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vertices 2x^2-1+5x-6

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vertices 2x²−1+5x−6

Solution

Minimum (−54 , −818 )

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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=2x²−1+5x−6

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=2x²−1+5x−6 in the form y=ax²+bx+c

y=2x²+5x−7

The parabola parameters are:

a=2, b=5, c=−7

x_v=−b2a

x_v=−52· 2

Simplify

x_v=−54

Plug in x_v=−54 to find the y_v value

y_v=−818

Therefore the parabola vertex is

(−54 , −818 )

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

Minimum (−54 , −818 )

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

vertices 2x^2-1+5x-6

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