course Mth 158
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17:25:13 2.5.22 (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 --> -2y=-x-5 y=(-x-5)/-2 y=-1/2x+2.5 y-0=-1/2(x-0) y=-1/2x moved the origional equation in to slope-intercept form and then substituted the given point (0,0) into slope point form to find a parallel equation. confidence assessment: 3
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17:25:29 ** 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 self critique assessment: 3
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17:29:54 2.5.28 (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 --> -2y=-x-5 y=1/2x-5/2 m=1/2 perpendicular line m=2 y-4=2(x-0) y=2x+4 confidence assessment: 2
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17:30:35 ** 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 --> Perpendicular lines slopes must equal -1 instead of 1 should have been y=-2x+4 self critique assessment: 3
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17:39:28 2.3.10 (was 2.4.30). (0,1) and (2,3) on diameter **** What are the center, radius and equation of the indicated circle?
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RESPONSE --> From the given graph the center is (1,2) sqrt[(2-1)^2+(3-2)^2]=r sqrt(1^2+1^2)=5 radius is 1 confidence assessment: 2
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17:39:47 ** The distance from (0,1) to (2,3) is sqrt( (2-0)^2 + (3-1)^2 ) = 2. This distance is a diameter so that the radius is 1/2 (2) = 1. The equation (x-h)^2 + (y-k)^2 = r^2 becomes (x-1)^2 + (y-2)^2 = r^2. Substituting the coordinates of the point (0, 1) we get (0-1)^2 + (1-2)^2 = r^2 so that r^2 = 2. Our equation is therefore (x-1)^2 + (y - 2)^2 = 2. You should double-check this solution by substituting the coordinates of the point (2, 3). **
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RESPONSE --> ok self critique assessment: 3
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17:43:31 2.3.16 (was 2.4.36). What is the standard form of a circle with (h, k) = (1, 0) with radius 3?
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RESPONSE --> r=3 (h,k)=(1,0) graphed to find (x,y) which was (1,3) (1-1)^2+(3-0)^2=9 confidence assessment: 3
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17:44:32 ** The standard form of a circle is (x - h)^2 + (y - k)^2 = r^2, where the center is at (h, k) and the radius is r. In this example we have (h, k) = (1, 0). We therefore have (x-1)^2 +(y - 0)^2 = 3^2. This is the requested standard form. This form can be expanded and simplified to a general quadratic form. Expanding (x-1)^2 and squaring the 3 we get x^2 - 2x +1+y^2 = 9 x^2 - 2x + y^2 = 8. However this is not the standard form.
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RESPONSE --> I did not need to find another point should have been written as (x-1)^2+(y-0)^2=9 self critique assessment: 3
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17:49:10 2.3.24 (was 2.4.40). x^2 + (y-1)^2 = 1 **** Give the center and radius of the circle and explain how they were obtained. In which quadrant(s) was your graph and where did it intercept x and/or y axes?
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RESPONSE --> r=1 (h,k)=(0, -1) intercepts at (0, -2) and (0,0) (x-0)^2+(y-1)^2=1 confidence assessment: 3
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17:49:27 ** The standard form of a circle is (x - h)^2 + (y - k)^2 = r^2, where the center is at (h, k) and the radius is r. In this example the equation can be written as (x - 0)^2 + (y-1)^2 = 1 So h = 0 and k = 1, and r^2 = 1. The center of the is therefore (0, -1) and r = sqrt(r^2) = sqrt(1) = 1. The x intercept occurs when y = 0: x^2 + (y-1)^2 = 1. I fy = 0 we get x^2 + (0-1)^2 = 1, which simplifies to x^2 +1=1, or x^2=0 so that x = 0. The x intercept is therefore (0, 0). The y intercept occurs when x = 0 so we obtain 0 + (y-1)^2 = 1, which is just (y - 1)^2 = 1. It follow that (y-1) = +-1. If y - 1 = 1 we get y = 2; if y - 1 = -1 we get y -2. So the y-intercepts are (0,0) and (0,-2)
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RESPONSE --> ok self critique assessment: 3
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17:57:22 2.3.34 (was 2.4.48). 2 x^2 + 2 y^2 + 8 x + 7 = 0 **** Give the center and radius of the circle and explain how they were obtained. In which quadrant(s) was your graph and where did it intercept x and/or y axes?
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RESPONSE --> Circle crosses all 4 quadrants r=4.5 intercepts at (13, 0) and (-5, 0) confidence assessment: 1
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17:58:10 ** Starting with 2x^2+ 2y^2 +8x+7=0 we group x and y terms to get 2x^2 +8x +2y^2 =-7. We then divide by the common factor 2 to get x^2 +4x + y^2 = -7/2. We complete the square on x^2 + 4x, adding to both sides (4/2)^2 = 4, to get x^2 + 4x + 4 + y^2 = -7/2 + 4. We factor the expression x^2 + 4x + 4 to obtain (x+2)^2 + y^2 = 1/2. From the standard form of the equation for a circle we see that the center is (-2,0) the radius is sqrt (1/2). To get the intercepts: We use (x+2)^2 + y^2 = 1/2 If y = 0 then we have (x+2)^2 + 0^2 = 1/2 (x+2)^2 = 1/2 (x+2) = +- sqrt(1/2) x + 2 = sqrt(1/2) yields x = sqrt(1/2) - 2 = -1.3 approx. x + 2 = -sqrt(1/2) yields x = -sqrt(1/2) - 2 = -2.7 approx If x = 0 we have (0+2)^2 + y^2 = 1/2 4 + y^2 = 1/2 y^2 = 1/2 - 4 = -7/2. y^2 cannot be negative so there is no y intercept. **
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RESPONSE --> Ok I understand I used the graph on this because I was unsure of how to work the standard equation but I can see from the answer. self critique assessment: 3
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18:06:26 2.3.30 (was 2.4.54). General equation if diameter contains (4, 3) and (0, 1). **** Give the general equation for your circle and explain how it was obtained.
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RESPONSE --> (h,k)=(2,2) M=(4+0)/2, 3+1/2)=(2,2) D=sqrt(0-4)^2 +(1-3)^2 D=sqrt20 =4.47 r=1/2(4.47) r=2.236 (x-2)^2+(y-2)^2=4.999 confidence assessment: 1
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18:07:21 ** The center of the circle is the midpoint between the two points, which is ((4+0)/2, (3+1)/2) = (2, 2). The radius of the circle is the distance from the center to either of the given points. The distance from (2, 2) to (0, 1) is sqrt( (2-0)^2 + (2-1)^2 ) = sqrt(5). The equation of the circle is therefore (x-2)^2 + (y-2)^2 = (sqrt(5))^2 or (x-2)^2 + (y-2)^2 = 5. **
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RESPONSE --> I should have done distance from center to point instead of half of point to point but it was very close. self critique assessment: 3
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