course z}̑~˧Cԑassignment #031
......!!!!!!!!...................................
14:53:53 4.4.18. Analyze the graph of y = (x^2 + x 12) / (x^2 4)
......!!!!!!!!...................................
RESPONSE --> The domain is x | x does not equal 2 or -2 x-intercepts are (using (-b =/- sqrt(b^2 - 4ac))/ 2a) 3(sqrt 11) - 1/2 and -3(sqrt 11) - 1/2 y-intercept is 3 There is no symmetry In lowest terms, the equation is x - 3 / x - 4, so the real 0 is 4, and the vertical asymptote will be at x = 4. using long division to find the horizontal asymptote, we get (x-8)/(x^2 - 4) + 1, so the horizontal asymptote will be on the line y = 1. The graph intersects the horizontal asymptote at (3, 0) from the interval (negative infinity, -4), the graph is below the x-axis, from the interval (-4,4) the graph is above, interval (4,8) the graph is above, and interval (8, infinity), graph is above. confidence assessment: 2
.................................................
......!!!!!!!!...................................
14:54:22 The factored form of the function is y = (x 3) ( x + 4) / [(x 2) ( x + 2)]. As x -> infinity the function is dominated by the highest-power terms in numerator and denominator, and the value approaches y = x^2 / x^2 = 1. The same occurs as x -> -infinity. So the graph has a horizontal asymptote at y = 1. The function has zeros where the numerator has zeros, at x = 3 and x = -4. The function is undefined and approaches vertical asymptotes when the denominator is zero, which occurs at x = 2 and x = -2. Since every factor is linear the function will change sign at every zero and vertical asymptote. So the function will alternate between positive and negative on the intervals (-infinity, -4), (-4, -2), (-2, 2), (2, 3) and (3, infinity). For large negative x, as we have seen, the function is positive (it approaches y = +1 as x -> -infinity). So on the interval (-infinity, -4) the function will be positive. Alternating between positive and negative, the function is negative on (-4, -2), positive on (-2, 2), negative on (2, 3) and positive on (3, infinity). It passes through the x axis at x = -4 and at x = 3. We can use these facts to determine the nature of the vertical asymptotes. As we approach x = -2 from the left we are in the interval (-4, -2) so function values will be negative, and we approach the asymptotes through negative values, descending toward the asymptote. To the right of x = -2 we are in the interval (-2, 2) so function values are positive, and the asymptote to the right of x = -2 descends from positive values. As we approach x = 2 from the left we are in the interval (-2, 2) so function values will be positive, and we approach the asymptotes through positive values, rising toward the asymptote. On the interval (-2, 2), then, the values of the function descend from a positive asymptote at the left and ascend toward a positive asymptote on the right. It does this without passing through the x axis, since there are no zeros in the interval (-2, 2), and therefore remains above the x axis on this interval. To the right of x = 2 we are in the interval (2, 3) so function values are negative, and the asymptote to the right of x = 2 ascends from negative values. At x = 3 we have a zero so the graph passes through the x axis from negative to positive, and thereafter remains positive while approaching y = 1 as a horizontal asymptote.
......!!!!!!!!...................................
RESPONSE --> wow. so the simple mistake of turning x^2 - 4 into (x + 4) and (x - 4) instead of (x - 2)(x + 2) messed everything else up completely. sorry! self critique assessment: 3
.................................................
......!!!!!!!!...................................
15:25:42 4.3.30. Analyze the graph of y = (x^2 - x 12) / (x + 1)
......!!!!!!!!...................................
RESPONSE --> The domain is x | x does not equal -1 x-int: 4 and 3 y-int: -12 There is no symmetry Real 0 is -1, and the vertical asymptote follows the line x = -1 the line x - 2 is the oblique asymptote, and it intersects the graph at (0, -12) The graph is below the x-axis on the intervals (-infinity, -3), (-3, 1), and (1,4), and above the x-axis on the interval (4, infinity) confidence assessment: 2
.................................................
......!!!!!!!!...................................
