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range of 1-sec^2(x)

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`x`²	`x`^□	log_□	√☐	^□√☐	≤	≥	□□	·	÷	`x`^◦	π
(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

☐²

x^☐

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log_□

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ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

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vertices y=x²−36x−20

Solution

صغرى (18, −344)

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Find vertex using polynomial form

Find vertex using parabola form

Find vertex using vertex form

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y=x²−36x−20

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=1, b=−36, c=−20

x_v=−b2a

x_v=−(−36)2· 1

بسّط

x_v=18

Plug in x_v=18 to find the y_v value

y_v=−344

لذلك رأس القطع المكافئ هو

(18, −344)

إذًا الرأس هو نقطة عظمى a<0 إذا تحقّق أنّ
إذًا الرأس هو نقطة صغرى a>0 إذا تحقّق أنّ
a=1

صغرى (18, −344)

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أهلاً, اطرح سؤالا حول vertices y=x²−36x−20 أو اختر من الخيارات أدناه

حل بواسطة: Find vertex using parabola form

حل بواسطة: Find vertex using vertex form

حل بواسطة: Find vertex using averaging the zeros

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

range of 1-sec^2(x)

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