{"id":3772,"date":"2021-05-12T20:24:07","date_gmt":"2021-05-12T20:24:07","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/review-for-section-2-2-the-limit-of-a-function\/"},"modified":"2021-05-12T20:24:07","modified_gmt":"2021-05-12T20:24:07","slug":"review-for-section-2-2-the-limit-of-a-function","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/review-for-section-2-2-the-limit-of-a-function\/","title":{"raw":"Skills Review for The Limit of a Function","rendered":"Skills Review for The Limit of a Function"},"content":{"raw":"\n<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n \t<li>Complete a table values with solutions to an equation<\/li>\n \t<li>Identify points on a graph<\/li>\n \t<li>Find function values (outputs) for specified numbers (inputs)<\/li>\n \t<li>Identify vertical and horizontal asymptotes<\/li>\n \t<li>Find vertical asymptotes of rational functions<\/li>\n<\/ul>\n<\/div>\nThe Limit of a Function section provides an introduction to how to find limits of functions. Ways to find limits will be discussed including a tabular approach and graphical approach. To prepare for this, we will review how to complete function value tables. We will also practice looking at the graph of a function and identifying the&nbsp;y-value or function value associated with a given&nbsp;<em>x<\/em>-value. Finally, we will review how to find vertical asymptotes of rational functions in order to get a head start with understanding when limit values are infinite.\n<h2>Complete a Table of Function Values<\/h2>\n<div>\n\nA table of values can be used to organize the&nbsp;<em>y<\/em>-values or function values that result from plugging specific&nbsp;<em>x<\/em>-values into a function's equation.\n\nSuppose we want to determine the various values of the equation [latex]f(x)=2x - 1[\/latex] at certain values of <em>x<\/em>. We can begin by substituting a value for <em>x<\/em> into the equation and determining the resulting value of the function. The table below&nbsp;lists some values of <em>x<\/em> from \u20133 to 3 and the resulting function values.\n<table style=\"width: 411px; height: 84px;\" summary=\"This is a table with 8 rows and 3 columns. The first row has columns labeled: x, y = 2x-1, (x, y). The entries in the second row are: negative 3; y = 2 times negative 3 minus 1 = negative 7; (-3, -7). The entries in the third row are: negative 2; y = 2 times negative 2 minus 1 = negative 5; (-2, -5). The entries in the fourth row are: negative1; y = 2 times negative 1 minus 1 = negative 3; (-1, -3). The entries in the fifth row are: 0; y = 2 times 0 minus 1 = negative 1; (0, -1). The entries in the sixth row are: 1; y = 2 times 1 minus 1 = 1; (1, 1). The entries in the seventh row are: 2; y = 2 times 2 minus 1 = 3; (2, 3). The entries in the eight row are: 3, y = 2 times 3 minus 1 = 5; (3,5)\">\n<tbody>\n<tr style=\"height: 12px;\">\n<td style=\"width: 116.5px; height: 12px;\">[latex]x[\/latex]<\/td>\n<td style=\"width: 269.5px; height: 12px;\">[latex]f(x)=2x - 1[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 116.5px; height: 12px;\">[latex]-3[\/latex]<\/td>\n<td style=\"width: 269.5px; height: 12px;\">[latex]f(-3)=2\\left(-3\\right)-1=-7[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 116.5px; height: 12px;\">[latex]-2[\/latex]<\/td>\n<td style=\"width: 269.5px; height: 12px;\">[latex]f(-2)=2\\left(-2\\right)-1=-5[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 116.5px; height: 12px;\">[latex]-1[\/latex]<\/td>\n<td style=\"width: 269.5px; height: 12px;\">[latex]f(-1)=2\\left(-1\\right)-1=-3[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 116.5px; height: 12px;\">[latex]0[\/latex]<\/td>\n<td style=\"width: 269.5px; height: 12px;\">[latex]f(0)=2\\left(0\\right)-1=-1[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 116.5px; height: 12px;\">[latex]2[\/latex]<\/td>\n<td style=\"width: 269.5px; height: 12px;\">[latex]f(2)=2\\left(2\\right)-1=3[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 116.5px; height: 12px;\">[latex]3[\/latex]<\/td>\n<td style=\"width: 269.5px; height: 12px;\">[latex]f(3)=2\\left(3\\right)-1=5[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\nWe can look for trends among our function values by looking at the table. For example, in this case, as&nbsp;<em>x<\/em>-values increase, so do the&nbsp;<em>y<\/em>-values. Also, it seems reasonable to assume, based on the table, an&nbsp;<em>x<\/em>-value of 1 would result in a function value of 1. You can verify this for yourself by plugging 1 into [latex]f(x)=2x - 1[\/latex].\n<div class=\"textbox exercises\">\n<h3>Example: Completing a table of function values<\/h3>\nCreate a table of function values for [latex]f(x)=-x+2[\/latex]. Use various integers&nbsp;from -5 to 5 as the&nbsp;<em>x<\/em>-values you plug into the function.