What is the vertices y=x^2+4x-12 ?

The vertices y=x^2+4x-12 is Minimum (-2,-16)

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vertices y=x^2+4x-12

vertices

y = x^{2} + 4 x - 12

Minimum (- 2, - 16)

Solution steps

y = x^{2} + 4 x - 12

The parabola parameters are:

a = 1, b = 4, c = - 12

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

x_{v} = - \frac{4}{2 \cdot 1}

Simplify

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = - 16

Therefore the parabola vertex is

(- 2, - 16)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 2, - 16)

2 (x + 1)^{2} - 3

2 y^{2} - 8 x = 0

x^{2} + 3 x + 1

x^{2} + 3 x - 1

x^{2} - 7 x - 8