{"id":314,"date":"2023-09-20T22:49:01","date_gmt":"2023-09-20T22:49:01","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/drawing-graphs-of-functions\/"},"modified":"2025-08-17T23:55:44","modified_gmt":"2025-08-17T23:55:44","slug":"limits-at-infinity-and-asymptotes-learn-it-6","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/limits-at-infinity-and-asymptotes-learn-it-6\/","title":{"raw":"Limits at Infinity and Asymptotes: Learn It 6","rendered":"Limits at Infinity and Asymptotes: Learn It 6"},"content":{"raw":"<h2>Drawing Graphs of Functions<\/h2>\r\n<h3>Guidelines for Graphing a Function<\/h3>\r\n<p id=\"fs-id1165042602938\">We now have enough analytical tools to draw graphs of a wide variety of algebraic and transcendental functions. Before showing how to graph specific functions, let\u2019s look at a general strategy to use when graphing any function.<\/p>\r\n<section class=\"textbox questionHelp\">\r\n<p><strong>How To: Draw the Graph of a Function<\/strong><\/p>\r\n<p id=\"fs-id1165042602951\">Given a function [latex]f[\/latex] use the following steps to sketch a graph of [latex]f[\/latex]:<\/p>\r\n<ul id=\"fs-id1165042602969\">\r\n\t<li><strong>Step 1:<\/strong> Determine the domain of the function.<\/li>\r\n\t<li><strong>Step 2: <\/strong>Locate the [latex]x[\/latex]- and [latex]y[\/latex]-intercepts.<\/li>\r\n\t<li><strong>Step 3: <\/strong>Evaluate [latex]\\underset{x\\to \\infty }{\\lim}f(x)[\/latex] and [latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)[\/latex] to determine the end behavior.*<\/li>\r\n\t<li><strong>Step 4: <\/strong>Determine whether [latex]f[\/latex] has any vertical asymptotes.<\/li>\r\n\t<li><strong>Step 5: <\/strong>Calculate [latex]f^{\\prime}[\/latex]. Find all critical points and determine the intervals where [latex]f[\/latex] is increasing and where [latex]f[\/latex] is decreasing. Determine whether [latex]f[\/latex] has any local extrema.<\/li>\r\n\t<li><strong>Step 6: <\/strong>Calculate [latex]f^{\\prime \\prime}[\/latex]. Determine the intervals where [latex]f[\/latex] is concave up and where [latex]f[\/latex] is concave down. Use this information to determine whether [latex]f[\/latex] has any inflection points. The second derivative can also be used as an alternate means to determine or verify that [latex]f[\/latex] has a local extremum at a critical point.<\/li>\r\n<\/ul>\r\n<p>*Note for Step 3: If either of these limits is a finite number [latex]L[\/latex], then [latex]y=L[\/latex] is a horizontal asymptote. If either of these limits is [latex]\\infty [\/latex] or [latex]\u2212\\infty[\/latex], determine whether [latex]f[\/latex] has an oblique asymptote. If [latex]f[\/latex] is a rational function such that [latex]f(x)=\\frac{p(x)}{q(x)}[\/latex], where the degree of the numerator is greater than the degree of the denominator, then [latex]f[\/latex] can be written as<\/p>\r\n<div id=\"fs-id1165042617521\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=\\dfrac{p(x)}{q(x)}=g(x)+\\dfrac{r(x)}{q(x)}[\/latex],<\/div>\r\n<p>where the degree of [latex]r(x)[\/latex] is less than the degree of [latex]q(x)[\/latex]. The values of [latex]f(x)[\/latex] approach the values of [latex]g(x)[\/latex] as [latex]x\\to \\pm \\infty[\/latex]. If [latex]g(x)[\/latex] is a linear function, it is known as an <em>oblique asymptote<\/em>.<\/p>\r\n<\/section>\r\n<p id=\"fs-id1165042617762\">Now let\u2019s use this strategy to graph several different functions. We start by graphing a polynomial function.<\/p>\r\n<section class=\"textbox example\">\r\n<p>Sketch a graph of [latex]f(x)=(x-1)^2 (x+2)[\/latex]<\/p>\r\n\r\n[reveal-answer q=\"fs-id1165042707761\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165042707761\"]\r\n\r\n<p id=\"fs-id1165042707761\"><strong>Step 1.<\/strong> Since [latex]f[\/latex] is a polynomial, the domain is the set of all real numbers.<\/p>\r\n<p id=\"fs-id1165042707768\"><strong>Step 2.<\/strong> When [latex]x=0, \\, f(x)=2[\/latex]. Therefore, the [latex]y[\/latex]-intercept is [latex](0,2)[\/latex]. To find the [latex]x[\/latex]-intercepts, we need to solve the equation [latex](x-1)^2 (x+2)=0[\/latex], which gives us the [latex]x[\/latex]-intercepts [latex](1,0)[\/latex] and [latex](-2,0)[\/latex]<\/p>\r\n<p id=\"fs-id1165042707901\"><strong>Step 3<\/strong>. We need to evaluate the end behavior of [latex]f[\/latex]. As [latex]x\\to \\infty[\/latex], [latex](x-1)^2 \\to \\infty [\/latex] and [latex](x+2)\\to \\infty[\/latex]. Therefore, [latex]\\underset{x\\to \\infty }{\\lim}f(x)=\\infty[\/latex]. As [latex]x\\to \u2212\\infty[\/latex], [latex](x-1)^2 \\to \\infty [\/latex] and [latex](x+2) \\to \u2212\\infty[\/latex]. Therefore, [latex]\\underset{x\\to -\\infty }{\\lim}f(x)=\u2212\\infty[\/latex]. To get even more information about the end behavior of [latex]f[\/latex], we can multiply the factors of [latex]f[\/latex]. When doing so, we see that,<\/p>\r\n<div id=\"fs-id1165042525463\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=(x-1)^2 (x+2)=x^3-3x+2[\/latex]<\/div>\r\n<p id=\"fs-id1165042525532\">Since the leading term of [latex]f[\/latex] is [latex]x^3[\/latex], we conclude that [latex]f[\/latex] behaves like [latex]y=x^3[\/latex] as [latex]x\\to \\pm \\infty[\/latex].<\/p>\r\n<p id=\"fs-id1165043382921\"><strong>Step 4.<\/strong> Since [latex]f[\/latex] is a polynomial function, it does not have any vertical asymptotes.<\/p>\r\n<p id=\"fs-id1165043382928\"><strong>Step 5.<\/strong> The first derivative of [latex]f[\/latex] is,<\/p>\r\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=3x^2-3[\/latex]<\/div>\r\n<p id=\"fs-id1165043382970\">Therefore, [latex]f[\/latex] has two critical points: [latex]x=1,-1[\/latex]. Divide the interval [latex](\u2212\\infty ,\\infty)[\/latex] into the three smaller intervals: [latex](\u2212\\infty ,-1)[\/latex], [latex](-1,1)[\/latex], and [latex](1,\\infty )[\/latex].<\/p>\r\n<p>Then, choose test points [latex]x=-2[\/latex], [latex]x=0[\/latex], and [latex]x=2[\/latex] from these intervals and evaluate the sign of [latex]f^{\\prime}(x)[\/latex] at each of these test points, as shown in the following table.<\/p>\r\n<table id=\"fs-id1165043383127\" class=\"unnumbered\" summary=\"This table has four rows and four columns. The first row is a header row, and it reads Interval, Test Point, Sign of Derivative f\u2019(x) = 3x2 \u2013 3 = 3(x \u2013 1)(x + 1), and Conclusion. Under the header row, the first column reads (\u2212\u221e, \u22121), (\u22121, 1), and (1, \u221e). The second column reads x = \u22122, x = 0, and x = 2. The third column reads (+)(\u2212)(\u2212) = +, (+)(\u2212)(+) = \u2212, and (+)(+)(+) = +. The fourth column reads f is increasing, f is decreasing, and f is increasing.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Interval<\/th>\r\n<th>Test Point<\/th>\r\n<th>Sign of Derivative [latex]f^{\\prime}(x)=3x^2-3=3(x-1)(x+1)[\/latex]<\/th>\r\n<th>Conclusion<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex](\u2212\\infty ,-1)[\/latex]<\/td>\r\n<td>[latex]x=-2[\/latex]<\/td>\r\n<td>[latex](+)(\u2212)(\u2212)=+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is increasing.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](-1,1)[\/latex]<\/td>\r\n<td>[latex]x=0[\/latex]<\/td>\r\n<td>[latex](+)(\u2212)(+)=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is decreasing.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](1,\\infty )[\/latex]<\/td>\r\n<td>[latex]x=2[\/latex]<\/td>\r\n<td>[latex](+)(+)(+)=+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is increasing.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165042539204\">From the table, we see that [latex]f[\/latex] has a local maximum at [latex]x=-1[\/latex] and a local minimum at [latex]x=1[\/latex]. Evaluating [latex]f(x)[\/latex] at those two points, we find that the local maximum value is [latex]f(-1)=4[\/latex] and the local minimum value is [latex]f(1)=0[\/latex].<\/p>\r\n<p id=\"fs-id1165042539284\"><strong>Step 6.<\/strong> The second derivative of [latex]f[\/latex] is,<\/p>\r\n<div id=\"fs-id1165042539292\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime \\prime}(x)=6x[\/latex]<\/div>\r\n<p id=\"fs-id1165042539318\">The second derivative is zero at [latex]x=0[\/latex].<\/p>\r\n<p>Therefore, to determine the concavity of [latex]f[\/latex], divide the interval [latex](\u2212\\infty ,\\infty)[\/latex] into the smaller intervals [latex](\u2212\\infty ,0)[\/latex] and [latex](0,\\infty )[\/latex], and choose test points [latex]x=-1[\/latex] and [latex]x=1[\/latex] to determine the concavity of [latex]f[\/latex] on each of these smaller intervals as shown in the following table.<\/p>\r\n<table id=\"fs-id1165042381927\" class=\"unnumbered\" summary=\"This table has three rows and four columns. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019\u2019(x) = 6x, and Conclusion. Under the header row, the first column reads (\u2212\u221e, 0) and (0, \u221e). The second column reads x = \u22121 and x = 1. The third column reads \u2212 and +. The fourth column reads f is concave down and f is concave up.