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directrix y=4x^2+1

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
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f(x)=(x−3)(x+1)

Solution

Domain: −∞ <x<∞

Range: f(x)≥ −4

X 가로채기: (3, 0), (−1, 0), Y 가로채기: (0, −3)

Vertex: 극히 적은 양 (1, −4)

Inverse: 1+√x+4, 1−√x+4

Interval Notation

Domain: (−∞ , ∞ )

Range: [−4, ∞ )

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도메인 (x−3)(x+1) : −∞ <x<∞

범위 (x−3)(x+1): f(x)≥ −4

축 가로채기 포인트 (x−3)(x+1): X 가로채기: (3, 0), (−1, 0), Y 가로채기: (0, −3)

꼭짓점 (x−3)(x+1): 극히 적은 양 (1, −4)

의 역 (x−3)(x+1): 1+√x+4, 1−√x+4

Solve instead

ddx ((x−3)(x+1))

동등성 (x−3)(x+1)

대칭 (x−3)(x+1)

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

directrix y=4x^2+1

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