{"id":8088,"date":"2021-11-13T21:36:39","date_gmt":"2021-11-13T16:06:39","guid":{"rendered":"https:\/\/mathemerize.com\/?p=8088"},"modified":"2021-11-14T00:11:44","modified_gmt":"2021-11-13T18:41:44","slug":"find-the-equation-of-the-tangent-to-curve-y-5x2-6x-7-at-the-point-1-2-35-4","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/find-the-equation-of-the-tangent-to-curve-y-5x2-6x-7-at-the-point-1-2-35-4\/","title":{"rendered":"Find the equation of the tangent to curve y = \\(-5x^2 + 6x + 7\\)\u00a0 at the point (1\/2, 35\/4)."},"content":{"rendered":"<h2>Solution :<\/h2>\n<p>The equation of the given curve is<\/p>\n<p>y = \\(-5x^2 + 6x + 7\\)<\/p>\n<p>\\(\\implies\\) \\(dy\\over dx\\) = -10x + 6<\/p>\n<p>\\(\\implies\\) \\(({dy\\over dx})_{(1\/2, 35\/4)}\\) = \\(-10\\over 4\\) + 6 = 1<\/p>\n<p>The required equation at (1\/2, 35\/4) is<\/p>\n<p>y &#8211; \\(35\\over 4\\) = \\(({dy\\over dx})_{(1\/2, 35\/4)}\\) \\((x &#8211; {1\\over 2})\\)<\/p>\n<p>\\(\\implies\\) y &#8211; 35\/4 = 1(x &#8211; 1\/2)<\/p>\n<p>\\(\\implies\\) Equation of tangent is y = x + 33\/4<\/p>\n<hr \/>\n<h3 style=\"text-align: center;\">Similar Questions<\/h3>\n<p><a class=\"row-title\" href=\"https:\/\/mathemerize.com\/find-the-equations-of-the-tangent-and-the-normal-at-the-point-t-on-the-curve-x-a-sin3-t-y-b-cos3-t\/\" aria-label=\"\u201cFind the equations of the tangent and the normal at the point \u2018t\u2019 on the curve x = \\(a sin^3 t\\), y = \\(b cos^3 t\\).\u201d (Edit)\">Find the equations of the tangent and the normal at the point \u2018t\u2019 on the curve x = \\(a sin^3 t\\), y = \\(b cos^3 t\\).<\/a><\/p>\n<p><a class=\"row-title\" href=\"https:\/\/mathemerize.com\/find-the-equation-of-the-normal-to-the-curve-y-2x2-3-sin-x-at-x-0\/\" aria-label=\"\u201cFind the equation of the normal to the curve y = \\(2x^2 + 3 sin x\\) at x = 0.\u201d (Edit)\">Find the equation of the normal to the curve y = \\(2x^2 + 3 sin x\\) at x = 0.<\/a><\/p>\n<p><a class=\"row-title\" href=\"https:\/\/mathemerize.com\/find-the-angle-between-the-curves-xy-6-and-x2-y-12\/\" aria-label=\"\u201cFind the angle between the curves xy = 6 and \\(x^2 y\\) =12.\u201d (Edit)\">Find the angle between the curves xy = 6 and \\(x^2 y\\) =12.<\/a><\/p>\n<p><a class=\"row-title\" href=\"https:\/\/mathemerize.com\/check-the-orthogonality-of-the-curves-y2-x-and-x2-y\/\" aria-label=\"\u201cCheck the orthogonality of the curves \\(y^2\\) = x and \\(x^2\\) = y.\u201d (Edit)\">Check the orthogonality of the curves \\(y^2\\) = x and \\(x^2\\) = y.<\/a><\/p>\n<p><a class=\"row-title\" href=\"https:\/\/mathemerize.com\/the-angle-of-intersection-between-the-curve-x2-32y-and-y2-4x-at-point-16-8-is\/\" aria-label=\"\u201cThe angle of intersection between the curve \\(x^2\\) = 32y and \\(y^2\\) = 4x at point (16, 8) is\u201d (Edit)\">The angle of intersection between the curve \\(x^2\\) = 32y and \\(y^2\\) = 4x at point (16, 8) is<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Solution : The equation of the given curve is y = \\(-5x^2 + 6x + 7\\) \\(\\implies\\) \\(dy\\over dx\\) = -10x + 6 \\(\\implies\\) \\(({dy\\over dx})_{(1\/2, 35\/4)}\\) = \\(-10\\over 4\\) + 6 = 1 The required equation at (1\/2, 35\/4) is y &#8211; \\(35\\over 4\\) = \\(({dy\\over dx})_{(1\/2, 35\/4)}\\) \\((x &#8211; {1\\over 2})\\) \\(\\implies\\) y &hellip;<\/p>\n<p class=\"read-more\"> <a class=\"\" href=\"https:\/\/mathemerize.com\/find-the-equation-of-the-tangent-to-curve-y-5x2-6x-7-at-the-point-1-2-35-4\/\"> <span class=\"screen-reader-text\">Find the equation of the tangent to curve y = \\(-5x^2 + 6x + 7\\)\u00a0 at the point (1\/2, 35\/4).<\/span> Read More &raquo;<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","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":""},"categories":[66,43],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v20.12 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Find the equation of the tangent to curve y = \\(-5x^2 + 6x + 7\\)\u00a0 at the point (1\/2, 35\/4).<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/find-the-equation-of-the-tangent-to-curve-y-5x2-6x-7-at-the-point-1-2-35-4\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Find the equation of the tangent to curve y = \\(-5x^2 + 6x + 7\\)\u00a0 at the point (1\/2, 35\/4).\" \/>\n<meta property=\"og:description\" content=\"Solution : The equation of the given curve is y = (-5x^2 + 6x + 7) (implies) (dyover dx) = -10x + 6 (implies) (({dyover dx})_{(1\/2, 35\/4)}) = (-10over 4) + 6 = 1 The required equation at (1\/2, 35\/4) is y &#8211; (35over 4) = (({dyover dx})_{(1\/2, 35\/4)}) ((x &#8211; {1over 2})) (implies) y &hellip; Find the equation of the tangent to curve y = (-5x^2 + 6x + 7)\u00a0 at the point (1\/2, 35\/4). 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