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Interval<\/th>\r\n<th>Test Point<\/th>\r\n<th>Sign of [latex]f^{\\prime \\prime}(x)=6x[\/latex]<\/th>\r\n<th>Conclusion<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex](\u2212\\infty ,0)[\/latex]<\/td>\r\n<td>[latex]x=-1[\/latex]<\/td>\r\n<td>[latex]-[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is concave down.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](0,\\infty )[\/latex]<\/td>\r\n<td>[latex]x=1[\/latex]<\/td>\r\n<td>[latex]+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is concave up.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165042382153\">We note that the information in the preceding table confirms the fact, found in step 5, that [latex]f[\/latex] has a local maximum at [latex]x=-1[\/latex] and a local minimum at [latex]x=1[\/latex]. In addition, the information found in step 5\u2014namely, [latex]f[\/latex] has a local maximum at [latex]x=-1[\/latex] and a local minimum at [latex]x=1[\/latex], and [latex]f^{\\prime}(x)=0[\/latex] at those points\u2014combined with the fact that [latex]f^{\\prime \\prime}[\/latex] changes sign only at [latex]x=0[\/latex] confirms the results found in step 6 on the concavity of [latex]f[\/latex].<\/p>\r\n<p id=\"fs-id1165042709981\">Combining this information, we arrive at the graph of [latex]f(x)=(x-1)^2 (x+2)[\/latex] shown in the following graph.<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"400\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211132\/CNX_Calc_Figure_04_06_015.jpg\" alt=\"The function f(x) = (x \u22121)2 (x + 2) is graphed. It crosses the x axis at x = \u22122 and touches the x axis at x = 1.\" width=\"400\" height=\"347\" \/> Figure 23. Graph of f(x)[\/caption]\r\n\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/QhOY9VfIefo?controls=0&amp;start=003&amp;end=317&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.6LimitsAtInfinityAndAsymptotesPart2_3to317_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.6 Limits at Infinity and Asymptotes (part 2 - curve sketching)\" here (opens in new window)<\/a>.[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>Sketch the graph of [latex]f(x)=\\dfrac{x^2}{1-x^2}[\/latex]<\/p>\r\n\r\n[reveal-answer q=\"fs-id1165042407414\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165042407414\"]\r\n\r\n<p id=\"fs-id1165042407414\"><strong>Step 1.<\/strong> The function [latex]f[\/latex] is defined as long as the denominator is not zero. Therefore, the domain is the set of all real numbers [latex]x[\/latex] except [latex]x=\\pm 1[\/latex].<\/p>\r\n<p id=\"fs-id1165042407440\"><strong>Step 2.<\/strong> Find the intercepts. If [latex]x=0[\/latex], then [latex]f(x)=0[\/latex], so 0 is an intercept. If [latex]y=0[\/latex], then [latex]\\frac{x^2}{1-x^2}=0[\/latex], which implies [latex]x=0[\/latex]. Therefore, [latex](0,0)[\/latex] is the only intercept.<\/p>\r\n<p id=\"fs-id1165042407553\"><strong>Step 3.<\/strong> Evaluate the limits at infinity. Since [latex]f[\/latex] is a rational function, divide the numerator and denominator by the highest power in the denominator: [latex]x^2[\/latex]. We obtain,<\/p>\r\n<div id=\"fs-id1165042407571\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\pm \\infty }{\\lim}\\frac{x^2}{1-x^2}=\\underset{x\\to \\pm \\infty }{\\lim}\\frac{1}{\\frac{1}{x^2}-1}=-1[\/latex].<\/div>\r\n<p>&nbsp;<\/p>\r\n<p id=\"fs-id1165042407655\">Therefore, [latex]f[\/latex] has a horizontal asymptote of [latex]y=-1[\/latex] as [latex]x\\to \\infty [\/latex] and [latex]x\\to \u2212\\infty[\/latex].<\/p>\r\n<p id=\"fs-id1165042491004\"><strong>Step 4.<\/strong> To determine whether [latex]f[\/latex] has any vertical asymptotes, first check to see whether the denominator has any zeroes. We find the denominator is zero when [latex]x=\\pm 1[\/latex].<\/p>\r\n<p>To determine whether the lines [latex]x=1[\/latex] or [latex]x=-1[\/latex] are vertical asymptotes of [latex]f[\/latex], evaluate [latex]\\underset{x\\to 1}{\\lim}f(x)[\/latex] and [latex]\\underset{x\\to \u22121}{\\lim}f(x)[\/latex].<\/p>\r\n<p>By looking at each one-sided limit as [latex]x\\to 1[\/latex], we see that,<\/p>\r\n<div id=\"fs-id1165042491126\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 1^+}{\\lim}\\frac{x^2}{1-x^2}=\u2212\\infty[\/latex] and [latex]\\underset{x\\to 1^-}{\\lim}\\frac{x^2}{1-x^2}=\\infty[\/latex].<\/div>\r\n<p id=\"fs-id1165042491222\">In addition, by looking at each one-sided limit as [latex]x\\to \u22121[\/latex], we find that,<\/p>\r\n<div id=\"fs-id1165042491242\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \u22121^+}{\\lim}\\frac{x^2}{1-x^2}=\\infty[\/latex] and [latex]\\underset{x\\to \u22121^-}{\\lim}\\frac{x^2}{1-x^2}=\u2212\\infty[\/latex].<\/div>\r\n<p id=\"fs-id1165043262266\"><strong>Step 5<\/strong>. Calculate the first derivative:<\/p>\r\n<div id=\"fs-id1165043262270\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{(1-x^2)(2x)-x^2(-2x)}{(1-x^2)^2}=\\frac{2x}{(1-x^2)^2}[\/latex].<\/div>\r\n<p id=\"fs-id1165043262390\">Critical points occur at points [latex]x[\/latex] where [latex]f^{\\prime}(x)=0[\/latex] or [latex]f^{\\prime}(x)[\/latex] is undefined. We see that [latex]f^{\\prime}(x)=0[\/latex] when [latex]x=0[\/latex]. The derivative [latex]f^{\\prime}[\/latex] is not undefined at any point in the domain of [latex]f[\/latex]. However, [latex]x=\\pm 1[\/latex] are not in the domain of [latex]f[\/latex].<\/p>\r\n<p>Therefore, to determine where [latex]f[\/latex] is increasing and where [latex]f[\/latex] is decreasing, divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into four smaller intervals: [latex](\u2212\\infty ,-1)[\/latex], [latex](-1,0)[\/latex], [latex](0,1)[\/latex], and [latex](1,\\infty )[\/latex], and choose a test point in each interval to determine the sign of [latex]f^{\\prime}(x)[\/latex] in each of these intervals. The values [latex]x=-2[\/latex], [latex]x=-\\frac{1}{2}[\/latex], [latex]x=\\frac{1}{2}[\/latex], and [latex]x=2[\/latex] are good choices for test points as shown in the following table.<\/p>\r\n<table id=\"fs-id1165043341618\" class=\"unnumbered\" summary=\"This table has four columns and five rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019(x) = 2x\/(1 \u2212 x2)2, and Conclusion. Under the header row, the first column reads (\u2212\u221e, \u22121), (\u22121, 0), (0, 1), and (1, \u221e). The second column reads x = \u22122, x = \u22121\/2, x = 1\/2, and x = 2. The third column reads \u2212\/+ = \u2212, \u2212\/+ = \u2212, +\/+ = +, and +\/+ = +. The fourth column reads f is decreasing, f is decreasing, f is increasing, and f is increasing.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Interval<\/th>\r\n<th>Test Point<\/th>\r\n<th>Sign of [latex]f^{\\prime}(x)=\\frac{2x}{(1-x^2)^2}[\/latex]<\/th>\r\n<th>Conclusion<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex](\u2212\\infty ,-1)[\/latex]<\/td>\r\n<td>[latex]x=-2[\/latex]<\/td>\r\n<td>[latex]\u2212\/+=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is decreasing.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](-1,0)[\/latex]<\/td>\r\n<td>[latex]x=-1\/2[\/latex]<\/td>\r\n<td>[latex]\u2212\/+=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is decreasing.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](0,1)[\/latex]<\/td>\r\n<td>[latex]x=1\/2[\/latex]<\/td>\r\n<td>[latex]+\/+=+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is increasing.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](1,\\infty )[\/latex]<\/td>\r\n<td>[latex]x=2[\/latex]<\/td>\r\n<td>[latex]+\/+=+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is increasing.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165043183348\">From this analysis, we conclude that [latex]f[\/latex] has a local minimum at [latex]x=0[\/latex] but no local maximum.<\/p>\r\n<p id=\"fs-id1165043183355\"><strong>Step 6.<\/strong> Calculate the second derivative:<\/p>\r\n<div id=\"fs-id1165043183358\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} f^{\\prime \\prime}(x) &amp; =\\frac{(1-x^2)^2(2)-2x(2(1-x^2)(-2x))}{(1-x^2)^4} \\\\ &amp; =\\frac{(1-x^2)[2(1-x^2)+8x^2]}{(1-x^2)^4} \\\\ &amp; =\\frac{2(1-x^2)+8x^2}{(1-x^2)^3} \\\\ &amp; =\\frac{6x^2+2}{(1-x^2)^3} \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165043384128\">To determine the intervals where [latex]f[\/latex] is concave up and where [latex]f[\/latex] is concave down, we first need to find all points [latex]x[\/latex] where [latex]f^{\\prime \\prime}(x)=0[\/latex] or [latex]f^{\\prime \\prime}(x)[\/latex] is undefined. Since the numerator [latex]6x^2+2 \\ne 0[\/latex] for any [latex]x[\/latex], [latex]f^{\\prime \\prime}(x)[\/latex] is never zero.<\/p>\r\n<p>Furthermore, [latex]f^{\\prime \\prime}[\/latex] is not undefined for any [latex]x[\/latex] in the domain of [latex]f[\/latex]. However, as discussed earlier, [latex]x=\\pm 1[\/latex] are not in the domain of [latex]f[\/latex].