focos -((y-3)^2}{25}+\frac{(x+2)^2)/9 =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 expand factor rationalize

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vertices |−3x|

Solution

(0, 0)

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Absolute value function vertex definition

Absolute value function vertex points are points where absolute values are zeros

Find the absolute values zeroes:

Solve |−3x|=0: x=0

x=0

Plug in x=0 to find the y value: (0, 0)

Therefore absolute value function vertices are

(0, 0)

Solve instead

domain |−3x|

range |−3x|

intercepts |−3x|

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

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focos -((y-3)^2}{25}+\frac{(x+2)^2)/9 =1

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focos -((y-3)^2}{25}+\frac{(x+2)^2)/9 =1

Solution

Solution steps

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