domain of 4x^2+1

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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

simplify solve for inverse tangent line

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vertices 7y²+3y²

Solution

(h, k)=(0, 0), p=140

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x=7y²+3y²

Parabola standard equation

4p(x−h)=(y−k)² is the standard equation for a right-left facing parabola with vertex at (h, k),
and a focal length |p|

Rewrite x=7y²+3y² in the standard form

4· 140 (x−0)=(y−0)²

Therefore parabola properties are:

(h, k)=(0, 0), p=140

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

domain of 4x^2+1

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