What is the vertices y=x^2+4x-8 ?

The vertices y=x^2+4x-8 is Minimum (-2,-12)

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vertices y=x^2+4x-8

vertices

y = x^{2} + 4 x - 8

Minimum (- 2, - 12)

Solution steps

y = x^{2} + 4 x - 8

The parabola parameters are:

a = 1, b = 4, c = - 8

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} = - 12

Therefore the parabola vertex is

(- 2, - 12)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 2, - 12)

y = x^{2} + 4 x - 3

f (x) = x^{2} - 12 x + 27

3 x^{2} + 2

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

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