What is the vertices f(x)=x^2-9 ?

The vertices f(x)=x^2-9 is Minimum (0,-9)

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vertices f(x)=x^2-9

vertices

f (x) = x^{2} - 9

Minimum (0, - 9)

Solution steps

y = x^{2} - 9

The parabola parameters are:

a = 1, b = 0, c = - 9

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} = - 9

Therefore the parabola vertex is

(0, - 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (0, - 9)

y^{2} = x - 3

\frac{( x - 5 ) ^{2}}{25} + \frac{( y + 3 ) ^{2}}{9} = 1

16 y^{2} - 25 x^{2} = 400

16 x^{2} + 25 y^{2} = 400

x^{2} + y^{2} = 25