{"id":28933,"date":"2021-04-27T13:00:20","date_gmt":"2021-04-27T13:00:20","guid":{"rendered":"http:\/\/toposuranos.com\/material\/?p=28933"},"modified":"2024-09-22T02:15:10","modified_gmt":"2024-09-22T02:15:10","slug":"caracterizacao-de-parabolas-e-seus-graficos","status":"publish","type":"post","link":"http:\/\/toposuranos.com\/material\/pt\/caracterizacao-de-parabolas-e-seus-graficos\/","title":{"rendered":"Caracteriza\u00e7\u00e3o de Par\u00e1bolas e seus Gr\u00e1ficos"},"content":{"rendered":"<p><center><\/p>\n<h1>Caracteriza\u00e7\u00e3o de Par\u00e1bolas e seus Gr\u00e1ficos<\/h1>\n<p><em><strong>Resumo:<\/strong><br \/>\n   Nesta aula revisaremos a caracteriza\u00e7\u00e3o das par\u00e1bolas a partir da sua equa\u00e7\u00e3o geral e forma can\u00f4nica, explicando como identificar elementos chave como o v\u00e9rtice, o foco, a diretriz, o eixo de simetria e as poss\u00edveis interse\u00e7\u00f5es com o eixo X.<br \/>\n   <\/em><br \/>\n   <strong>Objetivos de Aprendizagem:<\/strong><br \/>\n   Ao final desta aula, o estudante ser\u00e1 capaz de:<\/p>\n<ol style=\"text-align: left;\">\n<li><strong>Calcular<\/strong> a posi\u00e7\u00e3o do v\u00e9rtice, foco e diretriz da par\u00e1bola a partir da sua forma geral e can\u00f4nica.<\/li>\n<li><strong>Transformar<\/strong> a equa\u00e7\u00e3o can\u00f4nica para a forma geral para extrair informa\u00e7\u00f5es geom\u00e9tricas.<\/li>\n<li><strong>Esquematizar<\/strong> o gr\u00e1fico da par\u00e1bola com as informa\u00e7\u00f5es obtidas.<\/li>\n<\/ol>\n<p>   <strong>\u00cdNDICE DE CONTE\u00daDOS<\/strong><br \/>\n   <a href=\"#1\">Forma geral e can\u00f4nica das par\u00e1bolas<\/a><br \/>\n   <a href=\"#2\">Caracteriza\u00e7\u00e3o de Par\u00e1bolas a partir da Equa\u00e7\u00e3o Geral<\/a><br \/>\n   <a href=\"#3\">Caracteriza\u00e7\u00e3o de Par\u00e1bolas a partir da Equa\u00e7\u00e3o Can\u00f4nica<\/a><br \/>\n   <a href=\"#4\">Caracteriza\u00e7\u00e3o autom\u00e1tica com Excel<\/a>\n   <\/p>\n<p>   <\/center><br \/>\n   <center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/C6DbrJDiZTM\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center><br \/>\n   <a name=\"1\"><\/a><\/p>\n<h2>Forma geral e can\u00f4nica das par\u00e1bolas<\/h2>\n<p style=\"text-align: justify;\">Na aula anterior, vimos que as par\u00e1bolas podem ser expressas algebricamente atrav\u00e9s da equa\u00e7\u00e3o geral das par\u00e1bolas como.<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">(x-x_0)^2 = 4f(y-y_0)<\/span>\n<p style=\"text-align: justify;\">Onde o par <span class=\"katex-eq\" data-katex-display=\"false\">(x_0,y_0)<\/span> \u00e9 a posi\u00e7\u00e3o do v\u00e9rtice e <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> \u00e9 a dist\u00e2ncia focal. Se <span class=\"katex-eq\" data-katex-display=\"false\">f \\gt 0<\/span>, ent\u00e3o o foco est\u00e1 a uma dist\u00e2ncia <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> acima do v\u00e9rtice, e se <span class=\"katex-eq\" data-katex-display=\"false\">f \\lt 0<\/span>, ent\u00e3o o foco estar\u00e1 a uma dist\u00e2ncia <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> abaixo do v\u00e9rtice.