course mth 158 I don't know if I have submitted this before or not. It seems like I remember working the problems, but don't have any confirmation on it so I will submit it again. I went through it and tried to solve what I couldn't do before, but I don't know how much better it is now. 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: * * ** P = (x, y) is of the form (x, x^2 - 8). So the distance from P to (0, -1) is sqrt( (0 - x)^2 + (-1 - (x^2-8))^2) = sqrt(x^2 + (-7-x^2)^2) = sqrt( x^2 + 49 - 14 x^2 + x^4) = sqrt( x^4 - 13 x^2 + 49). ** What are the values of d for x=0 and x = -1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I really don’t know if it’s right. I think the values of x were right. confidence rating: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * If x = 0 we have sqrt( x^4 - 13 x^2 + 49) = sqrt(0^4 - 13 * 0 + 49) = sqrt(49) = 7. If x = -1 we have sqrt( x^4 - 13 x^2 + 49) = sqrt((-1)^4 - 13 * (-1) + 49) = sqrt( 64) = 8. Note that these results are the distances from the x = 0 and x = 1 points of the graph of y = x^2 - 8 to the point (0, -1). You should have a sketch of the function and you should vertify that these distances make sense. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I guess I got the whole thing wrong. ------------------------------------------------ Self-critique Rating: 0 ********************************************* Question: * 3.6.9 / 18 (was and remains 3.6.18). Circle inscribed in square. What is the expression for area A as a function of the radius r of the circle? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I was looking at the book trying to understand the problem and follow along with the given solution to get a better understanding of how to solve these problems and realized the book does not match the question. The book has: for 3.6.9 A rectangle is inscribed in a circle of radius 2. Let P = (x,y) be the point in quadrant I that is a vertex of the rectangle and is on the circle. a) express the area A of the rectangle as a function of x Even with that said, I cannot solve the problem. I just don’t understand them. confidence rating: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** A circle inscribed in a square touches the square at the midpoint of each of the square's edges; the circle is inside the square and its center coincides with the center of the square. A diameter of the circle is equal in length to the side of the square. If the circle has radius r then the square has sides of length 2 r and its area is (2r)^2 = 4 r^2. The area of the circle is pi r^2. So the area of the square which is not covered by the circle is 4 r^2 - pi r^2 = (4 - pi) r^2. ** What is the expression for perimeter p as a function of the radius r of the circle? The perimeter of the square is 4 times the length of a side which is 4 * 2r = 8r. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): How is the diameter of the circle equal in length to the side of the square? How do you determine that?
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Given Solution: * * ** At time t the position of one car is 2 miles south, increasing at 30 mph, so its position function is 2 + 30 t. The position function of the other is 3 + 40 t. If these are the x and the y coordinates of the position then the distance between the cars is distance = sqrt(x^2 + y^2) = sqrt( (2 + 30 t)^2 + (3 + 40t)^2 ) = sqrt( 4 + 120 t + 900 t^2 + 9 + 240 t + 1600 t^2) = sqrt( 2500 t^2 + 360 t + 13). ** qa college algebra part 2 030. * 30 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): There are no discrepancies ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 4.3.42 (4.1.42 7th edition) (was 4.1.30). Describe the graph of f(x)= x^2-2x-3 as instructed. