What is the vertices f(x)=x^2+2x+5 ?

The vertices f(x)=x^2+2x+5 is Minimum (-1,4)

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vertices f(x)=x^2+2x+5

vertices

f (x) = x^{2} + 2 x + 5

Minimum (- 1, 4)

Solution steps

y = x^{2} + 2 x + 5

The parabola parameters are:

a = 1, b = 2, c = 5

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} = 4

Therefore the parabola vertex is

(- 1, 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 1, 4)

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

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

f (x) = (x - 7)^{2} - 1

x^{2} + 6 x + 8 y = 7

x^{2} - 6 x + 5 y = - 34

What is the vertices f(x)=x^2+2x+5 ?
The vertices f(x)=x^2+2x+5 is Minimum (-1,4)