9 5/6 \div 2/3

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(☐)^′	`ddx`	∂∂`x`	∫	∫_□^□	lim	∑	∞	`θ`	(`f` ◦ `g`)	`f`(`x`)

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vertices y=3x· x

Solution

Minimum (0, 0)

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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=3xx

Parabola equation in polynomial form

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

Rewrite y=3x· x in the form y=ax²+bx+c

y=3x²

The parabola parameters are:

a=3, b=0, c=0

x_v=−b2a

x_v=−02· 3

Simplify

x_v=0

Plug in x_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=3

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`

9 5/6 \div 2/3

Solution

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