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頂点 3a^2+1-(4+2a^2)

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vertices 3a²+1−(4+2a²)

Solution

Minimum (0, −3)

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Find vertex using polynomial form

Find vertex using parabola form

Find vertex using vertex form

Find vertex using averaging the zeros

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y=3a²+1−(4+2a²)

Parabola equation in polynomial form

The vertex of an up-down facing parabola of the form y=ax²+bx+c is x_v=−b2a

Rewrite y=3a²+1−(4+2a²) in the form y=ax²+bx+c

y=a²−3

The parabola parameters are:

a=1, b=0, c=−3

x_v=−b2a

a_v=−02· 1

Simplify

a_v=0

Plug in a_v=0 to find the y_v value

y_v=−3

Therefore the parabola vertex is

(0, −3)

If a<0, then the vertex is a maximum value
If a>0, then the vertex is a minimum value
a=1

Minimum (0, −3)

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こんにちは, について質問する vertices 3a²+1−(4+2a²) または以下のオプションから選択してください

解答者： Find vertex using parabola form

解答者： Find vertex using vertex form

解答者： Find vertex using averaging the zeros

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

頂点 3a^2+1-(4+2a^2)

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頂点 3a^2+1-(4+2a^2)

Solution

Solution steps

Number Line

Graph