Question

19£®ÔÚÖ±½Ç×ø±êϵÖУ¬Èç¹û¶þ´Îº¯Êýy=ax²+bx+2£¨a¡Ù0£©µÄͼÏóÓëxÖá½»ÓÚA¡¢BÁ½µã£¬ÓëyÖá½»ÓÚµãC£¨0£¬2£©£¬ÇÒAB=OC£¬ÄÇôÎÒÃdzÆÕâ¸ö¶þ´Îº¯ÊýΪºÍºÏ¶þ´Îº¯Êý£®
£¨1£©Åж϶þ´Îº¯Êýy=x²+x+2ºÍy=$\frac{2}{3}$x²+$\frac{8}{3}$x+2ÊÇ·ñΪºÍºÏ¶þ´Îº¯Êý£¬²¢ËµÃ÷ÀíÓÉ£»
£¨2£©ºÍºÏ¶þ´Îº¯Êýy=ax²+bx+2µÄͼÏó¾­¹ýµã£¨-6£¬2£©£®
¢ÙÇóaÓëbµÄÖµ£»
¢Ú´Ëº¯ÊýͼÏó¿ÉÓÉÅ×ÎïÏßy=ax²¾­¹ýÔõÑùµÄƽÒƵõ½£¿

Answer 1

·ÖÎö £¨1£©¸ù¾ÝºÍºÏ¶þ´Îº¯ÊýµÄ¶¨ÒåÅжϼ´¿É£»
£¨2£©¢Ù¸ù¾ÝÅ×ÎïÏßÓëxÖáµÄ½»µãµÄÇ󷨺ͺͺ϶þ´Îº¯ÊýµÄ¶¨ÒåÁгö·½³Ì×éÇó³öa¡¢bµÄÖµ¼´¿É£»
¢Ú¸ù¾ÝÅ×ÎïÏßµÄƽÒƹæÂÉ×÷´ð£®

½â´ð ½â£º£¨1£©y=x²+x+2ÓëxÖáûÓн»µã£¬ËùÒÔ²»ÊǺͺ϶þ´Îº¯Êý£»
y=$\frac{2}{3}$x²+$\frac{8}{3}$x+2ÓëxÖáµÄ½»µã×ø±êΪA£¨-4£¬0£©£¬B£¨-2£¬0£©£¬
AB=2£¬
¡àAB=OC£¬
¡ày=$\frac{2}{3}$x²+$\frac{8}{3}$x+2ÊǺͺ϶þ´Îº¯Êý£»
£¨2£©¢Ùy=ax²+bx+2ÓëxÖá½»µãµÄºá×ø±êΪx₁£¬x₂£¬
Ôòx₁+x₂=-$\frac{b}{a}$£¬x₁•x₂=$\frac{2}{a}$£¬
ÓÉÌâÒâµÃ£¬
$\left\{\begin{array}{l}{\sqrt{\frac{{b}^{2}}{{a}^{2}}-\frac{8}{a}}=2}\\{36a-6b+2=2}\end{array}\right.$£¬
½âµÃa=$\frac{1}{4}$£¬b=$\frac{3}{2}$£»
¢Úy=$\frac{1}{4}$x²-$\frac{3}{2}$x+2
=$\frac{1}{4}$£¨x-3£©²-$\frac{1}{4}$£¬
Å×ÎïÏßy=$\frac{1}{4}$x²ÏòÓÒƽÒÆ3¸öµ¥Î»£¬ÔÙÏòÏÂƽÒÆ$\frac{1}{4}$¸öµ¥Î»µÃµ½y=$\frac{1}{4}$£¨x-3£©²-$\frac{1}{4}$£®

µãÆÀ ±¾Ì⿼²éµÄÊǺ¯ÊýµÄж¨Ò壬ÕÆÎÕÅ×ÎïÏßÓëxÖáµÄ½»µã¡¢¶þ´Îº¯ÊýµÄÐÔÖʺͶþ´Îº¯ÊýµÄƽÒƹæÂÉÊǽâÌâµÄ¹Ø¼ü£®