{"id":1458,"date":"2015-11-12T18:35:29","date_gmt":"2015-11-12T18:35:29","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=1458"},"modified":"2017-04-03T14:43:47","modified_gmt":"2017-04-03T14:43:47","slug":"solutions-39","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/solutions-39\/","title":{"raw":"Solutions","rendered":"Solutions"},"content":{"raw":"

Solutions to Try Its<\/h2>\r\n1.\u00a0End behavior: as [latex]x\\to \\pm \\infty , f\\left(x\\right)\\to 0[\/latex]; Local behavior: as [latex]x\\to 0, f\\left(x\\right)\\to \\infty [\/latex] (there are no x<\/em>- or y<\/em>-intercepts)\r\n\r\n2.\u00a0The function and the asymptotes are shifted 3 units right and 4 units down. As [latex]x\\to 3,f\\left(x\\right)\\to \\infty\\\\ [\/latex], and as [latex]x\\to \\pm \\infty ,f\\left(x\\right)\\to -4[\/latex].\r\n

The function is [latex]f\\left(x\\right)=\\frac{1}{{\\left(x - 3\\right)}^{2}}-4[\/latex].<\/p>\r\n\"Graph\r\n\r\n3.\u00a0[latex]\\frac{12}{11}[\/latex]\r\n\r\n4.\u00a0The domain is all real numbers except [latex]x=1[\/latex] and [latex]x=5[\/latex].\r\n\r\n5.\u00a0Removable discontinuity at [latex]x=5[\/latex]. Vertical asymptotes: [latex]x=0,\\text{ }x=1[\/latex].\r\n\r\n6.\u00a0Vertical asymptotes at [latex]x=2[\/latex] and [latex]x=-3[\/latex]; horizontal asymptote at [latex]y=4[\/latex].\r\n\r\n7.\u00a0For the transformed reciprocal squared function, we find the rational form. [latex]f\\left(x\\right)=\\frac{1}{{\\left(x - 3\\right)}^{2}}-4=\\frac{1 - 4{\\left(x - 3\\right)}^{2}}{{\\left(x - 3\\right)}^{2}}=\\frac{1 - 4\\left({x}^{2}-6x+9\\right)}{\\left(x - 3\\right)\\left(x - 3\\right)}=\\frac{-4{x}^{2}+24x - 35}{{x}^{2}-6x+9}[\/latex]\r\n

Because the numerator is the same degree as the denominator we know that as [latex]x\\to \\pm \\infty , f\\left(x\\right)\\to -4; \\text{so } y=-4[\/latex] is the horizontal asymptote. Next, we set the denominator equal to zero, and find that the vertical asymptote is [latex]x=3[\/latex], because as [latex]x\\to 3,f\\left(x\\right)\\to \\infty [\/latex]. We then set the numerator equal to 0 and find the x<\/em>-intercepts are at [latex]\\left(2.5,0\\right)[\/latex] and [latex]\\left(3.5,0\\right)[\/latex]. Finally, we evaluate the function at 0 and find the y<\/em>-intercept to be at [latex]\\left(0,\\frac{-35}{9}\\right)[\/latex].<\/p>\r\n8.\u00a0Horizontal asymptote at [latex]y=\\frac{1}{2}[\/latex]. Vertical asymptotes at [latex]x=1 \\text{and} x=3[\/latex]. y<\/em>-intercept at [latex]\\left(0,\\frac{4}{3}.\\right)[\/latex]\r\n

x<\/em>-intercepts at [latex]\\left(2,0\\right) \\text{ and }\\left(-2,0\\right)[\/latex]. [latex]\\left(-2,0\\right)[\/latex] is a zero with multiplicity 2, and the graph bounces off the x<\/em>-axis at this point. [latex]\\left(2,0\\right)[\/latex] is a single zero and the graph crosses the axis at this point.\r\n\"Graph<\/span><\/p>\r\n\r\n

