вершины x^2+(y+3)^2=16

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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 x²+(y+3)²=16

Solution

(0, 1), (0, −7)

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Ellipse vertices

The vertices are the two points on the ellipse that intersect the major axis
For an ellipse with major axis parallel to the y-axis, the vertices are (h, k+b), (h, k−b)

(h, k+b), (h, k−b)

Calculate ellipse properties

x²+(y+3)²=16: Ellipse with center (h, k)=(0, −3), b=4, a=4

(0, −3+4), (0, −3−4)

Refine

(0, 1), (0, −7)

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

вершины x^2+(y+3)^2=16

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