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 = x^{2} + 12 x + 36

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

f (x) = x (115 - 4 x)

x^{2} + 5 x + 1

f (x) = 2 x^{2} - 4 x + 4