漸近線 t^2

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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 25−x²

Solution

最大値 (0, 25)

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Solve by:

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

One step at a time

y=25−x²

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=25−x² を書き換える y=ax²+bx+c

y=−x²+25

放物線の媒介変数は:

a=−1, b=0, c=25

x_v=−b2a

x_v=−02(−1)

簡素化

x_v=0

x_v=0 を y_v 値に当てはめる

y_v=25

ゆえに放物線の頂点は

(0, 25)

a<0 ならば,頂点は最大値
a>0 ならば,頂点は最小値
a=−1

最大値 (0, 25)

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こんにちは, について質問する頂点 25−x² 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

解答者：頂点形式を使用して頂点を求める

解答者：ゼロの平均化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`

漸近線 t^2

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