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parity 3a^2+1-(4+2a^2)

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

☐²

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^□√☐

□□

log_□

θ

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ddx

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

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fatoração 2a⁴−4a²

Solution

2a²(a²−2)

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Solve by:

Encontrar o vértice usando a forma polinomial

Encontrar o vértice usando a forma da parábola

Encontrar o vértice usando a forma do vértice

One step at a time

y=7x²+6+2x²−4

Equação de parábola na forma polinomial

O vértice de uma parábola com concavidade para cima ou para baixo na forma y=ax²+bx+c é x_v=−b2a

Reescreva y=7x²+6+2x²−4 na forma y=ax²+bx+c

y=9x²+2

Os parâmetros da parábola são:

a=9, b=0, c=2

x_v=−b2a

x_v=−02· 9

Simplificar

x_v=0

Insira x_v=0 para encontrar o valor y_v

y_v=2

Portanto, o vértice da parábola é

(0, 2)

Se a<0, então o vértice é um valor máximo
Si a>0, então o vértice é um valor mínimo
a=9

Mínimo (0, 2)

Fractions Add, Subtract Like Denominators Examples

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

parity 3a^2+1-(4+2a^2)

Solution

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