Solutions to Try Its<\/h2>\r\n1.\u00a0The rational function will be represented by a quotient of polynomial functions.\r\n\r\n3.\u00a0The numerator and denominator must have a common factor.\r\n\r\n5.\u00a0Yes. The numerator of the formula of the functions would have only complex roots and\/or factors common to both the numerator and denominator.\r\n\r\n7.\u00a0[latex]\\text{All reals }x\\ne -1, 1[\/latex]\r\n\r\n9.\u00a0[latex]\\text{All reals }x\\ne -1, -2, 1, 2[\/latex]\r\n\r\n11.\u00a0V.A. at [latex]x=-\\frac{2}{5}[\/latex]; H.A. at [latex]y=0[\/latex]; Domain is all reals [latex]x\\ne -\\frac{2}{5}[\/latex]\r\n\r\n13.\u00a0V.A. at [latex]x=4, -9[\/latex]; H.A. at [latex]y=0[\/latex]; Domain is all reals [latex]x\\ne 4, -9[\/latex]\r\n\r\n15.\u00a0V.A. at [latex]x=0, 4, -4[\/latex]; H.A. at [latex]y=0[\/latex]; Domain is all reals [latex]x\\ne 0,4, -4[\/latex]\r\n\r\n17.\u00a0V.A. at [latex]x=-5[\/latex]; H.A. at [latex]y=0[\/latex]; Domain is all reals [latex]x\\ne 5,-5[\/latex]\r\n\r\n19.\u00a0V.A. at [latex]x=\\frac{1}{3}[\/latex]; H.A. at [latex]y=-\\frac{2}{3}[\/latex]; Domain is all reals [latex]x\\ne \\frac{1}{3}[\/latex].\r\n\r\n21.\u00a0none\r\n\r\n23.\u00a0[latex]x\\text{-intercepts none, }y\\text{-intercept }\\left(0,\\frac{1}{4}\\right)[\/latex]\r\n\r\n25.\u00a0Local behavior: [latex]x\\to -{\\frac{1}{2}}^{+},f\\left(x\\right)\\to -\\infty ,x\\to -{\\frac{1}{2}}^{-},f\\left(x\\right)\\to \\infty [\/latex]\r\n\r\nEnd behavior: [latex]x\\to \\pm \\infty ,f\\left(x\\right)\\to \\frac{1}{2}[\/latex]\r\n\r\n27.\u00a0Local behavior: [latex]x\\to {6}^{+},f\\left(x\\right)\\to -\\infty ,x\\to {6}^{-},f\\left(x\\right)\\to \\infty [\/latex], End behavior: [latex]x\\to \\pm \\infty ,f\\left(x\\right)\\to -2[\/latex]\r\n\r\n29.\u00a0Local behavior: [latex]x\\to -{\\frac{1}{3}}^{+},f\\left(x\\right)\\to \\infty ,x\\to -{\\frac{1}{3}}^{-}[\/latex], [latex]f\\left(x\\right)\\to -\\infty ,x\\to {\\frac{5}{2}}^{-},f\\left(x\\right)\\to \\infty ,x\\to {\\frac{5}{2}}^{+}[\/latex] ,\u00a0[latex]f\\left(x\\right)\\to -\\infty [\/latex]\r\n\r\nEnd behavior: [latex]x\\to \\pm \\infty\\\\ [\/latex], [latex]f\\left(x\\right)\\to \\frac{1}{3}[\/latex]\r\n\r\n31.\u00a0[latex]y=2x+4[\/latex]\r\n\r\n33.\u00a0[latex]y=2x[\/latex]\r\n\r\n35.\u00a0[latex]V.A.\\text{ }x=0,H.A.\\text{ }y=2[\/latex]\r\n\"Graph\r\n\r\n37.\u00a0[latex]V.A.\\text{ }x=2,\\text{ }H.A.\\text{ }y=0[\/latex]\r\n\"Graph\r\n\r\n39.\u00a0[latex]V.A.\\text{ }x=-4,\\text{ }H.A.\\text{ }y=2;\\left(\\frac{3}{2},0\\right);\\left(0,-\\frac{3}{4}\\right)[\/latex]\r\n\"Graph\r\n\r\n41.\u00a0[latex]V.A.\\text{ }x=2,\\text{ }H.A.\\text{ }y=0,\\text{ }\\left(0,1\\right)[\/latex]\r\n\"Graph\r\n\r\n43.\u00a0[latex]V.A.\\text{ }x=-4,\\text{ }x=\\frac{4}{3},\\text{ }H.A.\\text{ }y=1;\\left(5,0\\right);\\left(-\\frac{1}{3},0\\right);\\left(0,\\frac{5}{16}\\right)[\/latex]\r\n\r\n45.\u00a0[latex]V.A.\\text{ }x=-1,\\text{ }H.A.\\text{ }y=1;\\left(-3,0\\right);\\left(0,3\\right)[\/latex]\r\n\"Graph\r\n\r\n47.\u00a0[latex]V.A.\\text{ }x=4,\\text{ }S.A.\\text{ }y=2x+9;\\left(-1,0\\right);\\left(\\frac{1}{2},0\\right);\\left(0,\\frac{1}{4}\\right)[\/latex]\r\n\"Graph\r\n\r\n49.\u00a0[latex]V.A.\\text{ }x=-2,\\text{ }x=4,\\text{ }H.A.\\text{ }y=1,\\left(1,0\\right);\\left(5,0\\right);\\left(-3,0\\right);\\left(0,-\\frac{15}{16}\\right)[\/latex]\r\n\"Graph\r\n\r\n51.\u00a0[latex]y=50\\frac{{x}^{2}-x - 2}{{x}^{2}-25}[\/latex]\r\n\r\n53.\u00a0[latex]y=7\\frac{{x}^{2}+2x - 24}{{x}^{2}+9x+20}[\/latex]\r\n\r\n55.\u00a0[latex]y=\\frac{1}{2}\\frac{{x}^{2}-4x+4}{x+1}[\/latex]\r\n\r\n57.\u00a0[latex]y=4\\frac{x - 3}{{x}^{2}-x - 12}[\/latex]\r\n\r\n59.\u00a0[latex]y=-9\\frac{x - 2}{{x}^{2}-9}[\/latex]\r\n\r\n61.\u00a0[latex]y=\\frac{1}{3}\\frac{{x}^{2}+x - 6}{x - 1}[\/latex]\r\n\r\n63.\u00a0[latex]y=-6\\frac{{\\left(x - 1\\right)}^{2}}{\\left(x+3\\right){\\left(x - 2\\right)}^{2}}[\/latex]\r\n\r\n65.\r\n
x<\/em><\/td>\r\n2.01<\/td>\r\n2.001<\/td>\r\n2.0001<\/td>\r\n1.99<\/td>\r\n1.999<\/td>\r\n<\/tr>
y<\/em><\/td>\r\n100<\/td>\r\n1,000<\/td>\r\n10,000<\/td>\r\n\u2013100<\/td>\r\n\u20131,000<\/td>\r\n<\/tr><\/tbody><\/table>
x<\/em><\/td>\r\n10<\/td>\r\n100<\/td>\r\n1,000<\/td>\r\n10,000<\/td>\r\n100,000<\/td>\r\n<\/tr><\/tbody>
y<\/em><\/td>\r\n.125<\/td>\r\n.0102<\/td>\r\n.001<\/td>\r\n.0001<\/td>\r\n.00001<\/td>\r\n<\/tr><\/tbody><\/table>\r\nVertical asymptote [latex]x=2[\/latex], Horizontal asymptote [latex]y=0[\/latex]\r\n\r\n67.\r\n
x<\/em><\/td>\r\n\u20134.1<\/td>\r\n\u20134.01<\/td>\r\n\u20134.001<\/td>\r\n\u20133.99<\/td>\r\n\u20133.999<\/td>\r\n<\/tr>
y<\/em><\/td>\r\n82<\/td>\r\n802<\/td>\r\n8,002<\/td>\r\n\u2013798<\/td>\r\n\u20137998<\/td>\r\n<\/tr><\/tbody><\/table>
x<\/em><\/td>\r\n10<\/td>\r\n100<\/td>\r\n1,000<\/td>\r\n10,000<\/td>\r\n100,000<\/td>\r\n<\/tr>
y<\/em><\/td>\r\n1.4286<\/td>\r\n1.9331<\/td>\r\n1.992<\/td>\r\n1.9992<\/td>\r\n1.999992<\/td>\r\n<\/tr><\/tbody><\/table>

