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

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

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vertices f(x)=x^2+3

vertices

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

Minimum (0, 3)

Solution steps

y = x^{2} + 3

The parabola parameters are:

a = 1, b = 0, c = 3

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} = 3

Therefore the parabola vertex is

(0, 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (0, 3)

y = x^{2} + 3 x - 4

f (x) = - x^{2} + 6 x - 5

f (x) = x^{2} - 5 x + 6

- x^{2} + 4 x - 3

y = 2 x^{2} - 1 - (5 x - 6)