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since it is a quadratic function and a = 1 > 0 it will open upwards. To find the x coordinate of the vertex you take : a=1 b=-2 c=-3 -b/2a = -(-2)/2(1) = 1 and the y coordinate of the vertex is f(1) = 1^2 – 2(1) – 3 = -4 The axis of symmetry is x = 1 or the points (1, -4) If we factor x^2 -2x -3 we get (x - 3) ( x + 1) = 0 so that x - 3 = 0 or x + 1 = 0 x = 3 x = -1. The x intercepts are (-1, 0) and (3, 0) The y intercept is f(0) = -3 The y intercept is the point (0, -3) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * The function is of the form y = a x^2 + b x + c with a = 1, b = -2 and c = -3. The graph of this quadratic function will open upwards, since a > 0. The axis of symmetry is the line x = -b / (2 a) = -(-2) / (2 * 1) = 1. The vertex is on this line at y = f(1) = 1^2 - 2 * 1 - 3 = -4. So the the vertex is at the point (1, -4). The x intercepts occur where f(x) = x^2 - 2 x - 3 = 0. We can find the values of x by either factoring or by using the quadratic formula. Here will will factor to get (x - 3) ( x + 1) = 0 so that x - 3 = 0 OR x + 1 = 0, giving us x = 3 OR x = -1. So the x intercepts are (-1, 0) and (3, 0). The y intercept occurs when x = 0, giving us y = f(0) = -3. The y intercept is the point (0, -3). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): There are no discrepancies. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * 4.3.57. graph of parabola vertex (1, -3), point (3, 5) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph is a parabola that opens upwards with vertex (1, -3). If h = 1 and k = -3, then f(x) = a(x – h)^2 + k which is a(x – 1)^2 + (-3) To find the value of a we use the coordinates (3, 5) for the x value to get a(3 – 1))^2 + (-3) = 1 I don’t think this is right. There is an example of how to solve a problem like this in the book, but it has the y intercept as a given point on the graph, which can be used as y = f(x) to determine the value of a. Here the problem only gives one other point and I am not sure how to use that to get the quadratic function of the graph. confidence rating: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * The graph is a parabola with vertex (1, -3), so it has form {}{}(y - k) = a ( x - h)^2, with h = 1 and k = -3. Thus we have{}{}y - (-3) = a (x - 1)^2, {}{}and the only unknown is a. To find a we substitute the coordinates of (3, 5) for x and y, obtaining the equation(5 - (-3)) = a ( 3 - 1)^2, or{}(5 + 3) = a * 2^2, or{}8 = a * 4, with the obvious solution{}a = 2.{}{}Thus the equation of the parabola is{}{}y + 3 = 2 ( x - 1)^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I can see how it was solved, but I’m not sure where (y-k) come from. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: * Extra Problem (4.1.67 7th edition) (was 4.1.50). What are your quadratic functions whose x intercepts are -5 and 3, for the values a=1; a=2; a=-2; a=5? The quadratic function is a multiple. If you have x intercepts of -5 and 3, then x = – 5 would be (x+5) = 0 and x = 3 would be (x-3) = 0 So the values of a = 1 would be 1(x+5)(x-3) = x^2 + 2x – 15 a = 2 would be 2(x+5)(x-3) = 2x^2 + 4x - 30 a = -2 would be -2(x+5)(x-3) = -2x^2 -4x + 30 a = 5 would be 5(x+5)(x-3) = 5x^2 + 10x - 75 Since (x+5) = 0 when x = -5 and (x - 3) = 0 when x = 3, the quadratic function will be a multiple of (x+5)(x-3). If a = 1 the function is 1(x+5)(x-3) = x^2 + 2 x - 15. If a = 2 the function is 2(x+5)(x-3) = 2 x^2 + 4 x - 30. If a = -2 the function is -2(x+5)(x-3) = -2 x^2 -4 x + 30. If a = 5 the function is 5(x+5)(x-3) = 5 x^2 + 10 x - 75. Does the value of a affect the location of the vertex? In every case the vertex, which lies at -b / (2a), will be on the line x = -1. The value of a does not affect the x coordinate of the vertex, which lies halfway between the zeros at (-5 + 3) / 2 = -1. The y coordinate of the vertex does depend on a. Substituting x = -1 we obtain the following: For a = 1, we get 1 ( 1 + 5) ( 1 - 3) = -12. For a = 2, we get 2 ( 1 + 5) ( 1 - 3) = -24. For a = -2, we get -2 ( 1 + 5) ( 1 - 3) = 24. For a = 5, we get 5 ( 1 + 5) ( 1 - 3) = -60. So the vertices are (-1, -12), (-1, -24), (-1, 24) and (-1, -60). "