Vertical asymptote [latex]x=-4[\/latex], Horizontal asymptote [latex]y=2[\/latex]<\/p>\r\n69.\r\n
x<\/em><\/td>\r\n\u2013.9<\/td>\r\n\u2013.99<\/td>\r\n\u2013.999<\/td>\r\n\u20131.1<\/td>\r\n\u20131.01<\/td>\r\n<\/tr>
y<\/em><\/td>\r\n81<\/td>\r\n9,801<\/td>\r\n998,001<\/td>\r\n121<\/td>\r\n10,201<\/td>\r\n<\/tr><\/tbody><\/table>
x<\/em><\/td>\r\n10<\/td>\r\n100<\/td>\r\n1,000<\/td>\r\n10,000<\/td>\r\n100,000<\/td>\r\n<\/tr>
y<\/i><\/td>\r\n.82645<\/td>\r\n.9803<\/td>\r\n.998<\/td>\r\n.9998<\/td>\r\n\r\n<\/tr><\/tbody><\/table>\r\nVertical asymptote [latex]x=-1[\/latex], Horizontal asymptote [latex]y=1[\/latex]\r\n\r\n71.\u00a0[latex]\\left(\\frac{3}{2},\\infty \\right)[\/latex]\r\n\"Graph\r\n\r\n73.\u00a0[latex]\\left(-2,1\\right)\\cup \\left(4,\\infty \\right)[\/latex]\r\n\"Graph\r\n\r\n75.\u00a0[latex]\\left(2,4\\right)[\/latex]\r\n\r\n77.\u00a0[latex]\\left(2,5\\right)[\/latex]\r\n\r\n79.\u00a0[latex]\\left(-1,\\text{1}\\right)[\/latex]\r\n\r\n81.\u00a0[latex]C\\left(t\\right)=\\frac{8+2t}{300+20t}[\/latex]\r\n\r\n83.\u00a0After about 6.12 hours.\r\n\r\n85.\u00a0[latex]A\\left(x\\right)=50{x}^{2}+\\frac{800}{x}[\/latex]. 2 by 2 by 5 feet.\r\n\r\n87.\u00a0[latex]A\\left(x\\right)=\\pi {x}^{2}+\\frac{100}{x}[\/latex]. Radius = 2.52 meters.","rendered":"

