course Mth 158 4/8/10 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: D=sqrt[(0)^4 – 13(0) + 49] D= sqrt[ 0-0 + 49] D=sqrt[49] D=7 D=sqrt[(-1)^4 – 13(-1) + 49] D=sqrt[1+13+49] D=sqrt[63] confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * If x = 0 we have sqrt( x^4 - 13 x^2 + 49) = sqrt(0^4 - 13 * 0 + 49) = sqrt(49) = 7. If x = -1 we have sqrt( x^4 - 13 x^2 + 49) = sqrt((-1)^4 - 13 * (-1) + 49) = sqrt( 63). sqrt(64) = 8, so sqrt(63) is a little less than 8 (turns out that sqrt(63) is about 7.94). Note that these results are the distances from the x = 0 and x = 1 points of the graph of y = x^2 - 8 to the point (0, -1). You should have a sketch of the function and you should verify that these distances make sense. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*Ok ********************************************* 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: confidence rating #$&* 0 Looked at the problem in the book and it seemed to differ from yours. Not sure how to answer it. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** A circle inscribed in a square touches the square at the midpoint of each of the square's edges; the circle is inside the square and its center coincides with the center of the square. A diameter of the circle is equal in length to the side of the square. If the circle has radius r then the square has sides of length 2 r and its area is (2r)^2 = 4 r^2. The area of the circle is pi r^2. So the area of the square which is not covered by the circle is 4 r^2 - pi r^2 = (4 - pi) r^2. ** What is the expression for perimeter p as a function of the radius r of the circle? The perimeter of the square is 4 times the length of a side which is 4 * 2r = 8r. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): If the circle has radius r then the square has sides of length 2 r and its area is (2r)^2 = 4 r^2. The area of the circle is pi r^2. So the area of the square which is not covered by the circle is 4 r^2 - pi r^2 = (4 - pi) r^2. ** What is the expression for perimeter p as a function of the radius r of the circle? The perimeter of the square is 4 times the length of a side which is 4 * 2r = 8r. ------------------------------------------------ Self-critique rating #$&*2 ********************************************* Question: * 3.6.19 / 27 (was 3.6.30). one car 2 miles south of intersection at 30 mph, other 3 miles east at 40 mph Give your expression for the distance d between the cars as a function of time. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Sqrt(x^2 + y^2) = Sqrt[(2+ 30t)^2 + (3 + 40t)^2]= sqrt[(2 + 30t)(2 + 30t) + ( 3 + 40t) ( 3 + 40t) sqrt[4 + 60 + 900t^2 + 9 + 80t + 1600t^2]= sqrt[13 + 140t + 2500t^2] confidence rating #$&*2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * ** At time t the position of one car is 2 miles south, increasing at 30 mph, so its position function is 2 + 30 t. The position function of the other is 3 + 40 t. If these are the x and the y coordinates of the position then the distance between the cars is distance = sqrt(x^2 + y^2) = sqrt( (2 + 30 t)^2 + (3 + 40t)^2 ) = sqrt( 4 + 120 t + 900 t^2 + 9 + 240 t + 1600 t^2) = sqrt( 2500 t^2 + 360 t + 13). ** qa college algebra part 2 030. * 30 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*Ok ********************************************* 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: Graph opens up. a = 1 b=-2 c=-3 x-coordinate = x= [-(-2)]/2(1) = 1 y-coordinate= y=f(x)=x^2-2x-3 y=[(1)^2-2(1)-3 y=-4 vertex: (1,-4) x-intercept: x^2-2x-3=0 (x-3)(x+1)= X=(3,-1) Y-intercept: (0)^2-2(0)-3= (-3) [-(-2) + - sqrt (-2)^2 – 4(1)(-3)]/2(1) = [2 + - sqrt[4 + 12]/2 = [2 + - sqrt[16]/2 = [2 + - 4]/2 = 3, -1 Domain: { -inf, inf} Range: {-4<= x <= inf} Decreasing on the interval (-inf,-4). Increasing on interval (-4,inf) confidence rating #$&*2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * The function is of the form y = a x^2 + b x + c with a = 1, b = -2 and c = -3. The graph of this quadratic function will open upwards, since a > 0. The axis of symmetry is the line x = -b / (2 a) = -(-2) / (2 * 1) = 1. The vertex is on this line at y = f(1) = 1^2 - 2 * 1 - 3 = -4. So the the vertex is at the point (1, -4). The x intercepts occur where f(x) = x^2 - 2 x - 3 = 0. We can find the values of x by either factoring or by using the quadratic formula. Here will will factor to get (x - 3) ( x + 1) = 0 so that x - 3 = 0 OR x + 1 = 0, giving us x = 3 OR x = -1. So the x intercepts are (-1, 0) and (3, 0). The y intercept occurs when x = 0, giving us y = f(0) = -3. The y intercept is the point (0, -3). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): the x intercepts are (-1, 0) and (3, 0). The y intercept is the point (0, -3). ------------------------------------------------ Self-critique rating #$&*2 ********************************************* Question: * 4.3.57. graph of parabola vertex (1, -3), point (3, 5) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F(x)= a[x-(1)]^2 +(-3) 5 = a[(3) – (1)]^2 -3 5 = a[2]^2 -3 5 = 4a-3 8=4a 2=a (2)[(x-(1)]^2-3= (2)[(x-1)(x-1)]-3= 2x^2-4x+2-3= 2x^2-4x-1 confidence rating #$&*2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * The graph is a parabola with vertex (1, -3), so it has form {}{}(y - k) = a ( x - h)^2, with h = 1 and k = -3. Thus we have{}{}y - (-3) = a (x - 1)^2, {}{}and the only unknown is a. To find a we substitute the coordinates of (3, 5) for x and y, obtaining the equation(5 - (-3)) = a ( 3 - 1)^2, or{}(5 + 3) = a * 2^2, or{}8 = a * 4, with the obvious solution{}a = 2.{}{}Thus the equation of the parabola is{}{}y + 3 = 2 ( x - 1)^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&*OK ********************************************* 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? a) 1(x+5)(x-3)=0 X^2+2x-15 b) 2(x+5)(x-3)=0 2(x^2+2x-15)=0 2x^2+4x-30=0 c) -2(x+5)(x-3)=0 -2(x^2+2x-15)=0 -2x^2-2x+30=0 d) 5(x+5)(x-3)=0 5(x^2+2x-15)=0 5x^2+10x-75)=0 Solution: 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). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 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). ------------------------------------------------ Self-critique rating #$&*1 1/2 "