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

The vertices y=x^2+2x-12 is Minimum (-1,-13)

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vertices y=x^2+2x-12

vertices

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

Minimum (- 1, - 13)

Solution steps

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

The parabola parameters are:

a = 1, b = 2, c = - 12

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

x_{v} = - \frac{2}{2 \cdot 1}

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = - 13

Therefore the parabola vertex is

(- 1, - 13)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 1, - 13)

f (x) = 2 x^{2} - 80 x + 900

y = 5 x^{2}

f (x) = - (x + 3)^{2} + 5

4 t^{2}

x^{2} - 6 x - 4 y + 5 = 0