What is the vertices f(x)=x^2-6x-16 ?

The vertices f(x)=x^2-6x-16 is Minimum (3,-25)

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vertices f(x)=x^2-6x-16

vertices

f (x) = x^{2} - 6 x - 16

Minimum (3, - 25)

Solution steps

y = x^{2} - 6 x - 16

The parabola parameters are:

a = 1, b = - 6, c = - 16

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{( - 6 )}{2 \cdot 1}

Simplify

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = - 25

Therefore the parabola vertex is

(3, - 25)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (3, - 25)

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

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

y = 2 x^{2} - x

- x^{2} + 14 x - 45

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

What is the vertices f(x)=x^2-6x-16 ?
The vertices f(x)=x^2-6x-16 is Minimum (3,-25)