<\/p>\r\n<p>Therefore, to determine the concavity of [latex]f[\/latex], we divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into the three smaller intervals [latex](\u2212\\infty ,-1)[\/latex], [latex](-1,-1)[\/latex], and [latex](1,\\infty )[\/latex], and choose a test point in each of these intervals to evaluate the sign of [latex]f^{\\prime \\prime}(x)[\/latex] in each of these intervals. The values [latex]x=-2[\/latex], [latex]x=0[\/latex], and [latex]x=2[\/latex] are possible test points as shown in the following table.<\/p>\r\n<table id=\"fs-id1165043384406\" class=\"unnumbered\" summary=\"This table has four columns and four rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019\u2019(x) = (6x2 + 2)\/(1 \u2212 x2)3, and Conclusion. Under the header row, the first column reads (\u2212\u221e, \u22121), (\u22121, 1), and (1, \u221e). The second column reads x = \u22122, x = 0, and x = 2. The third column reads +\/\u2212 = \u2212, +\/+ = +, and +\/\u2212 = \u2212. The fourth column reads f is concave down, f is concave up, and f is concave down.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Interval<\/th>\r\n<th>Test Point<\/th>\r\n<th>Sign of [latex]f^{\\prime \\prime}(x)=\\frac{6x^2+2}{(1-x^2)^3}[\/latex]<\/th>\r\n<th>Conclusion<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex](\u2212\\infty ,-1)[\/latex]<\/td>\r\n<td>[latex]x=-2[\/latex]<\/td>\r\n<td>[latex]+\/-=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is concave down.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](-1,-1)[\/latex]<\/td>\r\n<td>[latex]x=0[\/latex]<\/td>\r\n<td>[latex]+\/+=+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is concave up.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](1,\\infty )[\/latex]<\/td>\r\n<td>[latex]x=2[\/latex]<\/td>\r\n<td>[latex]+\/-=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is concave down.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165042592688\">Combining all this information, we arrive at the graph of [latex]f[\/latex] shown below. Note that, although [latex]f[\/latex] changes concavity at [latex]x=-1[\/latex] and [latex]x=1[\/latex], there are no inflection points at either of these places because [latex]f[\/latex] is not continuous at [latex]x=-1[\/latex] or [latex]x=1[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"342\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211139\/CNX_Calc_Figure_04_06_016.jpg\" alt=\"The function f(x) = x2\/(1 \u2212 x2) is graphed. It has asymptotes y = \u22121, x = \u22121, and x = 1.\" width=\"342\" height=\"347\" \/> Figure 25. Graph of y[\/caption]\r\n\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>Sketch the graph of [latex]f(x)=\\dfrac{x^2}{x-1}[\/latex]<\/p>\r\n\r\n[reveal-answer q=\"fs-id1165042593006\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165042593006\"]\r\n\r\n<p id=\"fs-id1165042593006\"><strong>Step 1.<\/strong> The domain of [latex]f[\/latex] is the set of all real numbers [latex]x[\/latex] except [latex]x=1[\/latex].<\/p>\r\n<p id=\"fs-id1165042712534\"><strong>Step 2.<\/strong> Find the intercepts. We can see that when [latex]x=0[\/latex], [latex]f(x)=0[\/latex], so [latex](0,0)[\/latex] is the only intercept.<\/p>\r\n<p id=\"fs-id1165042712584\"><strong>Step 3<\/strong>. Evaluate the limits at infinity. Since the degree of the numerator is one more than the degree of the denominator, [latex]f[\/latex] must have an oblique asymptote. To find the oblique asymptote, use long division of polynomials to write,<\/p>\r\n<div id=\"fs-id1165042712594\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=\\frac{x^2}{x-1}=x+1+\\frac{1}{x-1}[\/latex].<\/div>\r\n<p id=\"fs-id1165042712649\">Since [latex]1\/(x-1)\\to 0[\/latex] as [latex]x\\to \\pm \\infty[\/latex], [latex]f(x)[\/latex] approaches the line [latex]y=x+1[\/latex] as [latex]x\\to \\pm \\infty[\/latex]. The line [latex]y=x+1[\/latex] is an oblique asymptote for [latex]f[\/latex].<\/p>\r\n<p id=\"fs-id1165042712758\"><strong>Step 4.<\/strong> To check for vertical asymptotes, look at where the denominator is zero. Here the denominator is zero at [latex]x=1[\/latex]. Looking at both one-sided limits as [latex]x\\to 1[\/latex], we find,<\/p>\r\n<div id=\"fs-id1165042712789\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 1^+}{\\lim}\\frac{x^2}{x-1}=\\infty[\/latex] and [latex]\\underset{x\\to 1^-}{\\lim}\\frac{x^2}{x-1}=\u2212\\infty[\/latex].<\/div>\r\n<p id=\"fs-id1165042403315\">Therefore, [latex]x=1[\/latex] is a vertical asymptote, and we have determined the behavior of [latex]f[\/latex] as [latex]x[\/latex] approaches 1 from the right and the left.<\/p>\r\n<p id=\"fs-id1165042403340\"><strong>Step 5.<\/strong> Calculate the first derivative:<\/p>\r\n<div id=\"fs-id1165042403344\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{(x-1)(2x)-x^2(1)}{(x-1)^2}=\\frac{x^2-2x}{(x-1)^2}[\/latex].<\/div>\r\n<p id=\"fs-id1165042403459\">We have [latex]f^{\\prime}(x)=0[\/latex] when [latex]x^2-2x=x(x-2)=0[\/latex]. Therefore, [latex]x=0[\/latex] and [latex]x=2[\/latex] are critical points. Since [latex]f[\/latex] is undefined at [latex]x=1[\/latex], we need to divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into the smaller intervals [latex](\u2212\\infty ,0)[\/latex], [latex](0,1)[\/latex], [latex](1,2)[\/latex], and [latex](2,\\infty )[\/latex], and choose a test point from each interval to evaluate the sign of [latex]f^{\\prime}(x)[\/latex] in each of these smaller intervals.<\/p>\r\n<p>For example, let [latex]x=-1[\/latex], [latex]x=\\frac{1}{2}[\/latex], [latex]x=\\frac{3}{2}[\/latex], and [latex]x=3[\/latex] be the test points as shown in the following table.<\/p>\r\n<table id=\"fs-id1165042710566\" class=\"unnumbered\" summary=\"This table has four columns and five rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019(x) = (x2 \u2212 2x)\/(x \u2212 1)2 = x(x \u2212 2)\/(x \u2212 1)2, and Conclusion. Under the header row, the first column reads (\u2212\u221e, 0), (0, 1), (1, 2), and (2, \u221e). The second column reads x = \u22121, x = 1\/2, x = 3\/2, and x = 3. The third column reads (\u2212)(\u2212)\/+ = +, (+)(\u2212)\/+ = \u2212, (+)(\u2212)\/+ = \u2212, and (+)(+)\/+ = +. The fourth column reads f is increasing, f is decreasing, f is decreasing, and f is increasing.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Interval<\/th>\r\n<th>Test Point<\/th>\r\n<th>Sign of [latex]f^{\\prime}(x)=\\frac{x^2-2x}{(x-1)^2}=\\frac{x(x-2)}{(x-1)^2}[\/latex]<\/th>\r\n<th>Conclusion<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex](\u2212\\infty ,0)[\/latex]<\/td>\r\n<td>[latex]x=-1[\/latex]<\/td>\r\n<td>[latex](\u2212)(\u2212)\/+=+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is increasing.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](0,1)[\/latex]<\/td>\r\n<td>[latex]x=1\/2[\/latex]<\/td>\r\n<td>[latex](+)(\u2212)\/+=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is decreasing.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](1,2)[\/latex]<\/td>\r\n<td>[latex]x=3\/2[\/latex]<\/td>\r\n<td>[latex](+)(\u2212)\/+=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is decreasing.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](2,\\infty )[\/latex]<\/td>\r\n<td>[latex]x=3[\/latex]<\/td>\r\n<td>[latex](+)(+)\/+=+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is increasing.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165042476453\">From this table, we see that [latex]f[\/latex] has a local maximum at [latex]x=0[\/latex] and a local minimum at [latex]x=2[\/latex]. The value of [latex]f[\/latex] at the local maximum is [latex]f(0)=0[\/latex] and the value of [latex]f[\/latex] at the local minimum is [latex]f(2)=4[\/latex]. Therefore, [latex](0,0)[\/latex] and [latex](2,4)[\/latex] are important points on the graph.<\/p>\r\n<p id=\"fs-id1165042464546\"><strong>Step 6.<\/strong> Calculate the second derivative:<\/p>\r\n<div id=\"fs-id1165042464549\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} f^{\\prime \\prime}(x) &amp; =\\frac{(x-1)^2(2x-2)-(x^2-2x)(2(x-1))}{(x-1)^4} \\\\ &amp; =\\frac{(x-1)[(x-1)(2x-2)-2(x^2-2x)]}{(x-1)^4} \\\\ &amp; =\\frac{(x-1)(2x-2)-2(x^2-2x)}{(x-1)^3} \\\\ &amp; =\\frac{2x^2-4x+2-(2x^2-4x)}{(x-1)^3} \\\\ &amp; =\\frac{2}{(x-1)^3} \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1165042652400\">We see that [latex]f^{\\prime \\prime}(x)[\/latex] is never zero or undefined for [latex]x[\/latex] in the domain of [latex]f[\/latex]. Since [latex]f[\/latex] is undefined at [latex]x=1[\/latex], to check concavity we just divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into the two smaller intervals [latex](\u2212\\infty ,1)[\/latex] and [latex](1,\\infty )[\/latex], and choose a test point from each interval to evaluate the sign of [latex]f^{\\prime \\prime}(x)[\/latex] in each of these intervals. The values [latex]x=0[\/latex] and [latex]x=2[\/latex] are possible test points as shown in the following table.