\n[reveal-answer q=\"792137\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"792137\"]\n\n&nbsp;\n<table style=\"width: 385px;\" summary=\"The table shows 8 rows and 3 columns. The entries in the first row are: x; y = negative x plus 2; and (x, y). The entries in the second row are: negative 5; y = the opposite of negative 5 plus 2 = 7; (-5, 7). The entries in the third row are: negative 3; y = the opposite of negative 3 plus 2 = 5; (-3, 5). The entries in the fourth row are: -1; y = the opposite of negative 1 plus 2 = 3; (-1, 3). The entries in the fifth row are: 0; y = opposite of zero plus 2 = 2; (0, 2). The entries in the sixth row are: 1; y = the opposite of 1 plus 2 = 1; (1, 1). The entries in the seventh row are: 3; y = the opposite of 3 plus 2 = negative 1; (3, -1). The entries in the eighth row are: 5; y = the opposite of 5 plus 2 = negative 3; (5, -3).\">\n<tbody>\n<tr>\n<td style=\"width: 117.656px;\">[latex]x[\/latex]<\/td>\n<td style=\"width: 244.656px;\">[latex]f(x)=-x+2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 117.656px;\">[latex]-5[\/latex]<\/td>\n<td style=\"width: 244.656px;\">[latex]f(-5)=-\\left(-5\\right)+2=7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 117.656px;\">[latex]-3[\/latex]<\/td>\n<td style=\"width: 244.656px;\">[latex]f(-3)=-\\left(-3\\right)+2=5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 117.656px;\">[latex]-1[\/latex]<\/td>\n<td style=\"width: 244.656px;\">[latex]f(-1)=-\\left(-1\\right)+2=3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 117.656px;\">[latex]0[\/latex]<\/td>\n<td style=\"width: 244.656px;\">[latex]f(0)=-\\left(0\\right)+2=2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 117.656px;\">[latex]1[\/latex]<\/td>\n<td style=\"width: 244.656px;\">[latex]f(1)=-\\left(1\\right)+2=1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 117.656px;\">[latex]3[\/latex]<\/td>\n<td style=\"width: 244.656px;\">[latex]f(3)=-\\left(3\\right)+2=-1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 117.656px;\">[latex]5[\/latex]<\/td>\n<td style=\"width: 244.656px;\">[latex]f(5)=-\\left(5\\right)+2=-3[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n[\/hidden-answer]\n\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n[ohm_question]219319[\/ohm_question]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n[ohm_question]219320[\/ohm_question]\n\n<\/div>\n<h2>Find Function Values from a Graph<\/h2>\nWe can view a function as a set of inputs and their corresponding outputs. That is, we can see a function as a set of ordered pairs, [latex]\\left(x, y \\right).[\/latex]\n\nRemember that, in function notation, [latex]y = f(x)[\/latex], so the ordered pairs containing inputs and outputs can be written in the form of (<em>input<\/em>,&nbsp;<em>output<\/em>) or [latex]\\left(x, f(x)\\right)[\/latex].\n\nEvaluating a function using a graph requires finding the corresponding output value for a given input value, only in this case, we find the output value (<em>y<\/em>-value) by looking at the graph.\n<div class=\"textbox exercises\">\n<h3>Example: Reading Function Values from a Graph<\/h3>\nGiven the graph below, evaluate [latex]f\\left(2\\right)[\/latex].\n\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18191001\/CNX_Precalc_Figure_01_01_0072.jpg\" alt=\"Graph of a positive parabola centered at (1, 0).\" width=\"487\" height=\"445\">\n[reveal-answer q=\"915833\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"915833\"]\n<ol>\n \t<li>To evaluate [latex]f\\left(2\\right)[\/latex], locate the point on the curve where [latex]x=2[\/latex], then read the [latex]y[\/latex]-coordinate of that point. The point has coordinates [latex]\\left(2,1\\right)[\/latex], so [latex]f\\left(2\\right)=1[\/latex].<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18191004\/CNX_Precalc_Figure_01_01_0082.jpg\" alt=\"Graph of a positive parabola centered at (1, 0) with the labeled point (2, 1) where f(2) =1.\" width=\"487\" height=\"445\"><\/li>\n<\/ol>\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n[ohm_question]2471[\/ohm_question]\n\n<\/div>\n<h2>Find Vertical Asymptotes of a Rational Function\u200b<\/h2>\nThe vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Essentially, you should factor and simplify the rational function first and then set the denominator equal to 0 to find vertical asymptotes.\n<div class=\"textbox\">\n<h3>How To: Given a rational function, identify any vertical asymptotes of its graph.<\/h3>\n<ol>\n \t<li>Factor the numerator and denominator.<\/li>\n \t<li>Reduce the expression by canceling common factors in the numerator and the denominator.