{"id":22,"date":"2023-09-17T11:07:37","date_gmt":"2023-09-17T11:07:37","guid":{"rendered":"https:\/\/mathority.org\/id\/asimtot-miring\/"},"modified":"2023-09-17T11:07:37","modified_gmt":"2023-09-17T11:07:37","slug":"asimtot-miring","status":"publish","type":"post","link":"https:\/\/mathority.org\/id\/asimtot-miring\/","title":{"rendered":"Asimtot miring"},"content":{"rendered":"<p>Pada artikel ini kami akan menjelaskan apa itu asimtot miring suatu fungsi. Anda akan mempelajari kapan suatu fungsi memiliki asimtot miring dan cara menghitungnya. Dan, sebagai tambahan, Anda akan dapat melihat contoh asimtot miring dan berlatih dengan latihan yang diselesaikan langkah demi langkah. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"%c2%bfque-es-una-asintota-oblicua\"><\/span> Apa yang dimaksud dengan asimtot miring?<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> <strong>Asimtot miring suatu fungsi adalah garis miring yang grafiknya mendekati tak terhingga tanpa pernah melintasinya.<\/strong> Akibatnya, semua asimtot miring adalah garis dengan persamaan <em>y=mx+n<\/em> .<\/p>\n<p> Kemiringan dan titik asal asimtot miring dihitung menggunakan rumus berikut: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/asymptote-oblique-dune-fonction.webp\" alt=\"asimtot miring suatu fungsi\" class=\"wp-image-1362\" width=\"290\" height=\"328\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"como-calcular-la-asintota-oblicua-de-una-funcion\"><\/span> Cara menghitung asimtot miring suatu fungsi<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Untuk menghitung asimtot miring suatu fungsi, langkah-langkah berikut harus dilakukan:<\/p>\n<ol style=\"color:#FF8A05; font-weight: bold;border:\">\n<li style=\"margin-bottom:20px\"> <span style=\"color:#101010;font-weight: normal;\">Hitung limit hingga tak terhingga dari fungsi tersebut dibagi x.<\/span><\/li>\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">Jika limit di atas menghasilkan bilangan real bukan nol, berarti fungsi tersebut mempunyai asimtot miring. Terlebih lagi, kemiringan asimtot miring tersebut akan menjadi nilai yang diperoleh pada batasnya.<\/span><\/li>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-004f6e72e10d1ba23da76d2fd8ea13f3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle m = \\lim_{x \\to \\pm\\infty}\\frac{f(x)}{x}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"125\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<li style=\"margin-bottom:12px\"> <span style=\"color:#101010;font-weight: normal;\">Dalam hal ini, yang tersisa hanyalah menghitung titik potong asimtot miring dengan menyelesaikan limit berikut:<\/span><\/li>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-9cc74ce0447b0a9148cae947674ad085_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle n = \\lim_{x \\to \\pm\\infty} [f(x)-mx]\" title=\"Rendered by QuickLaTeX.com\" height=\"26\" width=\"170\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<\/ol>\n<p> <strong>Catatan:<\/strong> limit harus dihitung pada plus dan minus tak terhingga, tetapi biasanya memberikan hasil yang sama dan itulah sebabnya kita menyederhanakannya dengan memasukkan \u00b1\u221e. Namun jika batas plus dan minus tak terhingga berbeda, asimtot miring kiri dan asimtot miring kanan harus dihitung secara terpisah. <\/p>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"ejemplo-de-asintota-oblicua\"><\/span> Contoh Asimtot Miring<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p> Selanjutnya, kita akan mengambil asimtot miring dari fungsi rasional berikut sehingga Anda dapat melihat contoh cara melakukannya:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b02f6283fd481e890a943badfa2c876f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"f(x)=\\cfrac{x^2+1}{x}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"109\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p> Asimtot miring termasuk dalam tipe ini<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ad313410fc976bc53709807aa8aed8e7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y=mx+n.