{"id":21,"date":"2025-09-24T17:08:52","date_gmt":"2025-09-24T17:08:52","guid":{"rendered":"https:\/\/hainn.id.vn\/?p=21"},"modified":"2026-02-04T04:42:03","modified_gmt":"2026-02-04T04:42:03","slug":"tim-dinh-cua-parabol-ham-bac-2","status":"publish","type":"post","link":"https:\/\/hainn.id.vn\/index.php\/2025\/09\/24\/tim-dinh-cua-parabol-ham-bac-2\/","title":{"rendered":"T\u00ecm \u0111\u1ec9nh c\u1ee7a parabol h\u00e0m b\u1eadc 2"},"content":{"rendered":"\n

M\u00ecnh ch\u1ee3t nh\u1eadn ra l\u00e0 trong c\u00e1c parabol c\u1ee7a h\u00e0m b\u1eadc 2 v\u1edbi d\u1ea1ng t\u1ed5ng qu\u00e1t f(x) = Ax^2+Bx+C (A\\neq0)<\/span>, n\u1ebfu A\\ge0<\/span> th\u00ec parabol c\u00f3 d\u1ea1ng “U th\u01b0\u1eddng” – \u0111\u00e1y \u1edf ph\u00eda d\u01b0\u1edbi, t\u1ee9c c\u00f3 c\u1ef1c ti\u1ec3u. C\u00f2n n\u1ebfu A<0<\/span> th\u00ec parabol c\u00f3 d\u1ea1ng “U l\u1ed9n ng\u01b0\u1ee3c” – \u0111\u1ec9nh \u1edf ph\u00eda tr\u00ean, t\u1ee9c c\u00f3 c\u1ef1c \u0111\u1ea1i.<\/p>\n\n\n\n

X\u00e9t th\u1ea5y \u0111\u00e2y l\u00e0 m\u1ed9t t\u00ednh ch\u1ea5t kh\u00e1 th\u00fa v\u1ecb n\u00ean trong b\u00e0i n\u00e0y, m\u00ecnh s\u1ebd tr\u00ecnh b\u00e0y chi ti\u1ebft c\u00e1ch ch\u1ee9ng minh t\u00ednh ch\u1ea5t n\u00e0y.<\/p>\n\n\n\n

