What is the vertices f(x)=x^2-12x+36 ?

The vertices f(x)=x^2-12x+36 is Minimum (6,0)

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vertices f(x)=x^2-12x+36

vertices

f (x) = x^{2} - 12 x + 36

Minimum (6, 0)

Solution steps

y = x^{2} - 12 x + 36

The parabola parameters are:

a = 1, b = - 12, c = 36

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} = 0

Therefore the parabola vertex is

(6, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (6, 0)

y = x^{2} + 4 x + 1

f (x) = x^{2} + 16 x + 63

y = 2 x^{2} - 4 x + 7

y = (x + 3) (x - 3)

f (x) = - 3 x^{2} - 18 x - 25

What is the vertices f(x)=x^2-12x+36 ?
The vertices f(x)=x^2-12x+36 is Minimum (6,0)