<\/p>\r\n<table id=\"fs-id1165042652542\" class=\"unnumbered\" summary=\"This table has four columns and three rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019\u2019(x) = 2\/(x \u2212 1)3, and Conclusion. Under the header row, the first column reads (\u2212\u221e, 1) and (1, \u221e). The column row reads x = 0 and x = 2. The third column reads +\/\u2212 = \u2212 and +\/+ = +. The fourth column reads f is concave down and f is concave up.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Interval<\/th>\r\n<th>Test Point<\/th>\r\n<th>Sign of [latex]f^{\\prime \\prime}(x)=\\frac{2}{(x-1)^3}[\/latex]<\/th>\r\n<th>Conclusion<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex](\u2212\\infty ,1)[\/latex]<\/td>\r\n<td>[latex]x=0[\/latex]<\/td>\r\n<td>[latex]+\/-=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is concave down.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](1,\\infty )[\/latex]<\/td>\r\n<td>[latex]x=2[\/latex]<\/td>\r\n<td>[latex]+\/+=+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is concave up.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165042606958\">From the information gathered, we arrive at the following graph for [latex]f.[\/latex]<\/p>\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"342\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211145\/CNX_Calc_Figure_04_06_017.jpg\" alt=\"The function f(x) = x2\/(x \u2212 1) is graphed. It has asymptotes y = x + 1 and x = 1.\" width=\"342\" height=\"346\" \/> Figure 27. Graph of y[\/caption]\r\n\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/QhOY9VfIefo?controls=0&amp;start=552&amp;end=777&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.6LimitsAtInfinityAndAsymptotesPart2_552to777_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.6 Limits at Infinity and Asymptotes (part 2 - curve sketching)\" here (opens in new window)<\/a>.[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>Sketch a graph of [latex]f(x)=(x-1)^{\\frac{2}{3}}[\/latex]<\/p>\r\n<div id=\"fs-id1165042607093\" class=\"exercise\">[reveal-answer q=\"fs-id1165042607145\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165042607145\"]\r\n\r\n<p id=\"fs-id1165042607145\"><strong>Step 1.<\/strong> Since the cube-root function is defined for all real numbers [latex]x[\/latex] and [latex](x-1)^{2\/3}=(\\sqrt[3]{x-1})^2[\/latex], the domain of [latex]f[\/latex] is all real numbers.<\/p>\r\n<p id=\"fs-id1165042607208\"><strong>Step 2:<\/strong> To find the [latex]y[\/latex]-intercept, evaluate [latex]f(0)[\/latex]. Since [latex]f(0)=1[\/latex], the [latex]y[\/latex]-intercept is [latex](0,1)[\/latex]. To find the [latex]x[\/latex]-intercept, solve [latex](x-1)^{2\/3}=0[\/latex]. The solution of this equation is [latex]x=1[\/latex], so the [latex]x[\/latex]-intercept is [latex](1,0)[\/latex].<\/p>\r\n<p id=\"fs-id1165042583208\"><strong>Step 3:<\/strong> Since [latex]\\underset{x\\to \\pm \\infty }{\\lim}(x-1)^{2\/3}=\\infty [\/latex], the function continues to grow without bound as [latex]x\\to \\infty [\/latex] and [latex]x\\to \u2212\\infty[\/latex].<\/p>\r\n<p id=\"fs-id1165042583285\"><strong>Step 4:<\/strong> The function has no vertical asymptotes.<\/p>\r\n<p id=\"fs-id1165042583288\"><strong>Step 5:<\/strong> To determine where [latex]f[\/latex] is increasing or decreasing, calculate [latex]f^{\\prime}[\/latex]. We find,<\/p>\r\n<div id=\"fs-id1165042583307\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{2}{3}(x-1)^{-1\/3}=\\frac{2}{3(x-1)^{1\/3}}[\/latex].<\/div>\r\n<p id=\"fs-id1165042583388\">This function is not zero anywhere, but it is undefined when [latex]x=1[\/latex]. Therefore, the only critical point is [latex]x=1[\/latex]. Divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into the smaller intervals [latex](\u2212\\infty ,1)[\/latex] and [latex](1,\\infty )[\/latex], and choose test points in each of these intervals to determine the sign of [latex]f^{\\prime}(x)[\/latex] in each of these smaller intervals.<\/p>\r\n<p>Let [latex]x=0[\/latex] and [latex]x=2[\/latex] be the test points as shown in the following table.<\/p>\r\n<table id=\"fs-id1165042504467\" class=\"unnumbered\" summary=\"This table has four columns and three rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019(x) = 2\/(3(x \u2212 1)1\/3), and Conclusion. Under the header row, the first column reads (\u2212\u221e, 1) and (1, \u221e). The second column reads x = 0 and x = 2. The third column reads +\/\u2212 = \u2212 and +\/+ = +. The fourth column reads f is decreasing and f is increasing.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Interval<\/th>\r\n<th>Test Point<\/th>\r\n<th>Sign of [latex]f^{\\prime}(x)=\\frac{2}{3(x-1)^{1\/3}}[\/latex]<\/th>\r\n<th>Conclusion<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex](\u2212\\infty ,1)[\/latex]<\/td>\r\n<td>[latex]x=0[\/latex]<\/td>\r\n<td>[latex]+\/-=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is decreasing.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](1,\\infty )[\/latex]<\/td>\r\n<td>[latex]x=2[\/latex]<\/td>\r\n<td>[latex]+\/+=+[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is increasing.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165042504734\">We conclude that [latex]f[\/latex] has a local minimum at [latex]x=1[\/latex]. Evaluating [latex]f[\/latex] at [latex]x=1[\/latex], we find that the value of [latex]f[\/latex] at the local minimum is zero. Note that [latex]f^{\\prime}(1)[\/latex] is undefined, so to determine the behavior of the function at this critical point, we need to examine [latex]\\underset{x\\to 1}{\\lim}f^{\\prime}(x)[\/latex].<\/p>\r\n<p>Looking at the one-sided limits, we have,<\/p>\r\n<div id=\"fs-id1165042510168\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 1^+}{\\lim}\\frac{2}{3(x-1)^{1\/3}}=\\infty[\/latex] and [latex]\\underset{x\\to 1^-}{\\lim}\\frac{2}{3(x-1)^{1\/3}}=\u2212\\infty[\/latex].<\/div>\r\n<p id=\"fs-id1165042510285\">Therefore, [latex]f[\/latex] has a cusp at [latex]x=1[\/latex].<\/p>\r\n<p id=\"fs-id1165042510304\"><strong>Step 6:<\/strong> To determine concavity, we calculate the second derivative of [latex]f[\/latex]:<\/p>\r\n<div id=\"fs-id1165042510313\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime \\prime}(x)=-\\frac{2}{9}(x-1)^{-4\/3}=\\frac{-2}{9(x-1)^{4\/3}}[\/latex].<\/div>\r\n<p id=\"fs-id1165042510398\">We find that [latex]f^{\\prime \\prime}(x)[\/latex] is defined for all [latex]x[\/latex], but is undefined when [latex]x=1[\/latex]. Therefore, divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into the smaller intervals [latex](\u2212\\infty ,1)[\/latex] and [latex](1,\\infty )[\/latex], and choose test points to evaluate the sign of [latex]f^{\\prime \\prime}(x)[\/latex] in each of these intervals.<\/p>\r\n<p>As we did earlier, let [latex]x=0[\/latex] and [latex]x=2[\/latex] be test points as shown in the following table.<\/p>\r\n<table id=\"fs-id1165042510530\" class=\"unnumbered\" summary=\"This table has four columns and three rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019\u2019(x) = \u22122\/(9(x \u2212 1)4\/3), and Conclusion. Under the header row, the first column reads (\u2212\u221e, 1) and (1, \u221e). The second column reads x = 0 and x = 2. The third column reads \u2212\/+ = \u2212 and \u2212\/+ = \u2212. The fourth column reads f is concave down and f is concave down.\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Interval<\/th>\r\n<th>Test Point<\/th>\r\n<th>Sign of [latex]f^{\\prime \\prime}(x)=\\frac{-2}{9(x-1)^{4\/3}}[\/latex]<\/th>\r\n<th>Conclusion<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>[latex](\u2212\\infty ,1)[\/latex]<\/td>\r\n<td>[latex]x=0[\/latex]<\/td>\r\n<td>[latex]\u2212\/+=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is concave down.<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>[latex](1,\\infty )[\/latex]<\/td>\r\n<td>[latex]x=2[\/latex]<\/td>\r\n<td>[latex]\u2212\/+=\u2212[\/latex]<\/td>\r\n<td>[latex]f[\/latex] is concave down.<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-id1165042643969\">From this table, we conclude that [latex]f[\/latex] is concave down everywhere. Combining all of this information, we arrive at the following graph for [latex]f[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"alignnone\" width=\"604\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211148\/CNX_Calc_Figure_04_06_018.jpg\" alt=\"The function f(x) = (x \u2212 1)2\/3 is graphed. It touches the x axis at x = 1, where it comes to something of a sharp point and then flairs out on either side.\" width=\"604\" height=\"272\" \/> Figure 28. Graph of f(x)[\/caption]\r\n\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/QhOY9VfIefo?controls=0&amp;start=781&amp;end=962&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.6LimitsAtInfinityAndAsymptotesPart2_781to962_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.6 Limits at Infinity and Asymptotes (part 2 - curve sketching)\" here (opens in new window)<\/a>.[\/hidden-answer]<\/p>\r\n<\/div>\r\n<\/section>","rendered":"<h2>Drawing Graphs of Functions<\/h2>\n<h3>Guidelines for Graphing a Function<\/h3>\n<p id=\"fs-id1165042602938\">We now have enough analytical tools to draw graphs of a wide variety of algebraic and transcendental functions. Before showing how to graph specific functions, let\u2019s look at a general strategy to use when graphing any function.