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vertices 3a^2+1-(4+2a^2)

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

☐²

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log_□

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ddx

≥	≤	·	÷	`x`^◦	(☐)	\|☐\|	(`f` ◦ `g`)	`f`(`x`)	ln	`e`^☐
(☐)^′	∂∂`x`	∫_□^□	lim	∑	sin	cos	tan	cot	csc	sec

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vertices x²+4

Solution

Minimum (0, 4)

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Find vertex using polynomial form

Find vertex using parabola form

Find vertex using vertex form

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y=x²+4

Parabola equation in polynomial form

The vertex of an up-down facing parabola of the form y=ax²+bx+c is x_v=−b2a

The parabola parameters are:

a=1, b=0, c=4

x_v=−b2a

x_v=−02· 1

Simplify

x_v=0

Plug in x_v=0 to find the y_v value

y_v=4

Therefore the parabola vertex is

(0, 4)

If a<0, then the vertex is a maximum value
If a>0, then the vertex is a minimum value
a=1

Minimum (0, 4)

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

vertices 3a^2+1-(4+2a^2)

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vertices 3a^2+1-(4+2a^2)

Solution

Solution steps

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