sequence type a_{n}=3n+2

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(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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vertices y=−2x²−2x−2

Solution

Maximum (−12 , −32 )

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Find vertex using polynomial form

Find vertex using parabola form

Find vertex using vertex form

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y=−2x²−2x−2

Parabola equation in polynomial form

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

The parabola parameters are:

a=−2, b=−2, c=−2

x_v=−b2a

x_v=−(−2)2(−2)

Simplify −−22(−2) : −12

x_v=−12

Plug in x_v=−12 to find the y_v value

y_v=−32

Therefore the parabola vertex is

(−12 , −32 )

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

Maximum (−12 , −32 )

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

sequence type a_{n}=3n+2

Solution

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