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vertices y=4x^2

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vertices y=4x²

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=4x²

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=4, b=0, c=0

x_v=−b2a

x_v=−02· 4

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=4

Minimum (0, 0)

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−▭▭	<	7	8	9	÷	`AC`
+▭▭	>	4	5	6	×	☐☐☐
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vertices y=4x^2

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