What is the vertices y=+x^2+8 ?

The vertices y=+x^2+8 is Minimum (0,8)

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vertices y=+x^2+8

vertices

y = + x^{2} + 8

Minimum (0, 8)

Solution steps

y = x^{2} + 8

The parabola parameters are:

a = 1, b = 0, c = 8

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} = 8

Therefore the parabola vertex is

(0, 8)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (0, 8)

- 3 x^{2} + 30 x - 8

f (x) = 4 x^{2} - 8 x + 7

f (x) = 2 x^{2} + 5 x + 2

f (x) = 8 x^{2}

y = 3 x - x^{2}