intercepts of (x^4)^5

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

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

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vertices x²+2x−1−(2x²−3x+6)

Solution

Maximum (52 , −34 )

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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=x²+2x−1−(2x²−3x+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=x²+2x−1−(2x²−3x+6) in the form y=ax²+bx+c

y=−x²+5x−7

The parabola parameters are:

a=−1, b=5, c=−7

x_v=−b2a

x_v=−52(−1)

Simplify

x_v=52

Plug in x_v=52 to find the y_v value

y_v=−34

Therefore the parabola vertex is

(52 , −34 )

If a<0, then the vertex is a maximum value
If a>0, then the vertex is a minimum value
a=−1

Maximum (52 , −34 )

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こんにちは, について質問する vertices x²+2x−1−(2x²−3x+6) または以下のオプションから選択してください

解答者： Find vertex using parabola form

解答者： Find vertex using vertex form

解答者： Find vertex using averaging the zeros

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

intercepts of (x^4)^5

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intercepts of (x^4)^5

Solution

Solution steps

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