What is the vertices y=x^2+2x+1 ?

The vertices y=x^2+2x+1 is Minimum (-1,0)

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vertices y=x^2+2x+1

vertices

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

Minimum (- 1, 0)

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} = 0

Therefore the parabola vertex is

(- 1, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 1, 0)

10 + 3 x^{2}

x^{2} - 5 x - 24

y = (2 x)^{2}

y = 2 x^{2} - 19 x + 9

4 x^{2} - 20 x - 24 y + 97 = 0