<\/li>\n \t<li>Note any values that cause the denominator to be zero in this simplified version. These are where the vertical asymptotes occur.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Identifying Vertical Asymptotes<\/h3>\nFind the vertical asymptotes of the graph of [latex]k\\left(x\\right)=\\dfrac{5+2{x}^{2}}{2-x-{x}^{2}}[\/latex].\n\n[reveal-answer q=\"787718\"]Show Solution[\/reveal-answer]\n[hidden-answer a=\"787718\"]\n\nFirst, factor the numerator and denominator.\n<p style=\"text-align: center;\">[latex]\\begin{align}k\\left(x\\right)&amp;=\\dfrac{5+2{x}^{2}}{2-x-{x}^{2}} \\\\[1mm] &amp;=\\dfrac{5+2{x}^{2}}{\\left(2+x\\right)\\left(1-x\\right)} \\end{align}[\/latex]<\/p>\nTo find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:\n<p style=\"text-align: center;\">[latex]\\left(2+x\\right)\\left(1-x\\right)=0[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x=-2,1[\/latex]<\/p>\nThese two values indicate two vertical asymptotes. The graph below confirms the location of the two vertical asymptotes.\n\n<img class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213927\/CNX_Precalc_Figure_03_07_0102.jpg\" alt=\"Graph of k(x)=(5+2x)^2\/(2-x-x^2) with its vertical asymptotes at x=-2 and x=1 and its horizontal asymptote at y=-2.\" width=\"487\" height=\"514\">\n\n[\/hidden-answer]\n\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n[ohm_question]219322[\/ohm_question]\n\n<\/div>\n","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Complete a table values with solutions to an equation<\/li>\n<li>Identify points on a graph<\/li>\n<li>Find function values (outputs) for specified numbers (inputs)<\/li>\n<li>Identify vertical and horizontal asymptotes<\/li>\n<li>Find vertical asymptotes of rational functions<\/li>\n<\/ul>\n<\/div>\n<p>The Limit of a Function section provides an introduction to how to find limits of functions. Ways to find limits will be discussed including a tabular approach and graphical approach. To prepare for this, we will review how to complete function value tables. We will also practice looking at the graph of a function and identifying the&nbsp;y-value or function value associated with a given&nbsp;<em>x<\/em>-value. Finally, we will review how to find vertical asymptotes of rational functions in order to get a head start with understanding when limit values are infinite.<\/p>\n<h2>Complete a Table of Function Values<\/h2>\n<div>\n<p>A table of values can be used to organize the&nbsp;<em>y<\/em>-values or function values that result from plugging specific&nbsp;<em>x<\/em>-values into a function&#8217;s equation.<\/p>\n<p>Suppose we want to determine the various values of the equation [latex]f(x)=2x - 1[\/latex] at certain values of <em>x<\/em>. We can begin by substituting a value for <em>x<\/em> into the equation and determining the resulting value of the function. The table below&nbsp;lists some values of <em>x<\/em> from \u20133 to 3 and the resulting function values.<\/p>\n<table style=\"width: 411px; height: 84px;\" summary=\"This is a table with 8 rows and 3 columns. The first row has columns labeled: x, y = 2x-1, (x, y). The entries in the second row are: negative 3; y = 2 times negative 3 minus 1 = negative 7; (-3, -7). The entries in the third row are: negative 2; y = 2 times negative 2 minus 1 = negative 5; (-2, -5). The entries in the fourth row are: negative1; y = 2 times negative 1 minus 1 = negative 3; (-1, -3). The entries in the fifth row are: 0; y = 2 times 0 minus 1 = negative 1; (0, -1). The entries in the sixth row are: 1; y = 2 times 1 minus 1 = 1; (1, 1). The entries in the seventh row are: 2; y = 2 times 2 minus 1 = 3; (2, 3). The entries in the eight row are: 3, y = 2 times 3 minus 1 = 5; (3,5)\">\n<tbody>\n<tr style=\"height: 12px;\">\n<td style=\"width: 116.5px; height: 12px;\">[latex]x[\/latex]<\/td>\n<td style=\"width: 269.5px; height: 12px;\">[latex]f(x)=2x - 1[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 116.5px; height: 12px;\">[latex]-3[\/latex]<\/td>\n<td style=\"width: 269.5px; height: 12px;\">[latex]f(-3)=2\\left(-3\\right)-1=-7[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 116.5px; height: 12px;\">[latex]-2[\/latex]<\/td>\n<td style=\"width: 269.5px; height: 12px;\">[latex]f(-2)=2\\left(-2\\right)-1=-5[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 116.5px; height: 12px;\">[latex]-1[\/latex]<\/td>\n<td style=\"width: 269.5px; height: 12px;\">[latex]f(-1)=2\\left(-1\\right)-1=-3[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 116.5px; height: 12px;\">[latex]0[\/latex]<\/td>\n<td style=\"width: 269.5px; height: 12px;\">[latex]f(0)=2\\left(0\\right)-1=-1[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 116.5px; height: 