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

The vertices y=x^2+2x-8 is Minimum (-1,-9)

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vertices y=x^2+2x-8

vertices

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

Minimum (- 1, - 9)

Solution steps

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

The parabola parameters are:

a = 1, b = 2, c = - 8

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} = - 9

Therefore the parabola vertex is

(- 1, - 9)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 1, - 9)

- 3 x^{2} + 5 x

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

y = 3 - x^{2}

y = 25 x^{2} - 36

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