\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"95\" style=\"vertical-align: -4px;\"><\/p>\n<p> Jadi kita hitung dulu kemiringan garisnya<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6b41df788161942c6f98604d37de8098_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"m\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"15\" style=\"vertical-align: 0px;\"><\/p>\n<p> dengan rumus yang sesuai:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-9b50ee5cbc3cf33f7fd42c3fe03a3d71_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle m = \\lim_{x \\to \\pm\\infty} \\frac{f(x)}{x}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"125\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-870fc158a1aabb54cb5f3b4296381512_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle m= \\lim_{x \\to \\pm\\infty} \\cfrac{\\cfrac{x^2+1}{x}}{x}\" title=\"Rendered by QuickLaTeX.com\" height=\"60\" width=\"141\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p> Untuk menyelesaikan limit ini kita harus menerapkan sifat-sifat pecahan:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-7f313d826cd1a2dd1ef66b1d0a40efb8_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\cfrac{\\cfrac{a}{b}}{\\cfrac{c}{d}}=\\cfrac{a\\cdot d}{b\\cdot c}\" title=\"Rendered by QuickLaTeX.com\" height=\"80\" width=\"69\" style=\"vertical-align: -39px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-dfc6b6aa917846535c6c4b6158961988_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle m= \\lim_{x \\to \\pm\\infty} \\cfrac{\\cfrac{x^2+1}{x}}{x}=\\lim_{x \\to \\pm\\infty} \\cfrac{x^2+1}{x^2}\" title=\"Rendered by QuickLaTeX.com\" height=\"60\" width=\"264\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p> Dan sekarang kita menghitung batasnya:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-542fc353481ddc465b7a40f665d3661d_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle m = \\lim_{x \\to \\pm\\infty} \\cfrac{x^2+1}{x^2} = \\cfrac{+\\infty}{+\\infty} = \\cfrac{1}{1} = \\bm{1}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"269\" style=\"vertical-align: -14px;\"><\/p>\n<\/p>\n<p> Dalam hal ini, hasil dari ketidakterbatasan antara tak terhingga adalah pembagian koefisien x pangkat tertinggi, karena pembilang dan penyebutnya berorde sama.<\/p>\n<p> Limit di atas menghasilkan bilangan real bukan nol, sehingga fungsi tersebut mempunyai asimtot miring. Sekarang kita akan menghitung perpotongan y<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-b170995d512c659d8668b4e42e1fef6b_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"n\" title=\"Rendered by QuickLaTeX.com\" height=\"8\" width=\"11\" style=\"vertical-align: 0px;\"><\/p>\n<p> dari asimtot menggunakan rumus yang sesuai:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-45119c7a74d77a92d7a6cfd5b5c3544f_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle n = \\lim_{x \\to \\pm\\infty} \\left[f(x)-mx\\right]\" title=\"Rendered by QuickLaTeX.com\" height=\"26\" width=\"173\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-9197669cc0e41aa22224b552b21b31ed_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle n = \\lim_{x \\to \\pm\\infty} \\left[\\cfrac{x^2+1}{x}-1x\\right]\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"191\" style=\"vertical-align: -23px;\"><\/p>\n<\/p>\n<p> Kami mencoba menghitung batasnya:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-4d7fa012eace37e82c243012c91f1a5c_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle n = \\lim_{x \\to \\pm\\infty} \\left[\\cfrac{x^2+1}{x}-x\\right] = \\cfrac{+\\infty}{+\\infty} - (+\\infty) = \\bm{+\\infty - \\infty}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"412\" style=\"vertical-align: -23px;\"><\/p>\n<\/p>\n<p> Tapi kita mendapatkan ketidakterbatasan tak terhingga dikurangi tak terhingga. Oleh karena itu, suku-suku tersebut perlu direduksi menjadi penyebut yang sama. Untuk melakukannya, kita mengalikan dan membagi x dengan penyebut pecahan:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a2355ed9411470b9fd20a50ebbd48726_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle n=\\lim_{x \\to \\pm\\infty} \\left[\\cfrac{x^2+1}{x}-\\cfrac{x\\cdot x}{x} \\right] = \\lim_{x \\to \\pm\\infty} \\left[\\cfrac{x^2+1}{x}-\\cfrac{x^2}{x}\\right]\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"391\" style=\"vertical-align: -23px;\"><\/p>\n<\/p>\n<p> Karena kedua suku tersebut memiliki penyebut yang sama, kita dapat mengelompokkannya:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f932ebc8728669c7c6b57e115c444fc7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle n = \\lim_{x \\to \\pm\\infty} \\left[\\cfrac{x^2+1}{x}-\\cfrac{x^2}{x} \\right] =  \\lim_{x \\to \\pm\\infty} \\cfrac{x^2+1-x^2}{x}\" title=\"Rendered by QuickLaTeX.com\" height=\"54\" width=\"358\" style=\"vertical-align: -23px;\"><\/p>\n<\/p>\n<p> Kami beroperasi pada pembilang:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c39259f829c9e99fc88819c6ae266e82_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle n = \\lim_{x \\to \\pm\\infty}  \\cfrac{\\phantom{2}1\\phantom{2}}{x}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"112\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p> Dan akhirnya, kami