12px;\">[latex]2[\/latex]<\/td>\n<td style=\"width: 269.5px; height: 12px;\">[latex]f(2)=2\\left(2\\right)-1=3[\/latex]<\/td>\n<\/tr>\n<tr style=\"height: 12px;\">\n<td style=\"width: 116.5px; height: 12px;\">[latex]3[\/latex]<\/td>\n<td style=\"width: 269.5px; height: 12px;\">[latex]f(3)=2\\left(3\\right)-1=5[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>We can look for trends among our function values by looking at the table. For example, in this case, as&nbsp;<em>x<\/em>-values increase, so do the&nbsp;<em>y<\/em>-values. Also, it seems reasonable to assume, based on the table, an&nbsp;<em>x<\/em>-value of 1 would result in a function value of 1. You can verify this for yourself by plugging 1 into [latex]f(x)=2x - 1[\/latex].<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Completing a table of function values<\/h3>\n<p>Create a table of function values for [latex]f(x)=-x+2[\/latex]. Use various integers&nbsp;from -5 to 5 as the&nbsp;<em>x<\/em>-values you plug into the function.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q792137\">Show Solution<\/span><\/p>\n<div id=\"q792137\" class=\"hidden-answer\" style=\"display: none\">\n<p>&nbsp;<\/p>\n<table style=\"width: 385px;\" summary=\"The table shows 8 rows and 3 columns. The entries in the first row are: x; y = negative x plus 2; and (x, y). The entries in the second row are: negative 5; y = the opposite of negative 5 plus 2 = 7; (-5, 7). The entries in the third row are: negative 3; y = the opposite of negative 3 plus 2 = 5; (-3, 5). The entries in the fourth row are: -1; y = the opposite of negative 1 plus 2 = 3; (-1, 3). The entries in the fifth row are: 0; y = opposite of zero plus 2 = 2; (0, 2). The entries in the sixth row are: 1; y = the opposite of 1 plus 2 = 1; (1, 1). The entries in the seventh row are: 3; y = the opposite of 3 plus 2 = negative 1; (3, -1). The entries in the eighth row are: 5; y = the opposite of 5 plus 2 = negative 3; (5, -3).\">\n<tbody>\n<tr>\n<td style=\"width: 117.656px;\">[latex]x[\/latex]<\/td>\n<td style=\"width: 244.656px;\">[latex]f(x)=-x+2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 117.656px;\">[latex]-5[\/latex]<\/td>\n<td style=\"width: 244.656px;\">[latex]f(-5)=-\\left(-5\\right)+2=7[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 117.656px;\">[latex]-3[\/latex]<\/td>\n<td style=\"width: 244.656px;\">[latex]f(-3)=-\\left(-3\\right)+2=5[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 117.656px;\">[latex]-1[\/latex]<\/td>\n<td style=\"width: 244.656px;\">[latex]f(-1)=-\\left(-1\\right)+2=3[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 117.656px;\">[latex]0[\/latex]<\/td>\n<td style=\"width: 244.656px;\">[latex]f(0)=-\\left(0\\right)+2=2[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 117.656px;\">[latex]1[\/latex]<\/td>\n<td style=\"width: 244.656px;\">[latex]f(1)=-\\left(1\\right)+2=1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 117.656px;\">[latex]3[\/latex]<\/td>\n<td style=\"width: 244.656px;\">[latex]f(3)=-\\left(3\\right)+2=-1[\/latex]<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 117.656px;\">[latex]5[\/latex]<\/td>\n<td style=\"width: 244.656px;\">[latex]f(5)=-\\left(5\\right)+2=-3[\/latex]<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm219319\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=219319&theme=oea&iframe_resize_id=ohm219319&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm219320\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=219320&theme=oea&iframe_resize_id=ohm219320&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Find Function Values from a Graph<\/h2>\n<p>We can view a function as a set of inputs and their corresponding outputs. That is, we can see a function as a set of ordered pairs, [latex]\\left(x, y \\right).[\/latex]<\/p>\n<p>Remember that, in function notation, [latex]y = f(x)[\/latex], so the ordered pairs containing inputs and outputs can be written in the form of (<em>input<\/em>,&nbsp;<em>output<\/em>) or [latex]\\left(x, f(x)\\right)[\/latex].<\/p>\n<p>Evaluating a function using a graph requires finding the corresponding output value for a given input value, only in this case, we find the output value (<em>y<\/em>-value) by looking at the graph.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: Reading Function Values from a Graph<\/h3>\n<p>Given the graph below, evaluate [latex]f\\left(2\\right)[\/latex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18191001\/CNX_Precalc_Figure_01_01_0072.jpg\" alt=\"Graph of a positive parabola centered at (1, 0).