What is the vertices f(x)=3x^2+1 ?

The vertices f(x)=3x^2+1 is Minimum (0,1)

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vertices f(x)=3x^2+1

vertices

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

Minimum (0, 1)

Solution steps

y = 3 x^{2} + 1

The parabola parameters are:

a = 3, b = 0, c = 1

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} = 1

Therefore the parabola vertex is

(0, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 3

Minimum (0, 1)

y^{2} + 2 y + x + 5 = 0

x^{2} + 6 x + 8

y = 2 x^{2} + 8 x - 34

f (x) = x^{2} + 6 x

3 x^{2} + 2 x + 1