menyelesaikan batasannya:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-16a0044416d02e77b05f65f1bb93d4cb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle n = \\lim_{x \\to \\pm\\infty}  \\cfrac{\\phantom{2}1\\phantom{2}}{x}= \\cfrac{1}{\\pm\\infty} = \\bm{0}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"201\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p> Jadi <em>n<\/em> =0. Oleh karena itu, asimtot miring adalah fungsi linier: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6fbe1cc5f3362ddbd80ed0b29c0bb4ef_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y = mx+n\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"91\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-a68ac5c51acd0f68bd022aee64cd9cd4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y = 1x+0\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"83\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-4909df7491ef54f0df1e922bc29417f3_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{y=x}\" title=\"Rendered by QuickLaTeX.com\" height=\"12\" width=\"43\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p> Fungsi yang dipelajari ditunjukkan pada grafik di bawah ini. Seperti yang Anda lihat, fungsinya sangat dekat dengan garis y=x tetapi tidak pernah menyentuhnya karena merupakan asimtot miring: <\/p>\n<div class=\"wp-block-image\">\n<figure class=\"aligncenter size-large is-resized\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/uploads\/2023\/07\/image-1.png\" alt=\"contoh asimtot miring\" class=\"wp-image-1374\" width=\"424\" height=\"478\" srcset=\"\" sizes=\"auto, \" data-src=\"\"><\/figure>\n<\/div>\n<h2 class=\"wp-block-heading\"><span class=\"ez-toc-section\" id=\"ejercicios-resueltos-de-asintotas-oblicuas\"><\/span> Latihan terpecahkan pada asimtot miring<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h3 class=\"wp-block-heading\"> Latihan 1<\/h3>\n<p> Temukan asimtot miring dari fungsi rasional berikut: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8ecc70adc78bf259cf6e36c0dcf1bee7_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle f(x)= \\frac{x^2+2x+3}{x+1}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"150\" style=\"vertical-align: -14px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E6F9EF\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E6F9EF\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>Lihat solusinya<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Asimtot miring berbentuk<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-8e4adcc4368f6296906b6231bf17a6a4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y=mx+n\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"91\" style=\"vertical-align: -4px;\"><\/p>\n<p> , oleh karena itu perlu untuk menghitung parameter <em>m<\/em> dan <em>n<\/em> . Pertama-tama kita menghitung <em>m<\/em> dengan menerapkan rumusnya:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-dc38f695cee95c4c60c6e2591345119e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle m = \\lim_{x \\to \\pm\\infty} \\frac{f(x)}{x} = \\lim_{x \\to \\pm\\infty} \\cfrac{\\cfrac{x^2+2x+3}{x+1}}{x}\" title=\"Rendered by QuickLaTeX.com\" height=\"62\" width=\"293\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Kita menyederhanakan pecahan dengan menerapkan sifat-sifat pecahan: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ef59ac0cd51c39c615896543993c12b6_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle m =\\lim_{x \\to \\pm\\infty} \\frac{x^2+2x+3}{(x+1)\\cdot x}\" title=\"Rendered by QuickLaTeX.com\" height=\"44\" width=\"180\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-c51e373fd07a821f8e75d63e38f252dd_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle m =\\lim_{x \\to \\pm\\infty} \\frac{x^2+2x+3}{x^2+x}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"180\" style=\"vertical-align: -14px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Dan kami memecahkan batasnya:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-653fa714bca94b5cc4f3ed715d7c1520_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle m =\\lim_{x \\to \\pm\\infty} \\frac{x^2+2x+3}{x^2+x}= \\frac{+\\infty}{+\\infty} = \\frac{1}{1} = \\bm{1}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"308\" style=\"vertical-align: -14px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Jadi <em>m<\/em> =1. Sekarang mari kita hitung titik potong asimtot miring dengan menerapkan rumusnya:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-e779b5ac239ae56c53427510dbd54dcb_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle n = \\lim_{x \\to \\pm\\infty} \\left[f(x)-mx\\right] = \\lim_{x \\to \\pm\\infty} \\left[ \\frac{x^2+2x+3}{x+1}-1x\\right]\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"395\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Kami mencoba menghitung batasnya:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-f95f290fbf258d45aa5765008d7aad13_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle n = \\lim_{x \\to \\pm\\infty} \\left[ \\frac{x^2+2x+3}{x+1}-x\\right]= \\bm{+\\infty - \\infty}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"320\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Tapi kita mendapatkan bentuk tak tentu tak terhingga dikurangi tak terhingga. Oleh karena itu, kita harus mereduksi suku-suku tersebut menjadi penyebut yang sama dan kemudian mengelompokkannya:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-0712d34ed442d9e12ef2490f04df078a_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}\\displaystyle n = \\lim_{x \\to \\pm\\infty} \\left[ \\frac{x^2+2x+3}{x+1}-x\\right] =\\\\[6ex]=\\displaystyle\\lim_{x \\to \\pm\\infty} \\left[ \\frac{x^2+2x+3}{x+1}-\\frac{x \\cdot (x+1)}{x+1} \\right] = \\\\[6ex]=\\displaystyle\\lim_{x \\to \\pm\\infty} \\left[ \\frac{x^2+2x+3}{x+1}-\\frac{x^2+x}{x+1} \\right]=\\\\[6ex]=\\displaystyle\\lim_{x \\to \\pm\\infty} \\frac{x^2+2x+3-(x^2+x)}{x+1}\\\\[6ex]\\displaystyle =\\lim_{x \\to \\pm\\infty} \\frac{x^2+2x+3-x^2-x}{x+1}=\\\\[6ex]=\\displaystyle \\lim_{x \\to \\pm\\infty} \\frac{x+3}{x+1}\\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"434\" width=\"300\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Dan akhirnya, kami menyelesaikan batasannya:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ee7e1fdd8e781abed322fed1182ddb15_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle n =\\lim_{x \\to \\pm\\infty} \\frac{x+3}{x+1} = \\frac{\\infty}{\\infty} = \\frac{1}{1} = \\bm{1}\" title=\"Rendered by QuickLaTeX.com\" height=\"38\" width=\"241\" style=\"vertical-align: -14px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Singkatnya, asimtot miring dari fungsi tersebut adalah: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6fbe1cc5f3362ddbd80ed0b29c0bb4ef_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y = mx+n\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"91\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-69c0f50795c1f6034c0cd04201f614d4_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y = 1x + 1\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"82\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-5ffe94db5ae8fa1abc72e6007c2c0586_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{y = x + 1}\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"73\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n<h3 class=\"wp-block-heading\">Latihan 2<\/h3>\n<p> Temukan semua asimtot miring dari fungsi rasional berikut: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-144807b8c72afbd43bb3f97d69cedb35_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle f(x)=\\frac{2x^2-5}{x+3}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"118\" style=\"vertical-align: -14px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-start otfm-sp__wrapper otfm-sp__box js-otfm-sp-box__closed otfm-sp__E6F9EF\" role=\"button\" tabindex=\"0\" aria-expanded=\"false\" data-otfm-spc=\"#E6F9EF\" style=\"text-align:center\">\n<div class=\"otfm-sp__title\"> <strong>Lihat solusinya<\/strong><\/div>\n<\/div>\n<p class=\"has-text-align-left\"> Pertama, kita menggunakan rumus kemiringan asimtot miring:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-bc900ded359235b2293ec151e715daea_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle m = \\lim_{x \\to \\pm\\infty} \\frac{f(x)}{x} = \\lim_{x \\to \\pm\\infty} \\cfrac{\\cfrac{2x^2-5}{x+3}}{x}\" title=\"Rendered by QuickLaTeX.com\" height=\"62\" width=\"261\" style=\"vertical-align: -12px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Kita menyederhanakan pecahan dengan menerapkan sifat-sifat pecahan: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-1c5afa9b1ca5f1c73e6b8e64c8fb9420_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle m =\\lim_{x \\to \\pm\\infty}\\frac{2x^2-5}{(x+3)\\cdot x}\" title=\"Rendered by QuickLaTeX.com\" height=\"44\" width=\"168\" style=\"vertical-align: -17px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-03a4b53a445bded103e8de4404620693_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle m =\\lim_{x \\to \\pm\\infty}\\frac{2x^2-5}{x^2+3x}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"149\" style=\"vertical-align: -14px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Dan kami menentukan batasnya:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-461c274fc210474eddaf061463e92aaf_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle m =\\lim_{x \\to \\pm\\infty}\\frac{2x^2-5}{x^2+3x}= \\frac{+\\infty}{+\\infty} = \\frac{2}{1} = \\bm{2}\" title=\"Rendered by QuickLaTeX.com\" height=\"41\" width=\"278\" style=\"vertical-align: -14px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Limit tersebut memberikan bilangan real selain nol, sehingga merupakan fungsi rasional dengan asimtot miring yang kemiringannya 2.