https://mathmanstudio.com/categories/%E5%A4%9A%E9%A0%85%E5%BC%8F%E5%87%BD%E6%95%B8/Recent content in 多項式函數 on MathManStudio.comHugozh-TWWed, 15 Jul 2026 19:42:56 +0800https://mathmanstudio.com/posts/114%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E5%AD%B8%E6%B8%AC%E6%95%B8a%E5%A4%9A%E9%81%B813/Wed, 15 Jul 2026 19:42:56 +0800https://mathmanstudio.com/posts/114%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E5%AD%B8%E6%B8%AC%E6%95%B8a%E5%A4%9A%E9%81%B813/<p><img loading="lazy" src="https://mathmanstudio.com/wp-content/uploads/2026/07/%E6%88%AA%E5%9C%96-2026-07-15-%E6%99%9A%E4%B8%8A7.30.00-1-1024x167.png"></p> <p>因為f(x)為三次多項式，可知q(x)為二次多項式，先把q(x)假設出來</p> <p>$$ q(x)=a(x+6)^2+8\quad(a&lt;0) $$ $$ 又f(x)=(x+6)q(x)+3 $$ $$ f(x)=(x+6)[a(x+6)^2+8]+3 $$ $$ f(x)=a(x+6)^3+8(x+6)+3 $$ $$ 對稱中心（-6,3) $$</p>https://mathmanstudio.com/posts/114%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E5%AD%B8%E6%B8%AC%E6%95%B8a%E5%A4%9A%E9%81%B89/Sat, 04 Jul 2026 17:32:13 +0800https://mathmanstudio.com/posts/114%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E5%AD%B8%E6%B8%AC%E6%95%B8a%E5%A4%9A%E9%81%B89/<p><img loading="lazy" src="https://mathmanstudio.com/wp-content/uploads/2026/07/%E6%88%AA%E5%9C%96-2026-07-02-%E6%99%9A%E4%B8%8A8.40.44-1024x333.png"> $$ 可得b^2-4c\ge0與(b+2)^2-4c&lt;0 $$ $$ 合併可得b^2\ge4c&gt;(b+2)^2 $$</p> <p>(1)由</p> <p>$$ 4c&gt;(b+2)^2 $$</p> <p>可得</p> <p>$$ c&gt;0 $$</p> <p>(2)整理一下</p> <p>$$ b^2&gt;(b+2)^2 $$</p> <p>可得</p> <p>$$ b&lt;-1 $$</p> <p>(3)整理</p> <p>$$ (b+1)^2-4c＝b^2-4c+2b+1其中(b^2-4c\ge0\quad b&lt;-1)無法確定正負 $$</p> <p>可得</p> <p>(4)</p> <p>$$ (b+2)^2+4c&gt;0 $$</p> <p>有兩相異實根。</p> <p>（5）</p> <p>$$ (b-2)^2-4c=b^2-4c-4b+4&gt;0 $$</p> <p>有兩相異實根。</p> <p>正解：（2）（4）（5）</p>https://mathmanstudio.com/posts/113%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E5%88%86%E7%A7%91%E6%B8%AC%E9%A9%97%E7%B7%B4%E5%8A%9F%E5%9D%8A-%E5%A4%9A%E9%A0%85%E5%BC%8F%E5%87%BD%E6%95%B8%E5%80%92%E6%95%B83%E5%A4%A9%E4%B8%8A%E6%88%B0/Tue, 09 Jul 2024 19:17:32 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