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diretriz y=x^2+2

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vertices 6a²+5a+10−(4a²+6a+12)

Solution

Minimum (14 , −178 )

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Find vertex using polynomial form

Find vertex using parabola form

Find vertex using vertex form

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y=6a²+5a+10−(4a²+6a+12)

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=6a²+5a+10−(4a²+6a+12) in the form y=ax²+bx+c

y=2a²−a−2

The parabola parameters are:

a=2, b=−1, c=−2

x_v=−b2a

a_v=−(−1)2· 2

Simplify

a_v=14

Plug in a_v=14 to find the y_v value

y_v=−178

Therefore the parabola vertex is

(14 , −178 )

If a<0, then the vertex is a maximum value
If a>0, then the vertex is a minimum value
a=2

Minimum (14 , −178 )

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

diretriz y=x^2+2

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diretriz y=x^2+2

Solution

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