Solutions to Try Its<\/h2>\n

1.\u00a0End behavior: as [latex]x\\to \\pm \\infty , f\\left(x\\right)\\to 0[\/latex]; Local behavior: as [latex]x\\to 0, f\\left(x\\right)\\to \\infty[\/latex] (there are no x<\/em>– or y<\/em>-intercepts)<\/p>\n

2.\u00a0The function and the asymptotes are shifted 3 units right and 4 units down. As [latex]x\\to 3,f\\left(x\\right)\\to \\infty\\\\[\/latex], and as [latex]x\\to \\pm \\infty ,f\\left(x\\right)\\to -4[\/latex].<\/p>\n

The function is [latex]f\\left(x\\right)=\\frac{1}{{\\left(x - 3\\right)}^{2}}-4[\/latex].<\/p>\n

\"Graph<\/p>\n

3.\u00a0[latex]\\frac{12}{11}[\/latex]<\/p>\n

4.\u00a0The domain is all real numbers except [latex]x=1[\/latex] and [latex]x=5[\/latex].<\/p>\n

5.\u00a0Removable discontinuity at [latex]x=5[\/latex]. Vertical asymptotes: [latex]x=0,\\text{ }x=1[\/latex].<\/p>\n

6.\u00a0Vertical asymptotes at [latex]x=2[\/latex] and [latex]x=-3[\/latex]; horizontal asymptote at [latex]y=4[\/latex].<\/p>\n

