domain of 3/(x^3-1)

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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

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Generated by AI

inverse (

24	0	1
8	−13	10
7	8	−5

)

Solution

(

341	−8205	−13205
−2241	127205	232205
−3141	192205	312205

)

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Find vertex using polynomial form

Find vertex using parabola form

Find vertex using vertex form

One step at a time

y=−2x²−2x−2

Parabola equation in polynomial form

The vertex of an up-down facing parabola of the form y=ax²+bx+c is x_v=−b2a

The parabola parameters are:

a=−2, b=−2, c=−2

x_v=−b2a

x_v=−(−2)2(−2)

Simplify −−22(−2) : −12

x_v=−12

Plug in x_v=−12 to find the y_v value

y_v=−32

Therefore the parabola vertex is

(−12 , −32 )

If a<0, then the vertex is a maximum value
If a>0, then the vertex is a minimum value
a=−2

Maximum (−12 , −32 )

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أهلاً, اطرح سؤالا حول inverse (

24	0	1
8	−13	10
7	8	−5

) أو اختر من الخيارات أدناه

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Notebook

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

domain of 3/(x^3-1)

Solution

Solution steps

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