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interceptos 2sqrt(4x)

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periodicidade sec²(x)

Solution

Decimal

3.14159…

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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=x²

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

x_v=−02· 1

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=1

Minimum (0, 0)

Solve instead

domínio sec²(x)

intervalo sec²(x)

amplitude sec²(x)

Order of operations two-step without parentheses Examples

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
×▭▭	(	1	2	3	−	`x`
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interceptos 2sqrt(4x)

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interceptos 2sqrt(4x)

Solution

Solution steps

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