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diameter x^2+(y+3)^2=16

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

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range 6a²+5a+10−(4a²+6a+12)

Solution

f(a)≥ −178

Interval Notation

[−178 , ∞ )

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Function range definition

The set of values of the dependent variable for which a function is defined

Vertex of 6a²+5a+10−(4a²+6a+12): Minimum (14 , −178 )

For a parabola ax²+bx+c with Vertex (x_v, y_v)
If a<0 the range is f(x)≤ y_v
If a>0 the range is f(x)≥ y_v
a=2, Vertex (x_v, y_v)=(14 , −178 )

f(a)≥ −178

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domain 6a²+5a+10−(4a²+6a+12)

vertices 6a²+5a+10−(4a²+6a+12)

inverse 6a²+5a+10−(4a²+6a+12)

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

diameter x^2+(y+3)^2=16

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diameter x^2+(y+3)^2=16

Solution

Solution steps

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