7.\u00a0For the transformed reciprocal squared function, we find the rational form. [latex]f\\left(x\\right)=\\frac{1}{{\\left(x - 3\\right)}^{2}}-4=\\frac{1 - 4{\\left(x - 3\\right)}^{2}}{{\\left(x - 3\\right)}^{2}}=\\frac{1 - 4\\left({x}^{2}-6x+9\\right)}{\\left(x - 3\\right)\\left(x - 3\\right)}=\\frac{-4{x}^{2}+24x - 35}{{x}^{2}-6x+9}[\/latex]<\/p>\n

Because the numerator is the same degree as the denominator we know that as [latex]x\\to \\pm \\infty , f\\left(x\\right)\\to -4; \\text{so } y=-4[\/latex] is the horizontal asymptote. Next, we set the denominator equal to zero, and find that the vertical asymptote is [latex]x=3[\/latex], because as [latex]x\\to 3,f\\left(x\\right)\\to \\infty[\/latex]. We then set the numerator equal to 0 and find the x<\/em>-intercepts are at [latex]\\left(2.5,0\\right)[\/latex] and [latex]\\left(3.5,0\\right)[\/latex]. Finally, we evaluate the function at 0 and find the y<\/em>-intercept to be at [latex]\\left(0,\\frac{-35}{9}\\right)[\/latex].<\/p>\n

8.\u00a0Horizontal asymptote at [latex]y=\\frac{1}{2}[\/latex]. Vertical asymptotes at [latex]x=1 \\text{and} x=3[\/latex]. y<\/em>-intercept at [latex]\\left(0,\\frac{4}{3}.\\right)[\/latex]<\/p>\n

x<\/em>-intercepts at [latex]\\left(2,0\\right) \\text{ and }\\left(-2,0\\right)[\/latex]. [latex]\\left(-2,0\\right)[\/latex] is a zero with multiplicity 2, and the graph bounces off the x<\/em>-axis at this point. [latex]\\left(2,0\\right)[\/latex] is a single zero and the graph crosses the axis at this point.
\n\"Graph<\/span><\/p>\n

Solutions to Try Its<\/h2>\n

1.\u00a0The rational function will be represented by a quotient of polynomial functions.<\/p>\n

3.\u00a0The numerator and denominator must have a common factor.<\/p>\n

5.\u00a0Yes. The numerator of the formula of the functions would have only complex roots and\/or factors common to both the numerator and denominator.<\/p>\n

7.\u00a0[latex]\\text{All reals }x\\ne -1, 1[\/latex]<\/p>\n

9.\u00a0[latex]\\text{All reals }x\\ne -1, -2, 1, 2[\/latex]<\/p>\n

11.\u00a0V.A. at [latex]x=-\\frac{2}{5}[\/latex]; H.A. at [latex]y=0[\/latex]; Domain is all reals [latex]x\\ne -\\frac{2}{5}[\/latex]<\/p>\n

13.\u00a0V.A. at [latex]x=4, -9[\/latex]; H.A. at [latex]y=0[\/latex]; Domain is all reals [latex]x\\ne 4, -9[\/latex]<\/p>\n

15.\u00a0V.A. at [latex]x=0, 4, -4[\/latex]; H.A. at [latex]y=0[\/latex]; Domain is all reals [latex]x\\ne 0,4, -4[\/latex]<\/p>\n

17.\u00a0V.A. at [latex]x=-5[\/latex]; H.A. at [latex]y=0[\/latex]; Domain is all reals [latex]x\\ne 5,-5[\/latex]<\/p>\n

19.\u00a0V.A. at [latex]x=\\frac{1}{3}[\/latex]; H.A. at [latex]y=-\\frac{2}{3}[\/latex]; Domain is all reals [latex]x\\ne \\frac{1}{3}[\/latex].<\/p>\n

21.\u00a0none<\/p>\n

23.\u00a0[latex]x\\text{-intercepts none, }y\\text{-intercept }\\left(0,\\frac{1}{4}\\right)[\/latex]<\/p>\n

25.\u00a0Local behavior: [latex]x\\to -{\\frac{1}{2}}^{+},f\\left(x\\right)\\to -\\infty ,x\\to -{\\frac{1}{2}}^{-},f\\left(x\\right)\\to \\infty[\/latex]<\/p>\n

End behavior: [latex]x\\to \\pm \\infty ,f\\left(x\\right)\\to \\frac{1}{2}[\/latex]<\/p>\n

