切片 x^4

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

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

simplify solve for inverse tangent line

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準線 y=(x+2)²

Solution

y=−14

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放物線の準線

放物線は複数の点の軌跡であり,1つの点（焦点）までの距離は1つの線（準線）までの距離に等しい

放物線の標準的なequation

4p(y−k)=(x−h)² は上下に面する放物線の標準的なequationで,頂点は (h, k),
焦点距離は |p| である

y=(x+2)² を標準的な形式で書き換える: 4· 14 (y−0)=(x−(−2))²

(h, k)=(−2, 0), p=14

放物線は y 軸に対して対称で,準線は x 軸に平行で中心 (−2, 0) の y 座標から距離 −p にある

y=0−p

y=0−14

改良

y=−14

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

切片 x^4

Solution

Solution steps

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