interceptos (2x-1)^3

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

☐²

x^☐

√☐

^□√☐

□□

log_□

θ

∞

∫

ddx

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

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simetria y=4x²+1

Solution

Symmetry about the line x=0

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y=4x²+1: Parábola com vértice em (h, k)=(0, 1), e distância focal |p|=116

Up-down facing parabola symmetries

Up-down facing parabola in form 4p(y−k)=(x−h)² exhibits the following symmetry:
Symmetry about the line x=h

Therefore the symmetry is:

Symmetry about the line x=0

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

interceptos (2x-1)^3

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interceptos (2x-1)^3

Solution

Solution steps

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Graph