27.\u00a0Local behavior: [latex]x\\to {6}^{+},f\\left(x\\right)\\to -\\infty ,x\\to {6}^{-},f\\left(x\\right)\\to \\infty[\/latex], End behavior: [latex]x\\to \\pm \\infty ,f\\left(x\\right)\\to -2[\/latex]<\/p>\n

29.\u00a0Local behavior: [latex]x\\to -{\\frac{1}{3}}^{+},f\\left(x\\right)\\to \\infty ,x\\to -{\\frac{1}{3}}^{-}[\/latex], [latex]f\\left(x\\right)\\to -\\infty ,x\\to {\\frac{5}{2}}^{-},f\\left(x\\right)\\to \\infty ,x\\to {\\frac{5}{2}}^{+}[\/latex] ,\u00a0[latex]f\\left(x\\right)\\to -\\infty[\/latex]<\/p>\n

End behavior: [latex]x\\to \\pm \\infty\\\\[\/latex], [latex]f\\left(x\\right)\\to \\frac{1}{3}[\/latex]<\/p>\n

31.\u00a0[latex]y=2x+4[\/latex]<\/p>\n

33.\u00a0[latex]y=2x[\/latex]<\/p>\n

35.\u00a0[latex]V.A.\\text{ }x=0,H.A.\\text{ }y=2[\/latex]
\n\"Graph<\/p>\n

37.\u00a0[latex]V.A.\\text{ }x=2,\\text{ }H.A.\\text{ }y=0[\/latex]
\n\"Graph<\/p>\n

39.\u00a0[latex]V.A.\\text{ }x=-4,\\text{ }H.A.\\text{ }y=2;\\left(\\frac{3}{2},0\\right);\\left(0,-\\frac{3}{4}\\right)[\/latex]
\n\"Graph<\/p>\n

41.\u00a0[latex]V.A.\\text{ }x=2,\\text{ }H.A.\\text{ }y=0,\\text{ }\\left(0,1\\right)[\/latex]
\n\"Graph<\/p>\n

43.\u00a0[latex]V.A.\\text{ }x=-4,\\text{ }x=\\frac{4}{3},\\text{ }H.A.\\text{ }y=1;\\left(5,0\\right);\\left(-\\frac{1}{3},0\\right);\\left(0,\\frac{5}{16}\\right)[\/latex]<\/p>\n

45.\u00a0[latex]V.A.\\text{ }x=-1,\\text{ }H.A.\\text{ }y=1;\\left(-3,0\\right);\\left(0,3\\right)[\/latex]
\n\"Graph<\/p>\n

47.\u00a0[latex]V.A.\\text{ }x=4,\\text{ }S.A.\\text{ }y=2x+9;\\left(-1,0\\right);\\left(\\frac{1}{2},0\\right);\\left(0,\\frac{1}{4}\\right)[\/latex]
\n\"Graph<\/p>\n

49.\u00a0[latex]V.A.\\text{ }x=-2,\\text{ }x=4,\\text{ }H.A.\\text{ }y=1,\\left(1,0\\right);\\left(5,0\\right);\\left(-3,0\\right);\\left(0,-\\frac{15}{16}\\right)[\/latex]
\n\"Graph<\/p>\n

51.\u00a0[latex]y=50\\frac{{x}^{2}-x - 2}{{x}^{2}-25}[\/latex]<\/p>\n

53.\u00a0[latex]y=7\\frac{{x}^{2}+2x - 24}{{x}^{2}+9x+20}[\/latex]<\/p>\n

55.\u00a0[latex]y=\\frac{1}{2}\\frac{{x}^{2}-4x+4}{x+1}[\/latex]<\/p>\n

57.\u00a0[latex]y=4\\frac{x - 3}{{x}^{2}-x - 12}[\/latex]<\/p>\n

59.\u00a0[latex]y=-9\\frac{x - 2}{{x}^{2}-9}[\/latex]<\/p>\n

61.\u00a0[latex]y=\\frac{1}{3}\\frac{{x}^{2}+x - 6}{x - 1}[\/latex]<\/p>\n

63.\u00a0[latex]y=-6\\frac{{\\left(x - 1\\right)}^{2}}{\\left(x+3\\right){\\left(x - 2\\right)}^{2}}[\/latex]<\/p>\n

