your graph. Give three points on your graph and tell to which basic points on the graph of your basic function each corresponds.
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RESPONSE --> basic function y= x^3 we then shift horizontallyto the right 2 unitswich gets us y=(x+2)^3 we the vertically shift down 3 units y=(x+2)^3 - 3 It first has a convex the a concave runs from negutive to positive. my 3 points (0,5) (1,24) (-2,0) (0,5)=(0,0) in the basic (1,24)=(1, 1) in the basic (-2,0)=(-2,-8) in the basic
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23:53:09 ** Starting with y = x^3 we replace x by x - 3 to shift the graph 3 units left. We obtain y = (x + 3)^3. We then shift this graph 2 units vertically by adding 2 to the value of the function, obtaining y = (x + 3 )^3 + 2. The basic points of the y = x^3 graph are (-1, -1), (0, 0) and (1, 1). Shifted 3 units left and 2 units up these points become (-1 - 3, -1 + 2) = (-4, 1), (0 - 3, 0 + 2) = (-3, 2) and (1 - 3, 1 + 2) = (-2, 3). The points you give might not be the same three points but should be obtained by a similar process, and you should specify both the original point and the point to which it is transformed. **
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RESPONSE --> o.k.
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00:22:27 query 3.5.58 (was 3.4.40). h(x) = 4 / x + 2. What basic function did you start with and, in order, what transformations were required to obtain the graph of the given function? Describe your graph. Give three points on your graph and tell to which basic points on the graph of your basic function each corresponds.
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RESPONSE --> basic function y=1/x or reciprocal function we then vertically stretch it 4 units it becomes y= 4/x we then horizantally shift it tho the left 2 units it becomes.
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00:22:46 ** We start with the basic reciprocal function y = 1 / x, which has vertical asymptote at the y axis and horizontal asymptotes at the right and left along the x axis and passes through the points (-1, -1) and (1, 1). To get y = 4 / x we must multiply y = 1 / x by 4. Multiplying a function by 4 results in a vertical stretch by factor 4, moving every point 4 times as far from the x axis. This will not affect the location of the vertical or horizontal asymptotes but for example the points (-1, -1) and (1, 1) will be transformed to (-1, -4) and (1, 4). At every point the new graph will at this point be 4 times as far from the x axis. At this point we have the graph of y = 4 / x. The function we wish to obtain is y = 4 / x + 2. Adding 2 in this mannerincreases the y value of each point by 2. The point (-1, -4) will therefore become (-1, -4 + 2) = (-1, -2). The point (1, 4) will similarly be raised to (1, 6). Since every point is raised vertically by 2 units, with no horizontal shift, the y axis will remain an asymptote. As the y = 4 / x graph approaches the x axis, where y = 0, the y = 4 / x + 2 graph will approach the line y = 0 + 2 = 2, so the horizontal asymptotes to the right and left will consist of the line y = 2. Our final graph will have asymptotes at the y axis and at the line y = 2, with basic points (-1, -2) and (1, 6).
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RESPONSE --> o.k.
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