What is the vertices y=x^2+6 ?

The vertices y=x^2+6 is Minimum (0,6)

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vertices y=x^2+6

vertices

y = x^{2} + 6

Minimum (0, 6)

Solution steps

y = x^{2} + 6

The parabola parameters are:

a = 1, b = 0, c = 6

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{0}{2 \cdot 1}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 6

Therefore the parabola vertex is

(0, 6)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (0, 6)

y = x^{2} + 2

x^{2} + 2 x - 8

f (x) = 3 x^{2} - 2 x + 1

- 2 x^{2} + 25 x

f (x) = (x - 3)^{2} - 4