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amplitude of cos(2t)

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domain x²+2

Solution

−∞ <x<∞

Interval Notation

(−∞ , ∞ )

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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

One step at a time

y=4x²

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=4, b=0, c=0

x_v=−b2a

x_v=−02· 4

بسّط

x_v=0

Plug in x_v=0 to find the y_v value

y_v=0

لذلك رأس القطع المكافئ هو

(0, 0)

إذًا الرأس هو نقطة عظمى a<0 إذا تحقّق أنّ
إذًا الرأس هو نقطة صغرى a>0 إذا تحقّق أنّ
a=4

صغرى (0, 0)

Solve instead

range x²+2

vertices x²+2

inverse x²+2

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

amplitude of cos(2t)

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