15:27:49 The factored form of the function is y = (x 4) ( x + 3) / (x + 1). As x -> infinity the function is dominated by the highest-power terms in numerator and denominator, and therefore approaches y = x^2 / x = x. So the graph is asymptotic to the line y = x at both left and right. The function has zeros where the numerator has zeros, at x = -3 and x = 4. The function is undefined and approaches vertical asymptotes when the denominator is zero, which occurs at x = -1. Since every factor is linear the function will change sign at every zero and vertical asymptote. So the function will alternate between positive and negative on the intervals (-infinity, -3), (-3, -1), (-1, 4) and (4, infinity). For large negative x, the function is close to y = x, which is negative. So on the interval (-infinity, -3) the function will be negative. Alternating between positive and negative, the function is positive on (-3, -1), negative on (-1, 4) and positive on (4, infinity). It passes through the x axis at x = 4 and at x = -3. We can use these facts to determine the nature of the vertical asymptote. As we approach x = -1 from the left we are in the interval (-3, -1) so function values will be positive, and we approach the asymptotes through positive values, ascending toward the asymptote. To the right of x = -1 we are in the interval (-1, 4) so function values are negative, and the asymptote to the right of x = -1 ascends from negative values. The function passes through the x axis at x = 4, and then approaches the line y = x as an asymptote, remaining positive from x = 4 on.
......!!!!!!!!...................................
RESPONSE --> What did I do wrong? self critique assessment: 1
.................................................
......!!!!!!!!...................................
15:45:32 4.3.42. Analyze the graph of y = 2 x^2 + 9 / x.
......!!!!!!!!...................................
RESPONSE --> 2 x^2 + 9 / x = (2x^3 +9) / x x-intercept: (- 9/2)^1/3 y-intercept: none There is no symmetry The vertical asymptote is at y = 0, since that is the only real zero of the denominator. y = 2x^2 isn't a line, so how can that be the oblique asymptote? and I don't know how to use (- 9/2)^1/3 in figuring out the intervals so I can find out where the graph is above and below the x-axis confidence assessment: 1
.................................................
......!!!!!!!!...................................
15:47:11 The denominator x indicates a vertical asymptote at x = 0, i.e., at the y axis. The function has zeros when 2 x^2 + 9 / x = 0 . Multiplying both sides by x we get 2 x^3 + 9 = 0 so that x^3 = -9/2 and x = -(9/2)^(1/3) = -1.65 approx.. The function therefore alternates between positive and negative on the intervals (-infinity, -1.65), (-1.65, 0) and (0, infinity). For large positive or negative values if x the term 9 / x is nearly zero and the term 2 x^2 dominates, so the graph is asymptotic to the y = 2 x^2 parabola. This function is positive for both large positive and large negative values of x. So the function is positive on (-infinity, -1.65), negative on (-1.65, 0) and positive on (0, infinity). Approaching the vertical asymptote from the left the function therefore approaches through negative y values, descending toward its vertical asymptote at the y axis. To the right of the vertical asymptote the function is positive, so it descends from its vertical asymptote. From left to right, therefore, the function starts close to the parabola y = 2 x^2, eventually curving away from this graph toward its zero at x = -1.65 and passing through the x axis at this point, then descending toward the y axis as a vertical asymptote. To the right of the y axis the graph descends from the y axis before turning back upward to become asymptotic to the graph of the parabola y = 2 x^2.
......!!!!!!!!...................................
RESPONSE --> Ok, I see that I should've just approximated. And I guess there just wasn't a horizontal nor oblique asymptote. self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:45:26 07-28-2008 16:45:26 4.4.56. Steel drum volume 100 ft^3, right circular cylinder. Find amount of material as a function of r and give amounts for r = 3, 4, 5 ft. Graph and indicate the min.
......!!!!!!!!...................................
NOTES -------> A = 2(pi)(r)(h) = 100, h = 50 / (pi)(r)
.................................................q[cM^C
assignment #031 031. Query 31 College Algebra 07-28-2008
......!!!!!!!!...................................
16:47:44 4.4.18. Analyze the graph of y = (x^2 + x 12) / (x^2 4)
......!!!!!!!!...................................
RESPONSE --> The domain is x | x does not equal 2 or -2 x-intercepts are (using (-b =/- sqrt(b^2 - 4ac))/ 2a) 3(sqrt 11) - 1/2 and -3(sqrt 11) - 1/2 y-intercept is 3 There is no symmetry In lowest terms, the equation is x - 3 / x - 4, so the real 0 is 4, and the vertical asymptote will be at x = 4. using long division to find the horizontal asymptote, we get (x-8)/(x^2 - 4) + 1, so the horizontal asymptote will be on the line y = 1. The graph intersects the horizontal asymptote at (3, 0) from the interval (negative infinity, -4), the graph is below the x-axis, from the interval (-4,4) the graph is above, interval (4,8) the graph is above, and interval (8, infinity), graph is above. confidence assessment: 1
.................................................