<\/p>\n<section class=\"textbox questionHelp\">\n<p><strong>How To: Draw the Graph of a Function<\/strong><\/p>\n<p id=\"fs-id1165042602951\">Given a function [latex]f[\/latex] use the following steps to sketch a graph of [latex]f[\/latex]:<\/p>\n<ul id=\"fs-id1165042602969\">\n<li><strong>Step 1:<\/strong> Determine the domain of the function.<\/li>\n<li><strong>Step 2: <\/strong>Locate the [latex]x[\/latex]&#8211; and [latex]y[\/latex]-intercepts.<\/li>\n<li><strong>Step 3: <\/strong>Evaluate [latex]\\underset{x\\to \\infty }{\\lim}f(x)[\/latex] and [latex]\\underset{x\\to \u2212\\infty }{\\lim}f(x)[\/latex] to determine the end behavior.*<\/li>\n<li><strong>Step 4: <\/strong>Determine whether [latex]f[\/latex] has any vertical asymptotes.<\/li>\n<li><strong>Step 5: <\/strong>Calculate [latex]f^{\\prime}[\/latex]. Find all critical points and determine the intervals where [latex]f[\/latex] is increasing and where [latex]f[\/latex] is decreasing. Determine whether [latex]f[\/latex] has any local extrema.<\/li>\n<li><strong>Step 6: <\/strong>Calculate [latex]f^{\\prime \\prime}[\/latex]. Determine the intervals where [latex]f[\/latex] is concave up and where [latex]f[\/latex] is concave down. Use this information to determine whether [latex]f[\/latex] has any inflection points. The second derivative can also be used as an alternate means to determine or verify that [latex]f[\/latex] has a local extremum at a critical point.<\/li>\n<\/ul>\n<p>*Note for Step 3: If either of these limits is a finite number [latex]L[\/latex], then [latex]y=L[\/latex] is a horizontal asymptote. If either of these limits is [latex]\\infty[\/latex] or [latex]\u2212\\infty[\/latex], determine whether [latex]f[\/latex] has an oblique asymptote. If [latex]f[\/latex] is a rational function such that [latex]f(x)=\\frac{p(x)}{q(x)}[\/latex], where the degree of the numerator is greater than the degree of the denominator, then [latex]f[\/latex] can be written as<\/p>\n<div id=\"fs-id1165042617521\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=\\dfrac{p(x)}{q(x)}=g(x)+\\dfrac{r(x)}{q(x)}[\/latex],<\/div>\n<p>where the degree of [latex]r(x)[\/latex] is less than the degree of [latex]q(x)[\/latex]. The values of [latex]f(x)[\/latex] approach the values of [latex]g(x)[\/latex] as [latex]x\\to \\pm \\infty[\/latex]. If [latex]g(x)[\/latex] is a linear function, it is known as an <em>oblique asymptote<\/em>.<\/p>\n<\/section>\n<p id=\"fs-id1165042617762\">Now let\u2019s use this strategy to graph several different functions. We start by graphing a polynomial function.<\/p>\n<section class=\"textbox example\">\n<p>Sketch a graph of [latex]f(x)=(x-1)^2 (x+2)[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165042707761\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165042707761\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042707761\"><strong>Step 1.<\/strong> Since [latex]f[\/latex] is a polynomial, the domain is the set of all real numbers.<\/p>\n<p id=\"fs-id1165042707768\"><strong>Step 2.<\/strong> When [latex]x=0, \\, f(x)=2[\/latex]. Therefore, the [latex]y[\/latex]-intercept is [latex](0,2)[\/latex]. To find the [latex]x[\/latex]-intercepts, we need to solve the equation [latex](x-1)^2 (x+2)=0[\/latex], which gives us the [latex]x[\/latex]-intercepts [latex](1,0)[\/latex] and [latex](-2,0)[\/latex]<\/p>\n<p id=\"fs-id1165042707901\"><strong>Step 3<\/strong>. We need to evaluate the end behavior of [latex]f[\/latex]. As [latex]x\\to \\infty[\/latex], [latex](x-1)^2 \\to \\infty[\/latex] and [latex](x+2)\\to \\infty[\/latex]. Therefore, [latex]\\underset{x\\to \\infty }{\\lim}f(x)=\\infty[\/latex]. As [latex]x\\to \u2212\\infty[\/latex], [latex](x-1)^2 \\to \\infty[\/latex] and [latex](x+2) \\to \u2212\\infty[\/latex]. Therefore, [latex]\\underset{x\\to -\\infty }{\\lim}f(x)=\u2212\\infty[\/latex]. To get even more information about the end behavior of [latex]f[\/latex], we can multiply the factors of [latex]f[\/latex]. When doing so, we see that,<\/p>\n<div id=\"fs-id1165042525463\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=(x-1)^2 (x+2)=x^3-3x+2[\/latex]<\/div>\n<p id=\"fs-id1165042525532\">Since the leading term of [latex]f[\/latex] is [latex]x^3[\/latex], we conclude that [latex]f[\/latex] behaves like [latex]y=x^3[\/latex] as [latex]x\\to \\pm \\infty[\/latex].<\/p>\n<p id=\"fs-id1165043382921\"><strong>Step 4.<\/strong> Since [latex]f[\/latex] is a polynomial function, it does not have any vertical asymptotes.<\/p>\n<p id=\"fs-id1165043382928\"><strong>Step 5.<\/strong> The first derivative of [latex]f[\/latex] is,<\/p>\n<div class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=3x^2-3[\/latex]<\/div>\n<p id=\"fs-id1165043382970\">Therefore, [latex]f[\/latex] has two critical points: [latex]x=1,-1[\/latex]. Divide the interval [latex](\u2212\\infty ,\\infty)[\/latex] into the three smaller intervals: [latex](\u2212\\infty ,-1)[\/latex], [latex](-1,1)[\/latex], and [latex](1,\\infty )[\/latex].<\/p>\n<p>Then, choose test points [latex]x=-2[\/latex], [latex]x=0[\/latex], and [latex]x=2[\/latex] from these intervals and evaluate the sign of [latex]f^{\\prime}(x)[\/latex] at each of these test points, as shown in the following table.<\/p>\n<table id=\"fs-id1165043383127\" class=\"unnumbered\" summary=\"This table has four rows and four columns. The first row is a header row, and it reads Interval, Test Point, Sign of Derivative f\u2019(x) = 3x2 \u2013 3 = 3(x \u2013 1)(x + 1), and Conclusion. Under the header row, the first column reads (\u2212\u221e, \u22121), (\u22121, 1), and (1, \u221e). The second column reads x = \u22122, x = 0, and x = 2. The third column reads (+)(\u2212)(\u2212) = +, (+)(\u2212)(+) = \u2212, and (+)(+)(+) = +. The fourth column reads f is increasing, f is decreasing, and f is increasing.\">\n<thead>\n<tr valign=\"top\">\n<th>Interval<\/th>\n<th>Test Point<\/th>\n<th>Sign of Derivative [latex]f^{\\prime}(x)=3x^2-3=3(x-1)(x+1)[\/latex]<\/th>\n<th>Conclusion<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex](\u2212\\infty ,-1)[\/latex]<\/td>\n<td>[latex]x=-2[\/latex]<\/td>\n<td>[latex](+)(\u2212)(\u2212)=+[\/latex]<\/td>\n<td>[latex]f[\/latex] is increasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](-1,1)[\/latex]<\/td>\n<td>[latex]x=0[\/latex]<\/td>\n<td>[latex](+)(\u2212)(+)=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is decreasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](1,\\infty )[\/latex]<\/td>\n<td>[latex]x=2[\/latex]<\/td>\n<td>[latex](+)(+)(+)=+[\/latex]<\/td>\n<td>[latex]f[\/latex] is increasing.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165042539204\">From the table, we see that [latex]f[\/latex] has a local maximum at [latex]x=-1[\/latex] and a local minimum at [latex]x=1[\/latex]. Evaluating [latex]f(x)[\/latex] at those two points, we find that the local maximum value is [latex]f(-1)=4[\/latex] and the local minimum value is [latex]f(1)=0[\/latex].<\/p>\n<p id=\"fs-id1165042539284\"><strong>Step 6.<\/strong> The second derivative of [latex]f[\/latex] is,<\/p>\n<div id=\"fs-id1165042539292\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime \\prime}(x)=6x[\/latex]<\/div>\n<p id=\"fs-id1165042539318\">The second derivative is zero at [latex]x=0[\/latex].<\/p>\n<p>Therefore, to determine the concavity of [latex]f[\/latex], divide the interval [latex](\u2212\\infty ,\\infty)[\/latex] into the smaller intervals [latex](\u2212\\infty ,0)[\/latex] and [latex](0,\\infty )[\/latex], and choose test points [latex]x=-1[\/latex] and [latex]x=1[\/latex] to determine the concavity of [latex]f[\/latex] on each of these smaller intervals as shown in the following table.<\/p>\n<table id=\"fs-id1165042381927\" class=\"unnumbered\" summary=\"This table has three rows and four columns. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019\u2019(x) = 6x, and Conclusion. Under the header row, the first column reads (\u2212\u221e, 0) and (0, \u221e). The second column reads x = \u22121 and x = 1. The third column reads \u2212 and +. The fourth column reads f is concave down and f is concave up.\">\n<thead>\n<tr valign=\"top\">\n<th>Interval<\/th>\n<th>Test Point<\/th>\n<th>Sign of [latex]f^{\\prime \\prime}(x)=6x[\/latex]<\/th>\n<th>Conclusion<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex](\u2212\\infty ,0)[\/latex]<\/td>\n<td>[latex]x=-1[\/latex]<\/td>\n<td>[latex]-[\/latex]<\/td>\n<td>[latex]f[\/latex] is concave down.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](0,\\infty )[\/latex]<\/td>\n<td>[latex]x=1[\/latex]<\/td>\n<td>[latex]+[\/latex]<\/td>\n<td>[latex]f[\/latex] is concave up.