What is the vertices f(x)=4x^2+2 ?

The vertices f(x)=4x^2+2 is Minimum (0,2)

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vertices f(x)=4x^2+2

vertices

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

Minimum (0, 2)

Solution steps

y = 4 x^{2} + 2

The parabola parameters are:

a = 4, b = 0, c = 2

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

x_{v} = - \frac{0}{2 \cdot 4}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 2

Therefore the parabola vertex is

(0, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (0, 2)

x^{2} + 3

(2 x - 1) (5 x - 6)

x^{2} + 2 x + 1

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

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