65.<\/p>\n\n\n\n\n
x<\/em><\/td>\n2.01<\/td>\n2.001<\/td>\n2.0001<\/td>\n1.99<\/td>\n1.999<\/td>\n<\/tr>\n
y<\/em><\/td>\n100<\/td>\n1,000<\/td>\n10,000<\/td>\n\u2013100<\/td>\n\u20131,000<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n\n\n\n\n
x<\/em><\/td>\n10<\/td>\n100<\/td>\n1,000<\/td>\n10,000<\/td>\n100,000<\/td>\n<\/tr>\n<\/tbody>\n
y<\/em><\/td>\n.125<\/td>\n.0102<\/td>\n.001<\/td>\n.0001<\/td>\n.00001<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Vertical asymptote [latex]x=2[\/latex], Horizontal asymptote [latex]y=0[\/latex]<\/p>\n

67.<\/p>\n\n\n\n\n
x<\/em><\/td>\n\u20134.1<\/td>\n\u20134.01<\/td>\n\u20134.001<\/td>\n\u20133.99<\/td>\n\u20133.999<\/td>\n<\/tr>\n
y<\/em><\/td>\n82<\/td>\n802<\/td>\n8,002<\/td>\n\u2013798<\/td>\n\u20137998<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n\n\n\n
x<\/em><\/td>\n10<\/td>\n100<\/td>\n1,000<\/td>\n10,000<\/td>\n100,000<\/td>\n<\/tr>\n
y<\/em><\/td>\n1.4286<\/td>\n1.9331<\/td>\n1.992<\/td>\n1.9992<\/td>\n1.999992<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Vertical asymptote [latex]x=-4[\/latex], Horizontal asymptote [latex]y=2[\/latex]<\/p>\n

69.<\/p>\n\n\n\n\n
x<\/em><\/td>\n\u2013.9<\/td>\n\u2013.99<\/td>\n\u2013.999<\/td>\n\u20131.1<\/td>\n\u20131.01<\/td>\n<\/tr>\n
y<\/em><\/td>\n81<\/td>\n9,801<\/td>\n998,001<\/td>\n121<\/td>\n10,201<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n\n\n\n\n
x<\/em><\/td>\n10<\/td>\n100<\/td>\n1,000<\/td>\n10,000<\/td>\n100,000<\/td>\n<\/tr>\n
y<\/i><\/td>\n.82645<\/td>\n.9803<\/td>\n.998<\/td>\n.9998<\/td>\n\n<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

Vertical asymptote [latex]x=-1[\/latex], Horizontal asymptote [latex]y=1[\/latex]<\/p>\n

71.\u00a0[latex]\\left(\\frac{3}{2},\\infty \\right)[\/latex]
\n\"Graph<\/p>\n

73.\u00a0[latex]\\left(-2,1\\right)\\cup \\left(4,\\infty \\right)[\/latex]
\n\"Graph<\/p>\n

75.\u00a0[latex]\\left(2,4\\right)[\/latex]<\/p>\n

77.\u00a0[latex]\\left(2,5\\right)[\/latex]<\/p>\n

79.\u00a0[latex]\\left(-1,\\text{1}\\right)[\/latex]<\/p>\n

81.\u00a0[latex]C\\left(t\\right)=\\frac{8+2t}{300+20t}[\/latex]<\/p>\n

83.\u00a0After about 6.12 hours.<\/p>\n

85.\u00a0[latex]A\\left(x\\right)=50{x}^{2}+\\frac{800}{x}[\/latex]. 2 by 2 by 5 feet.<\/p>\n

87.\u00a0[latex]A\\left(x\\right)=\\pi {x}^{2}+\\frac{100}{x}[\/latex]. Radius = 2.52 meters.<\/p>\n\n\t\t\t

\n\t\t\t

Candela Citations<\/h3>\n\t\t\t\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t\t\t\t
CC licensed content, Shared previously<\/div>