<\/p>\n<p style=\"text-align: justify;\">Tamb\u00e9m vimos que a equa\u00e7\u00e3o das par\u00e1bolas levada \u00e0 sua forma can\u00f4nica \u00e9 equivalente a um polin\u00f4mio de grau 2.<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">y(x) = ax^2 + bx + c,<\/span> com <span class=\"katex-eq\" data-katex-display=\"false\">a \\neq 0<\/span>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=C6DbrJDiZTM&amp;t=228s\" target=\"_blank\" rel=\"noopener\"><strong>Caracterizar uma par\u00e1bola consiste<\/strong><\/a> em revelar as seguintes informa\u00e7\u00f5es.<\/p>\n<ul style=\"text-align: justify;\">\n<li>As coordenadas do v\u00e9rtice<\/li>\n<li>As coordenadas do foco<\/li>\n<li>A equa\u00e7\u00e3o da diretriz<\/li>\n<li>A equa\u00e7\u00e3o do eixo de simetria<\/li>\n<li>As interse\u00e7\u00f5es com o eixo x (se existirem)<\/li>\n<li>Finalmente, construir um esbo\u00e7o do gr\u00e1fico com as informa\u00e7\u00f5es coletadas.<\/li>\n<\/ul>\n<p>   <img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/1.bp.blogspot.com\/-xAUdvfTRbjw\/YIbmIXDdT-I\/AAAAAAAAFAI\/8NH0t_EWbH0KIuFDnsRu2IyHdyN4WU54wCLcBGAsYHQ\/s0\/caracterizaci%25C3%25B3n%2Bde%2Bpar%25C3%25A1bolas.PNG\" alt=\"Caracteriza\u00e7\u00e3o de Par\u00e1bolas\" class=\" aligncenter lazyload\" width=\"537\" height=\"414\" \/><noscript><img decoding=\"async\" src=\"https:\/\/1.bp.blogspot.com\/-xAUdvfTRbjw\/YIbmIXDdT-I\/AAAAAAAAFAI\/8NH0t_EWbH0KIuFDnsRu2IyHdyN4WU54wCLcBGAsYHQ\/s0\/caracterizaci%25C3%25B3n%2Bde%2Bpar%25C3%25A1bolas.PNG\" alt=\"Caracteriza\u00e7\u00e3o de Par\u00e1bolas\" class=\" aligncenter lazyload\" width=\"537\" height=\"414\" \/><\/noscript><br \/>\n   <a name=\"2\"><\/a><\/p>\n<h2>Caracteriza\u00e7\u00e3o de Par\u00e1bolas a partir da Equa\u00e7\u00e3o Geral<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=C6DbrJDiZTM&amp;t=316s\" target=\"_blank\" rel=\"noopener\"><strong>Se voc\u00ea tem a par\u00e1bola descrita atrav\u00e9s <\/strong><\/a>da equa\u00e7\u00e3o geral, ent\u00e3o j\u00e1 tem quase toda a informa\u00e7\u00e3o necess\u00e1ria para completar a caracteriza\u00e7\u00e3o, apenas as interse\u00e7\u00f5es com o eixo x necessitar\u00e3o de an\u00e1lise adicional.<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">(x-x_0)^2 = 4f(y-y_0)<\/span>\n<p style=\"text-align: justify;\">A partir da\u00ed voc\u00ea j\u00e1 tem:<\/p>\n<ul style=\"text-align: justify;\">\n<li><strong>V\u00e9rtice:<\/strong> O ponto de coordenadas <span class=\"katex-eq\" data-katex-display=\"false\">(x_0,y_0)<\/span><\/li>\n<li><strong>Posi\u00e7\u00e3o focal:<\/strong> a <span class=\"katex-eq\" data-katex-display=\"false\">f<\/span> unidades acima do v\u00e9rtice<\/li>\n<li><strong>Foco:<\/strong> O ponto de coordenadas <span class=\"katex-eq\" data-katex-display=\"false\">(x_0,y_0 + f)<\/span><\/li>\n<li><strong>Diretriz:<\/strong> a reta de equa\u00e7\u00e3o <span class=\"katex-eq\" data-katex-display=\"false\">y = y_0 - f<\/span><\/li>\n<li><strong>Eixo de simetria:<\/strong> a reta de equa\u00e7\u00e3o <span class=\"katex-eq\" data-katex-display=\"false\">x = x_0<\/span><\/li>\n<\/ul>\n<p style=\"text-align: justify;\">Para encontrar as interse\u00e7\u00f5es com o eixo x, ser\u00e1 necess\u00e1rio passar a equa\u00e7\u00e3o geral para a sua forma can\u00f4nica, igualar o polin\u00f4mio de segundo grau resultante a zero. Se existirem solu\u00e7\u00f5es, tais ser\u00e3o as interse\u00e7\u00f5es com o eixo x.