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165042382153\">We note that the information in the preceding table confirms the fact, found in step 5, that [latex]f[\/latex] has a local maximum at [latex]x=-1[\/latex] and a local minimum at [latex]x=1[\/latex]. In addition, the information found in step 5\u2014namely, [latex]f[\/latex] has a local maximum at [latex]x=-1[\/latex] and a local minimum at [latex]x=1[\/latex], and [latex]f^{\\prime}(x)=0[\/latex] at those points\u2014combined with the fact that [latex]f^{\\prime \\prime}[\/latex] changes sign only at [latex]x=0[\/latex] confirms the results found in step 6 on the concavity of [latex]f[\/latex].<\/p>\n<p id=\"fs-id1165042709981\">Combining this information, we arrive at the graph of [latex]f(x)=(x-1)^2 (x+2)[\/latex] shown in the following graph.<\/p>\n<figure style=\"width: 400px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211132\/CNX_Calc_Figure_04_06_015.jpg\" alt=\"The function f(x) = (x \u22121)2 (x + 2) is graphed. It crosses the x axis at x = \u22122 and touches the x axis at x = 1.\" width=\"400\" height=\"347\" \/><figcaption class=\"wp-caption-text\">Figure 23. Graph of f(x)<\/figcaption><\/figure>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/QhOY9VfIefo?controls=0&amp;start=003&amp;end=317&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.6LimitsAtInfinityAndAsymptotesPart2_3to317_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.6 Limits at Infinity and Asymptotes (part 2 &#8211; curve sketching)&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Sketch the graph of [latex]f(x)=\\dfrac{x^2}{1-x^2}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165042407414\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165042407414\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042407414\"><strong>Step 1.<\/strong> The function [latex]f[\/latex] is defined as long as the denominator is not zero. Therefore, the domain is the set of all real numbers [latex]x[\/latex] except [latex]x=\\pm 1[\/latex].<\/p>\n<p id=\"fs-id1165042407440\"><strong>Step 2.<\/strong> Find the intercepts. If [latex]x=0[\/latex], then [latex]f(x)=0[\/latex], so 0 is an intercept. If [latex]y=0[\/latex], then [latex]\\frac{x^2}{1-x^2}=0[\/latex], which implies [latex]x=0[\/latex]. Therefore, [latex](0,0)[\/latex] is the only intercept.<\/p>\n<p id=\"fs-id1165042407553\"><strong>Step 3.<\/strong> Evaluate the limits at infinity. Since [latex]f[\/latex] is a rational function, divide the numerator and denominator by the highest power in the denominator: [latex]x^2[\/latex]. We obtain,<\/p>\n<div id=\"fs-id1165042407571\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \\pm \\infty }{\\lim}\\frac{x^2}{1-x^2}=\\underset{x\\to \\pm \\infty }{\\lim}\\frac{1}{\\frac{1}{x^2}-1}=-1[\/latex].<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1165042407655\">Therefore, [latex]f[\/latex] has a horizontal asymptote of [latex]y=-1[\/latex] as [latex]x\\to \\infty[\/latex] and [latex]x\\to \u2212\\infty[\/latex].<\/p>\n<p id=\"fs-id1165042491004\"><strong>Step 4.<\/strong> To determine whether [latex]f[\/latex] has any vertical asymptotes, first check to see whether the denominator has any zeroes. We find the denominator is zero when [latex]x=\\pm 1[\/latex].<\/p>\n<p>To determine whether the lines [latex]x=1[\/latex] or [latex]x=-1[\/latex] are vertical asymptotes of [latex]f[\/latex], evaluate [latex]\\underset{x\\to 1}{\\lim}f(x)[\/latex] and [latex]\\underset{x\\to \u22121}{\\lim}f(x)[\/latex].<\/p>\n<p>By looking at each one-sided limit as [latex]x\\to 1[\/latex], we see that,<\/p>\n<div id=\"fs-id1165042491126\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 1^+}{\\lim}\\frac{x^2}{1-x^2}=\u2212\\infty[\/latex] and [latex]\\underset{x\\to 1^-}{\\lim}\\frac{x^2}{1-x^2}=\\infty[\/latex].<\/div>\n<p id=\"fs-id1165042491222\">In addition, by looking at each one-sided limit as [latex]x\\to \u22121[\/latex], we find that,<\/p>\n<div id=\"fs-id1165042491242\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to \u22121^+}{\\lim}\\frac{x^2}{1-x^2}=\\infty[\/latex] and [latex]\\underset{x\\to \u22121^-}{\\lim}\\frac{x^2}{1-x^2}=\u2212\\infty[\/latex].<\/div>\n<p id=\"fs-id1165043262266\"><strong>Step 5<\/strong>. Calculate the first derivative:<\/p>\n<div id=\"fs-id1165043262270\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{(1-x^2)(2x)-x^2(-2x)}{(1-x^2)^2}=\\frac{2x}{(1-x^2)^2}[\/latex].<\/div>\n<p id=\"fs-id1165043262390\">Critical points occur at points [latex]x[\/latex] where [latex]f^{\\prime}(x)=0[\/latex] or [latex]f^{\\prime}(x)[\/latex] is undefined. We see that [latex]f^{\\prime}(x)=0[\/latex] when [latex]x=0[\/latex]. The derivative [latex]f^{\\prime}[\/latex] is not undefined at any point in the domain of [latex]f[\/latex]. However, [latex]x=\\pm 1[\/latex] are not in the domain of [latex]f[\/latex].<\/p>\n<p>Therefore, to determine where [latex]f[\/latex] is increasing and where [latex]f[\/latex] is decreasing, divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into four smaller intervals: [latex](\u2212\\infty ,-1)[\/latex], [latex](-1,0)[\/latex], [latex](0,1)[\/latex], and [latex](1,\\infty )[\/latex], and choose a test point in each interval to determine the sign of [latex]f^{\\prime}(x)[\/latex] in each of these intervals. The values [latex]x=-2[\/latex], [latex]x=-\\frac{1}{2}[\/latex], [latex]x=\\frac{1}{2}[\/latex], and [latex]x=2[\/latex] are good choices for test points as shown in the following table.<\/p>\n<table id=\"fs-id1165043341618\" class=\"unnumbered\" summary=\"This table has four columns and five rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019(x) = 2x\/(1 \u2212 x2)2, and Conclusion. Under the header row, the first column reads (\u2212\u221e, \u22121), (\u22121, 0), (0, 1), and (1, \u221e). The second column reads x = \u22122, x = \u22121\/2, x = 1\/2, and x = 2. The third column reads \u2212\/+ = \u2212, \u2212\/+ = \u2212, +\/+ = +, and +\/+ = +. The fourth column reads f is decreasing, f is decreasing, f is increasing, and f is increasing.\">\n<thead>\n<tr valign=\"top\">\n<th>Interval<\/th>\n<th>Test Point<\/th>\n<th>Sign of [latex]f^{\\prime}(x)=\\frac{2x}{(1-x^2)^2}[\/latex]<\/th>\n<th>Conclusion<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex](\u2212\\infty ,-1)[\/latex]<\/td>\n<td>[latex]x=-2[\/latex]<\/td>\n<td>[latex]\u2212\/+=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is decreasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](-1,0)[\/latex]<\/td>\n<td>[latex]x=-1\/2[\/latex]<\/td>\n<td>[latex]\u2212\/+=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is decreasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](0,1)[\/latex]<\/td>\n<td>[latex]x=1\/2[\/latex]<\/td>\n<td>[latex]+\/+=+[\/latex]<\/td>\n<td>[latex]f[\/latex] is increasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](1,\\infty )[\/latex]<\/td>\n<td>[latex]x=2[\/latex]<\/td>\n<td>[latex]+\/+=+[\/latex]<\/td>\n<td>[latex]f[\/latex] is increasing.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165043183348\">From this analysis, we conclude that [latex]f[\/latex] has a local minimum at [latex]x=0[\/latex] but no local maximum.<\/p>\n<p id=\"fs-id1165043183355\"><strong>Step 6.<\/strong> Calculate the second derivative:<\/p>\n<div id=\"fs-id1165043183358\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} f^{\\prime \\prime}(x) & =\\frac{(1-x^2)^2(2)-2x(2(1-x^2)(-2x))}{(1-x^2)^4} \\\\ & =\\frac{(1-x^2)[2(1-x^2)+8x^2]}{(1-x^2)^4} \\\\ & =\\frac{2(1-x^2)+8x^2}{(1-x^2)^3} \\\\ & =\\frac{6x^2+2}{(1-x^2)^3} \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165043384128\">To determine the intervals where [latex]f[\/latex] is concave up and where [latex]f[\/latex] is concave down, we first need to find all points [latex]x[\/latex] where [latex]f^{\\prime \\prime}(x)=0[\/latex] or [latex]f^{\\prime \\prime}(x)[\/latex] is undefined. Since the numerator [latex]6x^2+2 \\ne 0[\/latex] for any [latex]x[\/latex], [latex]f^{\\prime \\prime}(x)[\/latex] is never zero.<\/p>\n<p>Furthermore, [latex]f^{\\prime \\prime}[\/latex] is not undefined for any [latex]x[\/latex] in the domain of [latex]f[\/latex]. However, as discussed earlier, [latex]x=\\pm 1[\/latex] are not in the domain of [latex]f[\/latex].<\/p>\n<p>Therefore, to determine the concavity of [latex]f[\/latex], we divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into the three smaller intervals [latex](\u2212\\infty ,-1)[\/latex], [latex](-1,-1)[\/latex], and [latex](1,\\infty )[\/latex], and choose a test point in each of these intervals to evaluate the sign of [latex]f^{\\prime \\prime}(x)[\/latex] in each of these intervals. The values [latex]x=-2[\/latex], [latex]x=0[\/latex], and [latex]x=2[\/latex] are possible test points as shown in the following table.<\/p>\n<table id=\"fs-id1165043384406\" class=\"unnumbered\" summary=\"This table has four columns and four rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019\u2019(x) = (6x2 + 2)\/(1 \u2212 x2)3, and Conclusion. Under the header row, the first column reads (\u2212\u221e, \u22121), (\u22121, 1), and (1, \u221e). The second column reads x = \u22122, x = 0, and x = 2. The third column reads +\/\u2212 = \u2212, +\/+ = +, and +\/\u2212 = \u2212. The fourth column reads f is concave down, f is concave up, and f is concave down.