......!!!!!!!!...................................
16:48:09 The factored form of the function is y = (x 3) ( x + 4) / [(x 2) ( x + 2)]. As x -> infinity the function is dominated by the highest-power terms in numerator and denominator, and the value approaches y = x^2 / x^2 = 1. The same occurs as x -> -infinity. So the graph has a horizontal asymptote at y = 1. The function has zeros where the numerator has zeros, at x = 3 and x = -4. The function is undefined and approaches vertical asymptotes when the denominator is zero, which occurs at x = 2 and x = -2. Since every factor is linear the function will change sign at every zero and vertical asymptote. So the function will alternate between positive and negative on the intervals (-infinity, -4), (-4, -2), (-2, 2), (2, 3) and (3, infinity). For large negative x, as we have seen, the function is positive (it approaches y = +1 as x -> -infinity). So on the interval (-infinity, -4) the function will be positive. Alternating between positive and negative, the function is negative on (-4, -2), positive on (-2, 2), negative on (2, 3) and positive on (3, infinity). It passes through the x axis at x = -4 and at x = 3. We can use these facts to determine the nature of the vertical asymptotes. As we approach x = -2 from the left we are in the interval (-4, -2) so function values will be negative, and we approach the asymptotes through negative values, descending toward the asymptote. To the right of x = -2 we are in the interval (-2, 2) so function values are positive, and the asymptote to the right of x = -2 descends from positive values. As we approach x = 2 from the left we are in the interval (-2, 2) so function values will be positive, and we approach the asymptotes through positive values, rising toward the asymptote. On the interval (-2, 2), then, the values of the function descend from a positive asymptote at the left and ascend toward a positive asymptote on the right. It does this without passing through the x axis, since there are no zeros in the interval (-2, 2), and therefore remains above the x axis on this interval. To the right of x = 2 we are in the interval (2, 3) so function values are negative, and the asymptote to the right of x = 2 ascends from negative values. At x = 3 we have a zero so the graph passes through the x axis from negative to positive, and thereafter remains positive while approaching y = 1 as a horizontal asymptote.
......!!!!!!!!...................................
RESPONSE --> wow. so the simple mistake of turning x^2 - 4 into (x + 4) and (x - 4) instead of (x - 2)(x + 2) messed everything else up completely. sorry! self critique assessment: 3
.................................................
......!!!!!!!!...................................
16:48:43 4.3.30. Analyze the graph of y = (x^2 - x 12) / (x + 1)
......!!!!!!!!...................................
RESPONSE --> The domain is x | x does not equal -1 x-int: 4 and 3 y-int: -12 There is no symmetry Real 0 is -1, and the vertical asymptote follows the line x = -1 the line x - 2 is the oblique asymptote, and it intersects the graph at (0, -12) The graph is below the x-axis on the intervals (-infinity, -3), (-3, 1), and (1,4), and above the x-axis on the interval (4, infinity) confidence assessment: 2
.................................................
......!!!!!!!!...................................
16:49:05 The factored form of the function is y = (x 4) ( x + 3) / (x + 1). As x -> infinity the function is dominated by the highest-power terms in numerator and denominator, and therefore approaches y = x^2 / x = x. So the graph is asymptotic to the line y = x at both left and right. The function has zeros where the numerator has zeros, at x = -3 and x = 4. The function is undefined and approaches vertical asymptotes when the denominator is zero, which occurs at x = -1. Since every factor is linear the function will change sign at every zero and vertical asymptote. So the function will alternate between positive and negative on the intervals (-infinity, -3), (-3, -1), (-1, 4) and (4, infinity). For large negative x, the function is close to y = x, which is negative. So on the interval (-infinity, -3) the function will be negative. Alternating between positive and negative, the function is positive on (-3, -1), negative on (-1, 4) and positive on (4, infinity). It passes through the x axis at x = 4 and at x = -3. We can use these facts to determine the nature of the vertical asymptote. As we approach x = -1 from the left we are in the interval (-3, -1) so function values will be positive, and we approach the asymptotes through positive values, ascending toward the asymptote. To the right of x = -1 we are in the interval (-1, 4) so function values are negative, and the asymptote to the right of x = -1 ascends from negative values. The function passes through the x axis at x = 4, and then approaches the line y = x as an asymptote, remaining positive from x = 4 on.
......!!!!!!!!...................................
RESPONSE --> What did I do wrong? self critique assessment: 1
.................................................
