What is the vertices f(x)=x^2-12x ?

The vertices f(x)=x^2-12x is Minimum (6,-36)

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vertices f(x)=x^2-12x

vertices

f (x) = x^{2} - 12 x

Minimum (6, - 36)

Solution steps

y = x^{2} - 12 x

The parabola parameters are:

a = 1, b = - 12, c = 0

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

x_{v} = - \frac{( - 12 )}{2 \cdot 1}

Simplify

x_{v} = 6

Plug in

x_{v} = 6

to find the

y_{v}

value

y_{v} = - 36

Therefore the parabola vertex is

(6, - 36)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (6, - 36)

f (x) = x^{2} + 4 x - 21

5 + 4 x - x^{2}

8 - 5 x^{2}

f (x) = 2 x^{2} - 4 x - 6

f (x) = 2 x^{2} - 4 x - 3

What is the vertices f(x)=x^2-12x ?
The vertices f(x)=x^2-12x is Minimum (6,-36)