<\/p>\n<p>   <a name=\"3\"><\/a><\/p>\n<h2>Caracteriza\u00e7\u00e3o de Par\u00e1bolas a partir da Equa\u00e7\u00e3o Can\u00f4nica<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=C6DbrJDiZTM&amp;t=393s\" target=\"_blank\" rel=\"noopener\"><strong>Quando a equa\u00e7\u00e3o das par\u00e1bolas<\/strong><\/a> \u00e9 apresentada na forma can\u00f4nica, voc\u00ea tem duas op\u00e7\u00f5es: 1) Caracterizar transformando para a equa\u00e7\u00e3o geral ou 2) Usar a simetria e as interse\u00e7\u00f5es com o eixo x. Ambos os m\u00e9todos t\u00eam suas virtudes. O segundo \u00e9 geralmente mais r\u00e1pido, mas as par\u00e1bolas nem sempre interceptam o eixo X. O primeiro \u00e9 mais trabalhoso, mas tamb\u00e9m, como veremos mais adiante, \u00e9 simples de automatizar. N\u00f3s examinaremos ambas as alternativas para que voc\u00ea possa escolher de acordo com suas prefer\u00eancias e necessidades o caminho a seguir.<\/p>\n<h3>Transformando para a equa\u00e7\u00e3o geral<\/h3>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=C6DbrJDiZTM&amp;t=475s\" target=\"_blank\" rel=\"noopener\"><strong>A transforma\u00e7\u00e3o para a forma geral<\/strong><\/a> \u00e9 feita atrav\u00e9s do seguinte racioc\u00ednio, onde <span class=\"katex-eq\" data-katex-display=\"false\">a,b,c\\in\\mathbb{R}<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">a\\neq 0.<\/span>\n<table style=\"text-align: justify;\">\n<tbody>\n<tr>\n<td width=\"50\">(1)<\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">y=ax^2 + bx + c<\/span><\/td>\n<td>; Equa\u00e7\u00e3o can\u00f4nica das par\u00e1bolas<\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">y=a\\left[x^2 + \\dfrac{b}{a}x + \\dfrac{c}{a}\\right]<\/span><\/td>\n<td>; Fatorando por <span class=\"katex-eq\" data-katex-display=\"false\">a<\/span><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">y=a\\left[ \\left(x + \\dfrac{b}{2a}\\right)^2 - \\dfrac{b^2}{4a^2} + \\dfrac{c}{a}\\right]<\/span><\/td>\n<td>; Porque <span class=\"katex-eq\" data-katex-display=\"false\">\\left(x + \\dfrac{b}{2a}\\right)^2 = x^2 + \\dfrac{b}{a}x + \\dfrac{b^2}{4a^2}<\/span><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">y=a\\left[ \\left(x + \\dfrac{b}{2a}\\right)^2 + \\dfrac{4ac - b^2}{4a^2} \\right]<\/span><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">y=a \\left(x + \\dfrac{b}{2a}\\right)^2 + \\dfrac{4ac - b^2}{4a}<\/span><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\">y=a \\left(x + \\dfrac{b}{2a}\\right)^2 + \\left(c - \\dfrac{b^2}{4a}\\right)<\/span><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\"> \\left[x - \\left(- \\dfrac{b}{2a}\\right)\\right]^2 = \\dfrac{1}{a} \\left[y - \\left(c - \\dfrac{b^2}{4a}\\right)\\right]<\/span><\/td>\n<td><\/td>\n<\/tr>\n<tr>\n<td><\/td>\n<td><span class=\"katex-eq\" data-katex-display=\"false\"> \\left[x - \\left( -\\dfrac{b}{2a}\\right)\\right]^2 = 4\\left(\\dfrac{1}{4a}\\right) \\left[y - \\left(c - \\dfrac{b^2}{4a}\\right)\\right]<\/span><\/td>\n<td>; Equa\u00e7\u00e3o das par\u00e1bolas em forma geral<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p style=\"text-align: justify;\">A partir disso, podemos extrair todas as informa\u00e7\u00f5es que t\u00ednhamos da equa\u00e7\u00e3o geral relacionando seus par\u00e2metros com os da equa\u00e7\u00e3o can\u00f4nica. Assim, temos:<\/p>\n<ul style=\"text-align: justify;\">\n<li><strong>V\u00e9rtice:<\/strong> O ponto de coordenadas <span class=\"katex-eq\" data-katex-display=\"false\">(x_0,y_0) = \\left(-\\dfrac{b}{2a}, c -\\dfrac{b^2}{4a} \\right)<\/span><\/li>\n<li><strong>Posi\u00e7\u00e3o focal:<\/strong> a <span class=\"katex-eq\" data-katex-display=\"false\">f = \\dfrac{1}{4a}<\/span> unidades acima do v\u00e9rtice<\/li>\n<li><strong>Foco:<\/strong> O ponto de coordenadas <span class=\"katex-eq\" data-katex-display=\"false\">(x_0,y_0 + f) = \\left(-\\dfrac{b}{2a}, c -\\dfrac{b^2}{4a} + \\dfrac{1}{4a}\\right) = \\left(-\\dfrac{b}{2a}, c +\\dfrac{1-b^2}{4a}\\right) <\/span><\/li>\n<li><strong>Diretriz:<\/strong> A reta de equa\u00e7\u00e3o <span class=\"katex-eq\" data-katex-display=\"false\">y = y_0 - f = c -\\dfrac{b^2}{4a} - \\dfrac{1}{4a} = c -\\dfrac{1 + b^2}{4a}<\/span><\/li>\n<li><strong>Eixo de simetria:<\/strong> A reta de equa\u00e7\u00e3o <span class=\"katex-eq\" data-katex-display=\"false\">x = x_0 = -\\dfrac{b}{2a}<\/span><\/li>\n<\/ul>\n<p style=\"text-align: justify;\">E a partir daqui, a caracteriza\u00e7\u00e3o das par\u00e1bolas \u00e9 feita como j\u00e1 vimos, usando a equa\u00e7\u00e3o geral.<\/p>\n<h3>Usando a simetria e as interse\u00e7\u00f5es com o eixo x<\/h3>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=C6DbrJDiZTM&amp;t=769s\" target=\"_blank\" rel=\"noopener\"><strong>Quando temos a equa\u00e7\u00e3o<\/strong><\/a> das par\u00e1bolas escrita na forma can\u00f4nica <span class=\"katex-eq\" data-katex-display=\"false\">y=ax^2 + bx + c<\/span>, vemos que \u00e9 relativamente simples calcular as interse\u00e7\u00f5es com o eixo x. Basta resolver a equa\u00e7\u00e3o<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">ax^2 + bx + c = 0<\/span>\n<p style=\"text-align: justify;\">Quando isso \u00e9 poss\u00edvel, obtemos as interse\u00e7\u00f5es <span class=\"katex-eq\" data-katex-display=\"false\">x_1<\/span> e <span class=\"katex-eq\" data-katex-display=\"false\">x_2<\/span> dadas por<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">x_1 = \\dfrac{-b + \\sqrt{b^2-4ac}}{2a}<\/span>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">x_2 = \\dfrac{-b - \\sqrt{b^2-4ac}}{2a}<\/span>\n<p style=\"text-align: justify;\">Como as par\u00e1bolas s\u00e3o sim\u00e9tricas, teremos que o eixo de simetria ter\u00e1 equa\u00e7\u00e3o:<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">x = x_0 = \\dfrac{x_1 + x_2}{2}= -\\dfrac{b}{2a}<\/span>\n<p style=\"text-align: justify;\">O eixo de simetria passa necessariamente pelo v\u00e9rtice da par\u00e1bola, cujas coordenadas