What is the vertices f(x)=x^2+2x-1 ?

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

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vertices f(x)=x^2+2x-1

vertices

f (x) = x^{2} + 2 x - 1

Minimum (- 1, - 2)

Solution steps

y = x^{2} + 2 x - 1

The parabola parameters are:

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

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} = - 2

Therefore the parabola vertex is

(- 1, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 1, - 2)

y = 4 - (x - 3)^{2}

f (x) = 2 x^{2} - 10 x

f (x) = x^{2} - 6 x + 4

x^{2} + x - 4

x^{2} + x - 6

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