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inverse of (x^{11})/(x^{-3)}

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vértices x²−2x−3

Solution

Mínimo (1, −4)

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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=x²−2x−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=1, b=−2, c=−3

x_v=−b2a

x_v=−(−2)2· 1

Simplificar

x_v=1

Insira x_v=1 para encontrar o valor y_v

y_v=−4

Portanto, o vértice da parábola é

(1, −4)

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

Mínimo (1, −4)

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

inverse of (x^{11})/(x^{-3)}

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