ser\u00e3o<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">(x_0, y_0) = (x_0, y(x_0)) = \\left( -\\dfrac{b}{2a}, y\\left(-\\dfrac{b}{2a}\\right) \\right)<\/span>\n<p style=\"text-align: justify;\">Onde<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">y_0 = y\\left(-\\dfrac{b}{2a} \\right) = a\\left(-\\dfrac{b}{2a}\\right)^2 + b\\left(-\\dfrac{b}{2a}\\right) + c = \\dfrac{b^2}{4a} - \\dfrac{b^2}{2a} + c = c - \\dfrac{b^2}{4a}<\/span>\n<p style=\"text-align: justify;\">Assim \u00e9 como chegamos \u00e0s coordenadas do v\u00e9rtice que j\u00e1 conhec\u00edamos por outros meios<\/p>\n<p style=\"text-align: center;\"><span class=\"katex-eq\" data-katex-display=\"false\">(x_0, y_0) = \\left( -\\dfrac{b}{2a},c - \\dfrac{b^2}{4a} \\right)<\/span>\n<p style=\"text-align: justify;\">A posi\u00e7\u00e3o focal \u00e9, como j\u00e1 vimos, <span class=\"katex-eq\" data-katex-display=\"false\">f = \\dfrac{1}{4a},<\/span> e a partir disso j\u00e1 podemos calcular a posi\u00e7\u00e3o da diretriz, do foco e todas as informa\u00e7\u00f5es que j\u00e1 t\u00ednhamos desde a equa\u00e7\u00e3o geral.<\/p>\n<p>   <a name=\"4\"><\/a><\/p>\n<h2>Caracteriza\u00e7\u00e3o autom\u00e1tica com Excel<\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.youtube.com\/watch?v=C6DbrJDiZTM&amp;t=1086s\" target=\"_blank\" rel=\"noopener\"><strong>Tendo realizado<\/strong><\/a> todos esses racioc\u00ednios, agora \u00e9 muito simples automatizar a caracteriza\u00e7\u00e3o de qualquer par\u00e1bola atrav\u00e9s do Excel. Voc\u00ea pode encontrar um exemplo <a href=\"https:\/\/drive.google.com\/file\/d\/1LbNOKHHfzlPgHI3_NSzuB_6_b7KmXlTd\/view?usp=sharing\" rel=\"noopener\" target=\"_blank\">aqui.<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Caracteriza\u00e7\u00e3o de Par\u00e1bolas e seus Gr\u00e1ficos Resumo: Nesta aula revisaremos a caracteriza\u00e7\u00e3o das par\u00e1bolas a partir da sua equa\u00e7\u00e3o geral e forma can\u00f4nica, explicando como identificar elementos chave como o v\u00e9rtice, o foco, a diretriz, o eixo de simetria e as poss\u00edveis interse\u00e7\u00f5es com o eixo X. Objetivos de Aprendizagem: Ao final desta aula, o [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":28929,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":6,"footnotes":""},"categories":[587,571],"tags":[],"class_list":["post-28933","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-algebra-e-geometria","category-matematica-pt"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Caracteriza\u00e7\u00e3o de Par\u00e1bolas e seus Gr\u00e1ficos - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Aprenda a caracterizar par\u00e1bolas identificando v\u00e9rtice, foco, diretriz, eixo de simetria e interse\u00e7\u00f5es com o eixo x.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"http:\/\/toposuranos.com\/material\/pt\/caracterizacao-de-parabolas-e-seus-graficos\/\" \/>\n<meta property=\"og:locale\" content=\"es_ES\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Caracteriza\u00e7\u00e3o de Par\u00e1bolas e seus Gr\u00e1ficos\" \/>\n<meta property=\"og:description\" content=\"Aprenda a caracterizar par\u00e1bolas identificando v\u00e9rtice, foco, diretriz, eixo de simetria e interse\u00e7\u00f5es com o eixo x.