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

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

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

vertices

f (x) = 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)

y^{2} = 28 x

3 x^{2} - 3 y = 0

\frac{x ^{2}}{25} - \frac{y ^{2}}{16} = 1

f (x) = x^{2} - 4 x - 5

\frac{y ^{2}}{16} - \frac{x ^{2}}{4} = 1

What is the vertices f(x)=x^2+2x+1 ?
The vertices f(x)=x^2+2x+1 is Minimum (-1,0)