What is the vertices f(x)=x^2+2x-8 ?

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

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vertices f(x)=x^2+2x-8

vertices

f (x) = 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)

\frac{x ^{2}}{36} - \frac{y ^{2}}{9} = 1

y = x^{2}

f (x) = x^{2} - 4 x - 12

25 y^{2} - 36 (x - 8)^{2} = 900

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

What is the vertices f(x)=x^2+2x-8 ?
The vertices f(x)=x^2+2x-8 is Minimum (-1,-9)