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vertices -((x+3)^2)/(20)+1

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vertices −(x+3)²20 +1

Solution

Maximum (−3, 1)

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Find vertex using averaging the zeros

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Find vertex using parabola form

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y=−(x+3)²20 +1

Parabola equation in factored form

The vertex of an up-down facing parabola of the form y=a(x−m)(x−n) is the average of the zeros x_v=m+n2

y=−(x+3)²20 +1

The parabola parameters are:

a=−120 , m=2√5−3, n=−2√5−3

x_v=m+n2

x_v=(2√5−3)+(−2√5−3)2

Simplify (2√5−3)+(−2√5−3)2 : −3

x_v=−3

Plug in x_v=−3 to find the y_v value

y_v=1

Therefore the parabola vertex is

(−3, 1)

If a<0, then the vertex is a maximum value
If a>0, then the vertex is a minimum value
a=−120

Maximum (−3, 1)

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

vertices -((x+3)^2)/(20)+1

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