X\u00e9t h\u00e0m b\u1eadc 2:<\/p>\n\n\n\n

f(x) = Ax^2+Bx+C (A\\neq0)<\/pre><\/div>\n\n\n\n
Ta c\u00f3 ph\u01b0\u01a1ng tr\u00ecnh t\u01b0\u01a1ng \u0111\u01b0\u01a1ng:<\/p>\n\n\n\n
f(x)=A\\left(x^2+\\frac{B}{A}x\\right)+C\\, \\, (1)<\/pre><\/div>\n\n\n\n
X\u00e9t bi\u1ec3u th\u1ee9c trong ngo\u1eb7c:<\/p>\n\n\n\n
x^2+\\frac{B}{A}x \\\\\n\\begin{aligned}\n&= x^2 + 2x\\frac{B}{2A}+\\frac{B^2}{4A^2}-\\frac{B^2}{4A^2} \\\\\n&= \\left(x+\\frac{B}{2A}x\\right)^2-\\left(\\frac{B}{2A}\\right)^2\n\\end{aligned}<\/pre><\/div>\n\n\n\n
Thay v\u00e0o (1) ta c\u00f3:<\/p>\n\n\n\n
\\begin{aligned}\nf(x)&=A\\left(x^2+\\frac{B}{A}x\\right)+C\\\\\n\\Leftrightarrow f(x)&=A\\left[ \\left(x+\\frac{B}{2A}x\\right)^2-\\left(\\frac{B}{2A}\\right)^2 \\right] + C \\\\\n\\Leftrightarrow f(x)&=A\\left(x+\\frac{B}{2A}x\\right)^2 - \\frac{B^2}{4A} + C \\\\\n\\Leftrightarrow f(x)&=A\\left(x+\\frac{B}{2A}x\\right)^2 - \\frac{B^2-4AC}{4A} \\\\\n\\end{aligned}<\/pre><\/div>\n\n\n\n
V\u00ec \\frac{B^2-4AC}{4A}<\/span> l\u00e0 m\u1ed9t h\u1eb1ng s\u1ed1 kh\u00f4ng \u0111\u1ed5i (tr\u00ean th\u1ef1c t\u1ebf, c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh \u0111\u00e3 cho bi\u1ebft tr\u01b0\u1edbc gi\u00e1 tr\u1ecb c\u1ee7a A, B, C) n\u00ean \u0111\u1ed9 l\u1edbn\/ nh\u1ecf c\u1ee7a h\u00e0m f(x) ph\u1ee5 thu\u1ed9c v\u00e0o v\u1ebf A\\left(x+\\frac{B}{2A}x\\right)^2<\/span>.<\/p>\n\n\n\n
N\u1ebfu A>0 th\u00ec A\\left(x+\\frac{B}{2A}x\\right)^2<\/span> s\u1ebd c\u00f3 c\u1ef1c ti\u1ec3u t\u1ea1i 0. C\u1ee5 th\u1ec3:<\/p>\n\n\n\n
A\\left(x+\\frac{B}{2A}\\right)^2\\geq0\n\\rightarrow Min\\left(x+\\frac{B}{2A}\\right)=0\\, t\u1ea1i\\, x=\\frac{-B}{2A}<\/pre><\/div>\n\n\n\n
Ng\u01b0\u1ee3c l\u1ea1i, n\u1ebfu A<0 th\u00ec A\\left(x^2+\\frac{B}{2A}x\\right)^2<\/span> s\u1ebd c\u00f3 c\u1ef1c \u0111\u1ea1i t\u1ea1i 0. C\u1ee5 th\u1ec3:<\/p>\n\n\n\n
A\\left(x+\\frac{B}{2A}\\right)^2\\leq0\n\\rightarrow Max\\left(x+\\frac{B}{2A}\\right)=0\\, t\u1ea1i\\, x=\\frac{-B}{2A}<\/pre><\/div>\n\n\n\n
V\u00e0 \u0111i\u1ec3m \u0111\u1eb7c bi\u1ec7t l\u00e0 d\u00f9 c\u1ef1c \u0111\u1ea1i hay c\u1ef1c ti\u1ec3u, th\u00ec y=-\\frac{B^2-4AC}{4A}<\/span> t\u1ea1i x=\\frac{-B}{2A}<\/span>.<\/p>\n\n\n\n
\"\"<\/figure>
\n
\u0110\u1ed3 th\u1ecb h\u00e0m s\u1ed1 y=3x^2+1<\/span> (\u0111\u1ecf) v\u00e0 y'=-3x^2+1<\/span> (xanh) v\u1edbi \u0111i\u1ec3m c\u1ef1c \u0111\u1ea1i\/ c\u1ef1c ti\u1ec3u chung 1 t\u1ecda \u0111\u1ed9 (0,1).<\/p>\n<\/div><\/div>\n\n\n\n

<\/p>\n","protected":false},"excerpt":{"rendered":"
M\u00ecnh ch\u1ee3t nh\u1eadn ra l\u00e0 trong c\u00e1c parabol c\u1ee7a h\u00e0m b\u1eadc 2 v\u1edbi d\u1ea1ng t\u1ed5ng qu\u00e1t , n\u1ebfu th\u00ec parabol c\u00f3 d\u1ea1ng “U th\u01b0\u1eddng” – \u0111\u00e1y \u1edf ph\u00eda d\u01b0\u1edbi, t\u1ee9c c\u00f3 c\u1ef1c ti\u1ec3u. C\u00f2n n\u1ebfu th\u00ec parabol c\u00f3 d\u1ea1ng “U l\u1ed9n ng\u01b0\u1ee3c” – \u0111\u1ec9nh \u1edf ph\u00eda tr\u00ean, t\u1ee9c c\u00f3 c\u1ef1c \u0111\u1ea1i. X\u00e9t th\u1ea5y \u0111\u00e2y l\u00e0 […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[12,11,13,10,9],"class_list":["post-21","post","type-post","status-publish","format-standard","hentry","category-toan","tag-cuc-tieu","tag-cuc-tri","tag-do-thi-ham-so","tag-ham-bac-2","tag-parabol"],"_links":{"self":[{"href":"https:\/\/hainn.id.vn\/index.php\/wp-json\/wp\/v2\/posts\/21","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/hainn.id.vn\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/hainn.id.vn\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/hainn.id.vn\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/hainn.id.vn\/index.php\/wp-json\/wp\/v2\/comments?post=21"}],"version-history":[{"count":13,"href":"https:\/\/hainn.id.vn\/index.php\/wp-json\/wp\/v2\/posts\/21\/revisions"}],"predecessor-version":[{"id":39,"href":"https:\/\/hainn.id.vn\/index.php\/wp-json\/wp\/v2\/posts\/21\/revisions\/39"}],"wp:attachment":[{"href":"https:\/\/hainn.id.vn\/index.php\/wp-json\/wp\/v2\/media?parent=21"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/hainn.id.vn\/index.php\/wp-json\/wp\/v2\/categories?post=21"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/hainn.id.vn\/index.php\/wp-json\/wp\/v2\/tags?post=21"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}