範囲 1/(x^3)

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

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頂点 y=−3x²−2x+65

Solution

最大値 (−13 , 1915 )

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多項式形式を使用して頂点を求める

放物線形式を使用して頂点を求める

頂点形式を使用して頂点を求める

ゼロの平均化を使用して頂点を求める

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y=−3x²−2x+65

多項式形式の放物線equation

y=ax²+bx+c 形式の上下に面する放物線の頂点は x_v=−b2a

以下の形式で y=−3x²−2x+65 を書き換える y=ax²+bx+c

y=−3x²5 −2x5 +65

放物線の媒介変数は:

a=−35 , b=−25 , c=65

x_v=−b2a

x_v=−(−25 )2(−35 )

簡素化 −−25 2(−35 ) : −13

x_v=−13

x_v=−13 を y_v 値に当てはめる

y_v=1915

ゆえに放物線の頂点は

(−13 , 1915 )

a<0 ならば,頂点は最大値
a>0 ならば,頂点は最小値
a=−35

最大値 (−13 , 1915 )

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こんにちは, について質問する頂点 y=−3x²−2x+65 または以下のオプションから選択してください

解答者：放物線形式を使用して頂点を求める

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
▭ \|▭	)	.	0	=	+	`y`

範囲 1/(x^3)

Solution

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