domain of (x+5)^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

simplify solve for expand factor rationalize

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domain t²

Solution

−∞ <t<∞

Interval Notation

(−∞ , ∞ )

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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

Encontre o vértice usando o ponto médio das raízes

One step at a time

y=3a²+1−(4+2a²)

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=3a²+1−(4+2a²) na forma y=ax²+bx+c

y=a²−3

Os parâmetros da parábola são:

a=1, b=0, c=−3

x_v=−b2a

a_v=−02· 1

Simplificar

a_v=0

Insira a_v=0 para encontrar o valor y_v

y_v=−3

Portanto, o vértice da parábola é

(0, −3)

Se a<0, então o vértice é um valor máximo
Si a>0, então o vértice é um valor mínimo
a=1

Mínimo (0, −3)

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range t²

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

domain of (x+5)^2

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domain of (x+5)^2

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