\">\n<thead>\n<tr valign=\"top\">\n<th>Interval<\/th>\n<th>Test Point<\/th>\n<th>Sign of [latex]f^{\\prime \\prime}(x)=\\frac{6x^2+2}{(1-x^2)^3}[\/latex]<\/th>\n<th>Conclusion<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex](\u2212\\infty ,-1)[\/latex]<\/td>\n<td>[latex]x=-2[\/latex]<\/td>\n<td>[latex]+\/-=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is concave down.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](-1,-1)[\/latex]<\/td>\n<td>[latex]x=0[\/latex]<\/td>\n<td>[latex]+\/+=+[\/latex]<\/td>\n<td>[latex]f[\/latex] is concave up.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](1,\\infty )[\/latex]<\/td>\n<td>[latex]x=2[\/latex]<\/td>\n<td>[latex]+\/-=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is concave down.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165042592688\">Combining all this information, we arrive at the graph of [latex]f[\/latex] shown below. Note that, although [latex]f[\/latex] changes concavity at [latex]x=-1[\/latex] and [latex]x=1[\/latex], there are no inflection points at either of these places because [latex]f[\/latex] is not continuous at [latex]x=-1[\/latex] or [latex]x=1[\/latex].<\/p>\n<figure style=\"width: 342px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211139\/CNX_Calc_Figure_04_06_016.jpg\" alt=\"The function f(x) = x2\/(1 \u2212 x2) is graphed. It has asymptotes y = \u22121, x = \u22121, and x = 1.\" width=\"342\" height=\"347\" \/><figcaption class=\"wp-caption-text\">Figure 25. Graph of y<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Sketch the graph of [latex]f(x)=\\dfrac{x^2}{x-1}[\/latex]<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165042593006\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165042593006\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042593006\"><strong>Step 1.<\/strong> The domain of [latex]f[\/latex] is the set of all real numbers [latex]x[\/latex] except [latex]x=1[\/latex].<\/p>\n<p id=\"fs-id1165042712534\"><strong>Step 2.<\/strong> Find the intercepts. We can see that when [latex]x=0[\/latex], [latex]f(x)=0[\/latex], so [latex](0,0)[\/latex] is the only intercept.<\/p>\n<p id=\"fs-id1165042712584\"><strong>Step 3<\/strong>. Evaluate the limits at infinity. Since the degree of the numerator is one more than the degree of the denominator, [latex]f[\/latex] must have an oblique asymptote. To find the oblique asymptote, use long division of polynomials to write,<\/p>\n<div id=\"fs-id1165042712594\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(x)=\\frac{x^2}{x-1}=x+1+\\frac{1}{x-1}[\/latex].<\/div>\n<p id=\"fs-id1165042712649\">Since [latex]1\/(x-1)\\to 0[\/latex] as [latex]x\\to \\pm \\infty[\/latex], [latex]f(x)[\/latex] approaches the line [latex]y=x+1[\/latex] as [latex]x\\to \\pm \\infty[\/latex]. The line [latex]y=x+1[\/latex] is an oblique asymptote for [latex]f[\/latex].<\/p>\n<p id=\"fs-id1165042712758\"><strong>Step 4.<\/strong> To check for vertical asymptotes, look at where the denominator is zero. Here the denominator is zero at [latex]x=1[\/latex]. Looking at both one-sided limits as [latex]x\\to 1[\/latex], we find,<\/p>\n<div id=\"fs-id1165042712789\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 1^+}{\\lim}\\frac{x^2}{x-1}=\\infty[\/latex] and [latex]\\underset{x\\to 1^-}{\\lim}\\frac{x^2}{x-1}=\u2212\\infty[\/latex].<\/div>\n<p id=\"fs-id1165042403315\">Therefore, [latex]x=1[\/latex] is a vertical asymptote, and we have determined the behavior of [latex]f[\/latex] as [latex]x[\/latex] approaches 1 from the right and the left.<\/p>\n<p id=\"fs-id1165042403340\"><strong>Step 5.<\/strong> Calculate the first derivative:<\/p>\n<div id=\"fs-id1165042403344\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{(x-1)(2x)-x^2(1)}{(x-1)^2}=\\frac{x^2-2x}{(x-1)^2}[\/latex].<\/div>\n<p id=\"fs-id1165042403459\">We have [latex]f^{\\prime}(x)=0[\/latex] when [latex]x^2-2x=x(x-2)=0[\/latex]. Therefore, [latex]x=0[\/latex] and [latex]x=2[\/latex] are critical points. Since [latex]f[\/latex] is undefined at [latex]x=1[\/latex], we need to divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into the smaller intervals [latex](\u2212\\infty ,0)[\/latex], [latex](0,1)[\/latex], [latex](1,2)[\/latex], and [latex](2,\\infty )[\/latex], and choose a test point from each interval to evaluate the sign of [latex]f^{\\prime}(x)[\/latex] in each of these smaller intervals.<\/p>\n<p>For example, let [latex]x=-1[\/latex], [latex]x=\\frac{1}{2}[\/latex], [latex]x=\\frac{3}{2}[\/latex], and [latex]x=3[\/latex] be the test points as shown in the following table.<\/p>\n<table id=\"fs-id1165042710566\" class=\"unnumbered\" summary=\"This table has four columns and five rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019(x) = (x2 \u2212 2x)\/(x \u2212 1)2 = x(x \u2212 2)\/(x \u2212 1)2, and Conclusion. Under the header row, the first column reads (\u2212\u221e, 0), (0, 1), (1, 2), and (2, \u221e). The second column reads x = \u22121, x = 1\/2, x = 3\/2, and x = 3. The third column reads (\u2212)(\u2212)\/+ = +, (+)(\u2212)\/+ = \u2212, (+)(\u2212)\/+ = \u2212, and (+)(+)\/+ = +. The fourth column reads f is increasing, f is decreasing, f is decreasing, and f is increasing.\">\n<thead>\n<tr valign=\"top\">\n<th>Interval<\/th>\n<th>Test Point<\/th>\n<th>Sign of [latex]f^{\\prime}(x)=\\frac{x^2-2x}{(x-1)^2}=\\frac{x(x-2)}{(x-1)^2}[\/latex]<\/th>\n<th>Conclusion<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex](\u2212\\infty ,0)[\/latex]<\/td>\n<td>[latex]x=-1[\/latex]<\/td>\n<td>[latex](\u2212)(\u2212)\/+=+[\/latex]<\/td>\n<td>[latex]f[\/latex] is increasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](0,1)[\/latex]<\/td>\n<td>[latex]x=1\/2[\/latex]<\/td>\n<td>[latex](+)(\u2212)\/+=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is decreasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](1,2)[\/latex]<\/td>\n<td>[latex]x=3\/2[\/latex]<\/td>\n<td>[latex](+)(\u2212)\/+=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is decreasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](2,\\infty )[\/latex]<\/td>\n<td>[latex]x=3[\/latex]<\/td>\n<td>[latex](+)(+)\/+=+[\/latex]<\/td>\n<td>[latex]f[\/latex] is increasing.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165042476453\">From this table, we see that [latex]f[\/latex] has a local maximum at [latex]x=0[\/latex] and a local minimum at [latex]x=2[\/latex]. The value of [latex]f[\/latex] at the local maximum is [latex]f(0)=0[\/latex] and the value of [latex]f[\/latex] at the local minimum is [latex]f(2)=4[\/latex]. Therefore, [latex](0,0)[\/latex] and [latex](2,4)[\/latex] are important points on the graph.<\/p>\n<p id=\"fs-id1165042464546\"><strong>Step 6.<\/strong> Calculate the second derivative:<\/p>\n<div id=\"fs-id1165042464549\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{ll} f^{\\prime \\prime}(x) & =\\frac{(x-1)^2(2x-2)-(x^2-2x)(2(x-1))}{(x-1)^4} \\\\ & =\\frac{(x-1)[(x-1)(2x-2)-2(x^2-2x)]}{(x-1)^4} \\\\ & =\\frac{(x-1)(2x-2)-2(x^2-2x)}{(x-1)^3} \\\\ & =\\frac{2x^2-4x+2-(2x^2-4x)}{(x-1)^3} \\\\ & =\\frac{2}{(x-1)^3} \\end{array}[\/latex]<\/div>\n<p id=\"fs-id1165042652400\">We see that [latex]f^{\\prime \\prime}(x)[\/latex] is never zero or undefined for [latex]x[\/latex] in the domain of [latex]f[\/latex]. Since [latex]f[\/latex] is undefined at [latex]x=1[\/latex], to check concavity we just divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into the two smaller intervals [latex](\u2212\\infty ,1)[\/latex] and [latex](1,\\infty )[\/latex], and choose a test point from each interval to evaluate the sign of [latex]f^{\\prime \\prime}(x)[\/latex] in each of these intervals. The values [latex]x=0[\/latex] and [latex]x=2[\/latex] are possible test points as shown in the following table.<\/p>\n<table id=\"fs-id1165042652542\" class=\"unnumbered\" summary=\"This table has four columns and three rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019\u2019(x) = 2\/(x \u2212 1)3, and Conclusion. Under the header row, the first column reads (\u2212\u221e, 1) and (1, \u221e). The column row reads x = 0 and x = 2. The third column reads +\/\u2212 = \u2212 and +\/+ = +. The fourth column reads f is concave down and f is concave up.\">\n<thead>\n<tr valign=\"top\">\n<th>Interval<\/th>\n<th>Test Point<\/th>\n<th>Sign of [latex]f^{\\prime \\prime}(x)=\\frac{2}{(x-1)^3}[\/latex]<\/th>\n<th>Conclusion<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex](\u2212\\infty ,1)[\/latex]<\/td>\n<td>[latex]x=0[\/latex]<\/td>\n<td>[latex]+\/-=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is concave down.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](1,\\infty )[\/latex]<\/td>\n<td>[latex]x=2[\/latex]<\/td>\n<td>[latex]+\/+=+[\/latex]<\/td>\n<td>[latex]f[\/latex] is concave up.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165042606958\">From the information gathered, we arrive at the following graph for [latex]f.[\/latex]<\/p>\n<figure style=\"width: 342px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211145\/CNX_Calc_Figure_04_06_017.jpg\" alt=\"The function f(x) = x2\/(x \u2212 1) is graphed. It has asymptotes y = x + 1 and x = 1.