inverse of (x^2+1)(x^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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axis y=(x+5)²

Solution

x=−5

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y=(x+5)²

Parabola standard equation

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

Rewrite y=(x+5)² in the standard form: 4· 14 (y−0)=(x−(−5))²

Therefore parabola properties are:

(h, k)=(−5, 0), p=14

Parabola is of the form 4p(y−k)=(x−h)² and is symmetric around the y-axis
Axis of symmetry is a line parallel to the y-axis which intersects the vertex:

x=−5

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

inverse of (x^2+1)(x^2-1)

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