What is the vertices f(x)=8x^2 ?

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

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vertices f(x)=8x^2

vertices

f (x) = 8 x^{2}

Minimum (0, 0)

Solution steps

y = 8 x^{2}

The parabola parameters are:

a = 8, b = 0, c = 0

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

x_{v} = - \frac{0}{2 \cdot 8}

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)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 8

Minimum (0, 0)

y = 3 x - x^{2}

y = 3 x^{2} + 12 x - 3

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

f (x) = 3 x^{2} - 5

f (x) = 2 x^{2} - 12 x + 20