\" width=\"342\" height=\"346\" \/><figcaption class=\"wp-caption-text\">Figure 27. Graph of y<\/figcaption><\/figure>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/QhOY9VfIefo?controls=0&amp;start=552&amp;end=777&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.6LimitsAtInfinityAndAsymptotesPart2_552to777_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.6 Limits at Infinity and Asymptotes (part 2 &#8211; curve sketching)&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Sketch a graph of [latex]f(x)=(x-1)^{\\frac{2}{3}}[\/latex]<\/p>\n<div id=\"fs-id1165042607093\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165042607145\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165042607145\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042607145\"><strong>Step 1.<\/strong> Since the cube-root function is defined for all real numbers [latex]x[\/latex] and [latex](x-1)^{2\/3}=(\\sqrt[3]{x-1})^2[\/latex], the domain of [latex]f[\/latex] is all real numbers.<\/p>\n<p id=\"fs-id1165042607208\"><strong>Step 2:<\/strong> To find the [latex]y[\/latex]-intercept, evaluate [latex]f(0)[\/latex]. Since [latex]f(0)=1[\/latex], the [latex]y[\/latex]-intercept is [latex](0,1)[\/latex]. To find the [latex]x[\/latex]-intercept, solve [latex](x-1)^{2\/3}=0[\/latex]. The solution of this equation is [latex]x=1[\/latex], so the [latex]x[\/latex]-intercept is [latex](1,0)[\/latex].<\/p>\n<p id=\"fs-id1165042583208\"><strong>Step 3:<\/strong> Since [latex]\\underset{x\\to \\pm \\infty }{\\lim}(x-1)^{2\/3}=\\infty[\/latex], the function continues to grow without bound as [latex]x\\to \\infty[\/latex] and [latex]x\\to \u2212\\infty[\/latex].<\/p>\n<p id=\"fs-id1165042583285\"><strong>Step 4:<\/strong> The function has no vertical asymptotes.<\/p>\n<p id=\"fs-id1165042583288\"><strong>Step 5:<\/strong> To determine where [latex]f[\/latex] is increasing or decreasing, calculate [latex]f^{\\prime}[\/latex]. We find,<\/p>\n<div id=\"fs-id1165042583307\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(x)=\\frac{2}{3}(x-1)^{-1\/3}=\\frac{2}{3(x-1)^{1\/3}}[\/latex].<\/div>\n<p id=\"fs-id1165042583388\">This function is not zero anywhere, but it is undefined when [latex]x=1[\/latex]. Therefore, the only critical point is [latex]x=1[\/latex]. Divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into the smaller intervals [latex](\u2212\\infty ,1)[\/latex] and [latex](1,\\infty )[\/latex], and choose test points in each of these intervals to determine the sign of [latex]f^{\\prime}(x)[\/latex] in each of these smaller intervals.<\/p>\n<p>Let [latex]x=0[\/latex] and [latex]x=2[\/latex] be the test points as shown in the following table.<\/p>\n<table id=\"fs-id1165042504467\" class=\"unnumbered\" summary=\"This table has four columns and three rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019(x) = 2\/(3(x \u2212 1)1\/3), and Conclusion. Under the header row, the first column reads (\u2212\u221e, 1) and (1, \u221e). The second column reads x = 0 and x = 2. The third column reads +\/\u2212 = \u2212 and +\/+ = +. The fourth column reads f is decreasing and f is increasing.\">\n<thead>\n<tr valign=\"top\">\n<th>Interval<\/th>\n<th>Test Point<\/th>\n<th>Sign of [latex]f^{\\prime}(x)=\\frac{2}{3(x-1)^{1\/3}}[\/latex]<\/th>\n<th>Conclusion<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex](\u2212\\infty ,1)[\/latex]<\/td>\n<td>[latex]x=0[\/latex]<\/td>\n<td>[latex]+\/-=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is decreasing.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](1,\\infty )[\/latex]<\/td>\n<td>[latex]x=2[\/latex]<\/td>\n<td>[latex]+\/+=+[\/latex]<\/td>\n<td>[latex]f[\/latex] is increasing.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165042504734\">We conclude that [latex]f[\/latex] has a local minimum at [latex]x=1[\/latex]. Evaluating [latex]f[\/latex] at [latex]x=1[\/latex], we find that the value of [latex]f[\/latex] at the local minimum is zero. Note that [latex]f^{\\prime}(1)[\/latex] is undefined, so to determine the behavior of the function at this critical point, we need to examine [latex]\\underset{x\\to 1}{\\lim}f^{\\prime}(x)[\/latex].<\/p>\n<p>Looking at the one-sided limits, we have,<\/p>\n<div id=\"fs-id1165042510168\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\underset{x\\to 1^+}{\\lim}\\frac{2}{3(x-1)^{1\/3}}=\\infty[\/latex] and [latex]\\underset{x\\to 1^-}{\\lim}\\frac{2}{3(x-1)^{1\/3}}=\u2212\\infty[\/latex].<\/div>\n<p id=\"fs-id1165042510285\">Therefore, [latex]f[\/latex] has a cusp at [latex]x=1[\/latex].<\/p>\n<p id=\"fs-id1165042510304\"><strong>Step 6:<\/strong> To determine concavity, we calculate the second derivative of [latex]f[\/latex]:<\/p>\n<div id=\"fs-id1165042510313\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime \\prime}(x)=-\\frac{2}{9}(x-1)^{-4\/3}=\\frac{-2}{9(x-1)^{4\/3}}[\/latex].<\/div>\n<p id=\"fs-id1165042510398\">We find that [latex]f^{\\prime \\prime}(x)[\/latex] is defined for all [latex]x[\/latex], but is undefined when [latex]x=1[\/latex]. Therefore, divide the interval [latex](\u2212\\infty ,\\infty )[\/latex] into the smaller intervals [latex](\u2212\\infty ,1)[\/latex] and [latex](1,\\infty )[\/latex], and choose test points to evaluate the sign of [latex]f^{\\prime \\prime}(x)[\/latex] in each of these intervals.<\/p>\n<p>As we did earlier, let [latex]x=0[\/latex] and [latex]x=2[\/latex] be test points as shown in the following table.<\/p>\n<table id=\"fs-id1165042510530\" class=\"unnumbered\" summary=\"This table has four columns and three rows. The first row is a header row, and it reads Interval, Test Point, Sign of f\u2019\u2019(x) = \u22122\/(9(x \u2212 1)4\/3), and Conclusion. Under the header row, the first column reads (\u2212\u221e, 1) and (1, \u221e). The second column reads x = 0 and x = 2. The third column reads \u2212\/+ = \u2212 and \u2212\/+ = \u2212. The fourth column reads f is concave down and f is concave down.\">\n<thead>\n<tr valign=\"top\">\n<th>Interval<\/th>\n<th>Test Point<\/th>\n<th>Sign of [latex]f^{\\prime \\prime}(x)=\\frac{-2}{9(x-1)^{4\/3}}[\/latex]<\/th>\n<th>Conclusion<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>[latex](\u2212\\infty ,1)[\/latex]<\/td>\n<td>[latex]x=0[\/latex]<\/td>\n<td>[latex]\u2212\/+=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is concave down.<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>[latex](1,\\infty )[\/latex]<\/td>\n<td>[latex]x=2[\/latex]<\/td>\n<td>[latex]\u2212\/+=\u2212[\/latex]<\/td>\n<td>[latex]f[\/latex] is concave down.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1165042643969\">From this table, we conclude that [latex]f[\/latex] is concave down everywhere. Combining all of this information, we arrive at the following graph for [latex]f[\/latex].<\/p>\n<figure style=\"width: 604px\" class=\"wp-caption alignnone\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11211148\/CNX_Calc_Figure_04_06_018.jpg\" alt=\"The function f(x) = (x \u2212 1)2\/3 is graphed. It touches the x axis at x = 1, where it comes to something of a sharp point and then flairs out on either side.\" width=\"604\" height=\"272\" \/><figcaption class=\"wp-caption-text\">Figure 28. Graph of f(x)<\/figcaption><\/figure>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/QhOY9VfIefo?controls=0&amp;start=781&amp;end=962&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.6LimitsAtInfinityAndAsymptotesPart2_781to962_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.6 Limits at Infinity and Asymptotes (part 2 &#8211; curve sketching)&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":6,"menu_order":10,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"4.6 Limits at Infinity and Asymptotes (part 2 - curve sketching)\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":292,"module-header":"learn_it","content_attributions":[{"type":"cc","description":"Calculus Volume 1","author":"Gilbert Strang, Edwin (Jed) Herman","organization":"OpenStax","url":"https:\/\/openstax.org\/details\/books\/calculus-volume-1","project":"","license":"cc-by-nc-sa","license_terms":"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction"},{"type":"original","description":"4.6 Limits at Infinity and Asymptotes (part 2 - curve sketching)","author":"Ryan Melton","organization":"","url":"","project":"","license":"cc-by","license_terms":""}],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/314"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/314\/revisions"}],"predecessor-version":[{"id":4782,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/314\/revisions\/4782"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/292"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/314\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=314"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=314"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=314"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=314"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}