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intervalo (z/5)^2

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

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

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vértices −(x+3)²20 +1

Solution

Máximo (−3, 1)

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Solve by:

Encontre o vértice usando o ponto médio das raízes

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=−(x+3)²20 +1

Equação de parábola na forma fatorada

O vértice de uma parábola com concavidade para cima ou para baixo na forma y=a(x−m)(x−n) é a média das raízes x_v=m+n2

y=−(x+3)²20 +1

Os parâmetros da parábola são:

a=−120 , m=2√5−3, n=−2√5−3

x_v=m+n2

x_v=(2√5−3)+(−2√5−3)2

Simplificar (2√5−3)+(−2√5−3)2 : −3

x_v=−3

Insira x_v=−3 para encontrar o valor y_v

y_v=1

Portanto, o vértice da parábola é

(−3, 1)

Se a<0, então o vértice é um valor máximo
Si a>0, então o vértice é um valor mínimo
a=−120

Máximo (−3, 1)

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

intervalo (z/5)^2

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