What is the vertices 3x^2+2 ?

The vertices 3x^2+2 is Minimum (0,2)

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vertices 3x^2+2

vertices

3 x^{2} + 2

Minimum (0, 2)

Solution steps

y = 3 x^{2} + 2

The parabola parameters are:

a = 3, b = 0, c = 2

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

x_{v} = - \frac{0}{2 \cdot 3}

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 = 3

Minimum (0, 2)

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

y = x^{2} + 4 x + 7

- x^{2} - 4 x + 5

f (x) = x^{2} + 16 x + 64

y = 2 x^{2} - 4 x + 5