\" \/>\n<meta property=\"og:url\" content=\"http:\/\/toposuranos.com\/material\/pt\/caracterizacao-de-parabolas-e-seus-graficos\/\" \/>\n<meta property=\"og:site_name\" content=\"toposuranos.com\/material\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/groups\/toposuranos\" \/>\n<meta property=\"article:published_time\" content=\"2021-04-27T13:00:20+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2024-09-22T02:15:10+00:00\" \/>\n<meta property=\"og:image\" content=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/focodirectriz-1024x467.png\" \/>\n<meta name=\"author\" content=\"giorgio.reveco\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:title\" content=\"Caracteriza\u00e7\u00e3o de Par\u00e1bolas e seus Gr\u00e1ficos\" \/>\n<meta name=\"twitter:description\" content=\"Aprenda a caracterizar par\u00e1bolas identificando v\u00e9rtice, foco, diretriz, eixo de simetria e interse\u00e7\u00f5es com o eixo x.\" \/>\n<meta name=\"twitter:image\" content=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/focodirectriz.png\" \/>\n<meta name=\"twitter:creator\" content=\"@topuranos\" \/>\n<meta name=\"twitter:site\" content=\"@topuranos\" \/>\n<meta name=\"twitter:label1\" content=\"Escrito por\" \/>\n\t<meta name=\"twitter:data1\" content=\"giorgio.reveco\" \/>\n\t<meta name=\"twitter:label2\" content=\"Tiempo de lectura\" \/>\n\t<meta name=\"twitter:data2\" content=\"5 minutos\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"http:\/\/toposuranos.com\/material\/pt\/caracterizacao-de-parabolas-e-seus-graficos\/#article\",\"isPartOf\":{\"@id\":\"http:\/\/toposuranos.com\/material\/pt\/caracterizacao-de-parabolas-e-seus-graficos\/\"},\"author\":{\"name\":\"giorgio.reveco\",\"@id\":\"http:\/\/toposuranos.com\/material\/#\/schema\/person\/e15164361c3f9a2a02cf6c234cf7fdc1\"},\"headline\":\"Caracteriza\u00e7\u00e3o de Par\u00e1bolas e seus Gr\u00e1ficos\",\"datePublished\":\"2021-04-27T13:00:20+00:00\",\"dateModified\":\"2024-09-22T02:15:10+00:00\",\"mainEntityOfPage\":{\"@id\":\"http:\/\/toposuranos.com\/material\/pt\/caracterizacao-de-parabolas-e-seus-graficos\/\"},\"wordCount\":1297,\"commentCount\":0,\"publisher\":{\"@id\":\"http:\/\/toposuranos.com\/material\/#organization\"},\"image\":{\"@id\":\"http:\/\/toposuranos.com\/material\/pt\/caracterizacao-de-parabolas-e-seus-graficos\/#primaryimage\"},\"thumbnailUrl\":\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2024\/09\/focodirectriz.png\",\"articleSection\":[\"\u00c1lgebra e Geometria\",\"Matem\u00e1tica\"],\"inLanguage\":\"es\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"http:\/\/toposuranos.com\/material\/pt\/caracterizacao-de-parabolas-e-seus-graficos\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"http:\/\/toposuranos.com\/material\/pt\/caracterizacao-de-parabolas-e-seus-graficos\/\",\"url\":\"http:\/\/toposuranos.com\/material\/pt\/caracterizacao-de-parabolas-e-seus-graficos\/\",\"name\":\"Caracteriza\u00e7\u00e3o de Par\u00e1bolas e seus Gr\u00e1ficos - 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