\" width=\"487\" height=\"445\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q915833\">Show Solution<\/span><\/p>\n<div id=\"q915833\" class=\"hidden-answer\" style=\"display: none\">\n<ol>\n<li>To evaluate [latex]f\\left(2\\right)[\/latex], locate the point on the curve where [latex]x=2[\/latex], then read the [latex]y[\/latex]-coordinate of that point. The point has coordinates [latex]\\left(2,1\\right)[\/latex], so [latex]f\\left(2\\right)=1[\/latex].<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/10\/18191004\/CNX_Precalc_Figure_01_01_0082.jpg\" alt=\"Graph of a positive parabola centered at (1, 0) with the labeled point (2, 1) where f(2) =1.\" width=\"487\" height=\"445\" \/><\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm2471\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=2471&theme=oea&iframe_resize_id=ohm2471&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<h2>Find Vertical Asymptotes of a Rational Function\u200b<\/h2>\n<p>The vertical asymptotes of a rational function may be found by examining the factors of the denominator that are not common to the factors in the numerator. Essentially, you should factor and simplify the rational function first and then set the denominator equal to 0 to find vertical asymptotes.<\/p>\n<div class=\"textbox\">\n<h3>How To: Given a rational function, identify any vertical asymptotes of its graph.<\/h3>\n<ol>\n<li>Factor the numerator and denominator.<\/li>\n<li>Reduce the expression by canceling common factors in the numerator and the denominator.<\/li>\n<li>Note any values that cause the denominator to be zero in this simplified version. These are where the vertical asymptotes occur.<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: Identifying Vertical Asymptotes<\/h3>\n<p>Find the vertical asymptotes of the graph of [latex]k\\left(x\\right)=\\dfrac{5+2{x}^{2}}{2-x-{x}^{2}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q787718\">Show Solution<\/span><\/p>\n<div id=\"q787718\" class=\"hidden-answer\" style=\"display: none\">\n<p>First, factor the numerator and denominator.<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{align}k\\left(x\\right)&=\\dfrac{5+2{x}^{2}}{2-x-{x}^{2}} \\\\[1mm] &=\\dfrac{5+2{x}^{2}}{\\left(2+x\\right)\\left(1-x\\right)} \\end{align}[\/latex]<\/p>\n<p>To find the vertical asymptotes, we determine where this function will be undefined by setting the denominator equal to zero:<\/p>\n<p style=\"text-align: center;\">[latex]\\left(2+x\\right)\\left(1-x\\right)=0[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]x=-2,1[\/latex]<\/p>\n<p>These two values indicate two vertical asymptotes. The graph below confirms the location of the two vertical asymptotes.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/896\/2016\/11\/02213927\/CNX_Precalc_Figure_03_07_0102.jpg\" alt=\"Graph of k(x)=(5+2x)^2\/(2-x-x^2) with its vertical asymptotes at x=-2 and x=1 and its horizontal asymptote at y=-2.\" width=\"487\" height=\"514\" \/><\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try it<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm219322\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=219322&theme=oea&iframe_resize_id=ohm219322&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-3772\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>Modification and Revision. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>College Algebra Corequisite. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\">https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Precalculus. <strong>Provided by<\/strong>: Lumen Learning. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/courses.lumenlearning.com\/precalculus\/\">https:\/\/courses.lumenlearning.com\/precalculus\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"College Algebra Corequisite\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/waymakercollegealgebracorequisite\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Precalculus\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"https:\/\/courses.lumenlearning.com\/precalculus\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Modification and Revision\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3772","chapter","type-chapter","status-publish","hentry"],"part":3770,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3772","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":0,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3772\/revisions"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/3770"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3772\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=3772"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=3772"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=3772"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=3772"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}