What is the vertices f(x)=x^2-5 ?

The vertices f(x)=x^2-5 is Minimum (0,-5)

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vertices f(x)=x^2-5

vertices

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

Minimum (0, - 5)

Solution steps

y = x^{2} - 5

The parabola parameters are:

a = 1, b = 0, c = - 5

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} = - 5

Therefore the parabola vertex is

(0, - 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (0, - 5)

y = x^{2} + 3 x - 5

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

y = 4 x^{2} - 16 x + 22

x^{2} + 8 x - 12 y + 88 = 0

4 x^{2} - 16 x + 11