What is the vertices y=x^2+12x+36 ?

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

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vertices y=x^2+12x+36

vertices

y = 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)

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

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

x^{2} + 5 x + 1

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

f (x) = 2 x^{2} - 4 x + 1