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domain of 1/(x^2)

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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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inverser (

1	2	1
6	−1	0
−1	−2	−1

)

Solution

La matrice est singulière

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Solve by:

Trouver le sommet à l'aide de la formule du polynôme

Trouver le sommet à l'aide de la formule de la parabole

Trouver le sommet à l'aide de la formule du sommet

Trouver le sommet en faisant la moyenne des valeurs nulles

One step at a time

y=4x²−x−3

Equation d'une parabole sous une forme polynomiale

Le sommet d'une parabole tournée de haut en bas de la forme y=ax²+bx+c est x_v=−b2a

Les paramètres de la parabole sont :

a=4, b=−1, c=−3

x_v=−b2a

x_v=−(−1)2· 4

Simplifier

x_v=18

Intégrer x_v=18 pour trouver la valeur y_v

y_v=−4916

Par conséquent le vertex de la parabole est

(18 , −4916 )

Si a<0, alors le vertex est une valeur maximale
Si a>0, alors le vertex est une valeur minimale
a=4

Minimum (18 , −4916 )

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Notebook

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

domain of 1/(x^2)

Solution

Solution steps

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