What is the vertices+x^2-36 ?

The vertices+x^2-36 is Minimum (0,-36)

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vertices+x^2-36

vertices

+ x^{2} - 36

Minimum (0, - 36)

Solution steps

y = x^{2} - 36

The parabola parameters are:

a = 1, b = 0, c = - 36

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} = - 36

Therefore the parabola vertex is

(0, - 36)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (0, - 36)

f (x) = 2 (x + 14)^{2} - 6

y = 3 x^{2} - 6 x + 5

x^{2} - 2

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

y = (x + 4)^{2}