asse y=x^2+4

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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 t²

Solution

Minimum (0, 0)

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Solve by:

Find vertex using polynomial form

Find vertex using parabola form

Find vertex using vertex form

Find vertex using averaging the zeros

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y=t²

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=0

x_v=−b2a

t_v=−02· 1

Simplify

t_v=0

Plug in t_v=0 to find the y_v value

y_v=0

Therefore the parabola vertex is

(0, 0)

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

Minimum (0, 0)

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Solve by: Find vertex using parabola form

Solve by: Find vertex using vertex form

Solve by: Find vertex using averaging the zeros

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

asse y=x^2+4

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