{"id":247358,"date":"2025-07-07T16:16:22","date_gmt":"2025-07-07T16:16:22","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=247358"},"modified":"2025-07-07T16:16:24","modified_gmt":"2025-07-07T16:16:24","slug":"graph-y-tan-x-4","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/07\/graph-y-tan-x-4\/","title":{"rendered":"Graph y = tan (x + 4)"},"content":{"rendered":"\n<p>Graph y = tan (x + 4) 13) Graph y = cot (x)<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/image-282.png\" alt=\"\" class=\"wp-image-247359\"\/><\/figure>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>Of course. Here is the correct answer and a detailed explanation for graphing the two trigonometric functions from the image.<\/p>\n\n\n\n<p><strong>Correct Answer and Explanation<\/strong><\/p>\n\n\n\n<p>The problems require graphing two transformed trigonometric functions:&nbsp;y = tan(x + \u03c0\/4)&nbsp;and&nbsp;y = cot(1\/3 * x). This involves understanding the parent functions and applying transformations such as phase shifts and period changes.<\/p>\n\n\n\n<p><strong>12) Graphing y = tan(x + \u03c0\/4)<\/strong><\/p>\n\n\n\n<p>To graph this function, we start with the parent function&nbsp;y = tan(x)&nbsp;and apply the given transformation. The equation is in the form&nbsp;y = tan(x &#8211; c), where the value of&nbsp;c&nbsp;determines the horizontal shift, also known as the phase shift. In this case, the equation is&nbsp;y = tan(x + \u03c0\/4), which can be rewritten as&nbsp;y = tan(x &#8211; (-\u03c0\/4)). This means&nbsp;c = -\u03c0\/4, indicating a phase shift of \u03c0\/4 units to the left.<\/p>\n\n\n\n<p>The key features of the parent function&nbsp;y = tan(x)&nbsp;are its period of \u03c0 and its vertical asymptotes at&nbsp;x = \u03c0\/2 + n\u03c0, where n is any integer. A typical cycle is centered at the origin, passing through&nbsp;(0, 0).<\/p>\n\n\n\n<p>Applying the leftward shift of \u03c0\/4:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Period:<\/strong>\u00a0The period is not affected by a horizontal shift, so it remains \u03c0.<\/li>\n\n\n\n<li><strong>Asymptotes:<\/strong>\u00a0Each asymptote of the parent function is shifted \u03c0\/4 units to the left. The new asymptotes are located at\u00a0x = (\u03c0\/2 &#8211; \u03c0\/4) + n\u03c0, which simplifies to\u00a0x = \u03c0\/4 + n\u03c0. For example, two consecutive asymptotes are at\u00a0x = -3\u03c0\/4\u00a0and\u00a0x = \u03c0\/4.<\/li>\n\n\n\n<li><strong>Key Points:<\/strong>\u00a0The center of a cycle, originally at\u00a0(0, 0), shifts to\u00a0(-\u03c0\/4, 0). This point is now an x-intercept.<\/li>\n<\/ul>\n\n\n\n<p>To sketch the graph, you would first draw the vertical asymptotes at&nbsp;x = \u03c0\/4,&nbsp;x = 5\u03c0\/4, etc. Then, plot the x-intercept halfway between two asymptotes, for example at&nbsp;x = -\u03c0\/4. The graph retains its increasing S-shape, passing through&nbsp;(-\u03c0\/2, -1)&nbsp;and&nbsp;(0, 1)&nbsp;within one cycle, and repeating this pattern every \u03c0 units.<\/p>\n\n\n\n<p><strong>13) Graphing y = cot(1\/3 * x)<\/strong><\/p>\n\n\n\n<p>This problem involves a transformation of the parent function&nbsp;y = cot(x). The equation&nbsp;y = cot(1\/3 * x)&nbsp;is in the form&nbsp;y = cot(bx), where&nbsp;b&nbsp;affects the period of the function.<\/p>\n\n\n\n<p>The key features of the parent function&nbsp;y = cot(x)&nbsp;are its period of \u03c0 and its vertical asymptotes at&nbsp;x = n\u03c0. It is a decreasing function between its asymptotes.<\/p>\n\n\n\n<p>The transformation here is a horizontal stretch caused by&nbsp;b = 1\/3.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Period:<\/strong>\u00a0The new period is calculated by the formula\u00a0Period = \u03c0 \/ |b|. So, the new period is\u00a0\u03c0 \/ (1\/3) = 3\u03c0. This means the graph is stretched horizontally by a factor of 3, and one full cycle now spans a distance of 3\u03c0.<\/li>\n\n\n\n<li><strong>Asymptotes:<\/strong>\u00a0The original asymptotes at\u00a0x = n\u03c0\u00a0are stretched by a factor of 3. We find their new locations by setting the argument of cotangent to\u00a0n\u03c0:\u00a0(1\/3)x = n\u03c0, which gives\u00a0x = 3n\u03c0. Therefore, the new vertical asymptotes are at\u00a0x = 0,\u00a0x = 3\u03c0,\u00a0x = -3\u03c0, and so on.<\/li>\n\n\n\n<li><strong>Key Points:<\/strong>\u00a0The x-intercept, originally at\u00a0x = \u03c0\/2, is now at\u00a0x = (\u03c0\/2) * 3 = 3\u03c0\/2.<\/li>\n<\/ul>\n\n\n\n<p>To sketch this graph, first draw the vertical asymptotes at&nbsp;x=0&nbsp;and&nbsp;x=3\u03c0. Plot the x-intercept at&nbsp;(3\u03c0\/2, 0). The graph is a decreasing curve that passes through points like&nbsp;(3\u03c0\/4, 1)&nbsp;and&nbsp;(9\u03c0\/4, -1), approaching the asymptotes on either side. This stretched cotangent shape repeats every 3\u03c0 units along the x-axis.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner5-1002.jpeg\" alt=\"\" class=\"wp-image-247360\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Graph y = tan (x + 4) 13) Graph y = cot (x) The Correct Answer and Explanation is: Of course. Here is the correct answer and a detailed explanation for graphing the two trigonometric functions from the image. Correct Answer and Explanation The problems require graphing two transformed trigonometric functions:&nbsp;y = tan(x + \u03c0\/4)&nbsp;and&nbsp;y [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-247358","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/247358","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=247358"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/247358\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=247358"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=247358"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=247358"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}