vertices 2x^2-2

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

☐²

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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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f(x)=(x+3)(x−3)

Solution

Domain: −∞ <x<∞

Range: f(x)≥ −9

X Intercepts: (−3, 0), (3, 0), Y Intercepts: (0, −9)

Vertex: Minimum (0, −9)

Inverse: √x+9, −√x+9

Interval Notation

Domain: (−∞ , ∞ )

Range: [−9, ∞ )

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Solution steps

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Domain of (x+3)(x−3) : −∞ <x<∞

Range of (x+3)(x−3): f(x)≥ −9

Axis interception points of (x+3)(x−3): X Intercepts: (−3, 0), (3, 0), Y Intercepts: (0, −9)

Vertex of (x+3)(x−3): Minimum (0, −9)

Inverse of (x+3)(x−3): √x+9, −√x+9

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ddx ((x+3)(x−3))

parity (x+3)(x−3)

symmetry (x+3)(x−3)

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

vertices 2x^2-2

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