center x^2-6x+8y+y^2=0

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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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vértices y=4x²−x−3

Solution

Mínimo (18 , −4916 )

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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=4x²−x−3

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

Os parâmetros da parábola são:

a=4, b=−1, c=−3

x_v=−b2a

x_v=−(−1)2· 4

Simplificar

x_v=18

Insira x_v=18 para encontrar o valor y_v

y_v=−4916

Portanto, o vértice da parábola é

(18 , −4916 )

Se a<0, então o vértice é um valor máximo
Si a>0, então o vértice é um valor mínimo
a=4

Mínimo (18 , −4916 )

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

`x`²	`x`<−2
ln(`x`+5)	`x`≥−2

sin(`x`)	`x`<0
2`x`	`x`≥0

√`x`+2	`x`<1
`xx`²−6`x`+8	`x`≥1

√`x`²−4	`x`<0
`x`³−`x`	`x`≥0

center x^2-6x+8y+y^2=0

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center x^2-6x+8y+y^2=0

Solution

Solution steps

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