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)

16 x^{2} - 9 y^{2} = 144

y^{2} = - 9 x

x^{2} + 20 x + y^{2} + 16 y = - 20

\frac{x ^{2}}{64} - \frac{y ^{2}}{36} = 1

x^{2} + y^{2} - 6 x + 2 y + 9 = 0