<\/p>\n<p class=\"has-text-align-left\"> Sekarang mari kita hitung intersepnya dengan menerapkan rumus yang sesuai:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-4a04a1abaebfc5e1781dd7d98399888e_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle n = \\lim_{x \\to \\pm\\infty} \\left[f(x)-mx\\right] = \\lim_{x \\to \\pm\\infty} \\left[\\frac{2x^2-5}{x+3}-2x\\right]\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"364\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Kami mencoba menghitung batasnya:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-00f35703d153fe6911328d143588e1cc_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle n = \\lim_{x \\to \\pm\\infty} \\left[\\frac{2x^2-5}{x+3}-2x\\right]= \\bm{+\\infty - \\infty}\" title=\"Rendered by QuickLaTeX.com\" height=\"43\" width=\"298\" style=\"vertical-align: -16px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Tapi kita mendapatkan perbedaan ketidakpastian dari ketidakterbatasan. Oleh karena itu, kami mengurangi suku-suku tersebut menjadi penyebut yang sama dan kemudian mengoperasikannya:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-4920e8b21b180c4f2740ce712d9f30d0_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\begin{array}{l}\\displaystyle n = \\lim_{x \\to \\pm\\infty} \\left[\\frac{2x^2-5}{x+3}-2x\\right]=\\\\[6ex]=\\displaystyle\\lim_{x \\to \\pm\\infty} \\left[\\frac{2x^2-5}{x+3}-\\frac{2x\\cdot (x+3)}{x+3} \\right] = \\\\[6ex]=\\displaystyle\\lim_{x \\to \\pm\\infty} \\left[ \\frac{2x^2-5}{x+3}-\\frac{2x^2+6x}{x+3}\\right]=\\\\[6ex]=\\displaystyle\\lim_{x \\to \\pm\\infty}\\frac{2x^2-5-(2x^2+6x)}{x+3}\\\\[6ex]\\displaystyle =\\lim_{x \\to \\pm\\infty}\\frac{2x^2-5-2x^2-6x}{x+3}=\\\\[6ex]=\\displaystyle \\lim_{x \\to \\pm\\infty} \\frac{-6x-5}{x+3}\\end{array}\" title=\"Rendered by QuickLaTeX.com\" height=\"434\" width=\"277\" style=\"vertical-align: 0px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Dan akhirnya, kami menyelesaikan batasannya:<\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-00b75da44399a44a4e215fd4baccf214_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\displaystyle n =\\lim_{x \\to \\pm\\infty} \\frac{-6x-5}{x+3}= \\frac{\\infty}{\\infty}=\\frac{-6}{1} = \\bm{-6}\" title=\"Rendered by QuickLaTeX.com\" height=\"39\" width=\"292\" style=\"vertical-align: -14px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-left\"> Ringkasnya, asimtot miring dari fungsi pecahan adalah: <\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-6fbe1cc5f3362ddbd80ed0b29c0bb4ef_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"y = mx+n\" title=\"Rendered by QuickLaTeX.com\" height=\"15\" width=\"91\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<p class=\"has-text-align-center\"><img decoding=\"async\" loading=\"lazy\" src=\"https:\/\/mathority.org\/wp-content\/ql-cache\/quicklatex.com-ac6ac25ec7b85209d4d7d855e3d0b501_l3.png\" class=\"ql-img-inline-formula quicklatex-auto-format\" alt=\"\\bm{y=2x-6}\" title=\"Rendered by QuickLaTeX.com\" height=\"16\" width=\"83\" style=\"vertical-align: -4px;\"><\/p>\n<\/p>\n<div class=\"wp-block-otfm-box-spoiler-end otfm-sp_end\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>Pada artikel ini kami akan menjelaskan apa itu asimtot miring suatu fungsi. Anda akan mempelajari kapan suatu fungsi memiliki asimtot miring dan cara menghitungnya. Dan, sebagai tambahan, Anda akan dapat melihat contoh asimtot miring dan berlatih dengan latihan yang diselesaikan langkah demi langkah. Apa yang dimaksud dengan asimtot miring? Asimtot miring suatu fungsi adalah garis &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathority.org\/id\/asimtot-miring\/\"> <span class=\"screen-reader-text\">Asimtot miring<\/span> Selengkapnya &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"","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":"","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":"","footnotes":""},"categories":[43],"tags":[],"class_list":["post-22","post","type-post","status-publish","format-standard","hentry","category-batasan-fungsi"],"yoast_head":"<!-- This site is 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