......!!!!!!!!...................................
16:49:30 4.3.42. Analyze the graph of y = 2 x^2 + 9 / x.
......!!!!!!!!...................................
RESPONSE --> 2 x^2 + 9 / x = (2x^3 +9) / x x-intercept: (- 9/2)^1/3 y-intercept: none There is no symmetry The vertical asymptote is at y = 0, since that is the only real zero of the denominator. y = 2x^2 isn't a line, so how can that be the oblique asymptote? and I don't know how to use (- 9/2)^1/3 in figuring out the intervals so I can find out where the graph is above and below the x-axis confidence assessment: 1
.................................................
......!!!!!!!!...................................
16:49:51 The denominator x indicates a vertical asymptote at x = 0, i.e., at the y axis. The function has zeros when 2 x^2 + 9 / x = 0 . Multiplying both sides by x we get 2 x^3 + 9 = 0 so that x^3 = -9/2 and x = -(9/2)^(1/3) = -1.65 approx.. The function therefore alternates between positive and negative on the intervals (-infinity, -1.65), (-1.65, 0) and (0, infinity). For large positive or negative values if x the term 9 / x is nearly zero and the term 2 x^2 dominates, so the graph is asymptotic to the y = 2 x^2 parabola. This function is positive for both large positive and large negative values of x. So the function is positive on (-infinity, -1.65), negative on (-1.65, 0) and positive on (0, infinity). Approaching the vertical asymptote from the left the function therefore approaches through negative y values, descending toward its vertical asymptote at the y axis. To the right of the vertical asymptote the function is positive, so it descends from its vertical asymptote. From left to right, therefore, the function starts close to the parabola y = 2 x^2, eventually curving away from this graph toward its zero at x = -1.65 and passing through the x axis at this point, then descending toward the y axis as a vertical asymptote. To the right of the y axis the graph descends from the y axis before turning back upward to become asymptotic to the graph of the parabola y = 2 x^2.
......!!!!!!!!...................................
RESPONSE --> Ok, I see that I should've just approximated. And I guess there just wasn't a horizontal nor oblique asymptote. self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:22:49 4.4.56. Steel drum volume 100 ft^3, right circular cylinder. Find amount of material as a function of r and give amounts for r = 3, 4, 5 ft. Graph and indicate the min.
......!!!!!!!!...................................
RESPONSE --> V = (pi)(r^2)(h) = 100 h = 100/(pi)(r^2) A(r) = 2(pi)(r) * 100/(pi)r^2 = 200/r If radius is 3, A(3) = 200/3 = 66.67 ft^3 If radius is 4, A(4) = 200/4 = 50 ft^3 If radius is 5, A(5) = 200/5 = 40 ft^3 I can't graph because I don't have a graphing calculator confidence assessment: 1, because I just realized that all the answers are ""ft^2"" and not ""ft^3""
.................................................
......!!!!!!!!...................................
17:25:27 If the radius of the cylinder is r then the area of its circular base is pi r^2. The volume of the drum is area of base * height = 100, so that pi r^2 * height = 100 and height = 100 / (pi r^2). The surface area is the sum of the surface areas of the bases, which is 2 pi r^2, and the surface area of the sides, which is circumference * height = 2 pi r * height = 2 pi r * (100 / ( pi r^2 )) = 200 / r. So the total surface area is Surface Area = 2 pi r^2 + 200 / r. For r = 3 we get 2 pi * 3^2 + 200 / 3 = 123.2. Similarly for r = 4 and r = 5 we get areas 150.5 and 197.1. Analysis of the function tells us that the graph descends from the positive vertical axis as an asymptote, reaches a minimum then begins ascending toward the 2 pi r^2 parabola, to which it is asymptotic. There must therefore be a minimum in there somewhere. Our areas 123.2, 150.5 and 197.1 are increasing, so the minimum lies either to the left of r = 3 or between r = 3 and r = 4. Evaluating the function half a unit to the left and right of r = 3 gives us values 119.2699081, 134.1118771 at r = 2.5 and r = 3.5. We conclude that the minimum lies to the left of r = 3. Evaluating at r = 2.6 and r = 2.4 we get areas 119.3974095 and 119.5244807, both greater than the 119.27 we got at r = 2.5. So our minimum will lie close to r = 2.5.
......!!!!!!!!...................................
RESPONSE --> It didn't say to include the surface areas of the bases in the problem. A steel drum just has the sides, so I don't know what this is talking about. self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:25:41