What is the vertices f(x)=x^2+6x+11 ?

The vertices f(x)=x^2+6x+11 is Minimum (-3,2)

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vertices f(x)=x^2+6x+11

vertices

f (x) = x^{2} + 6 x + 11

Minimum (- 3, 2)

Solution steps

y = x^{2} + 6 x + 11

The parabola parameters are:

a = 1, b = 6, c = 11

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{6}{2 \cdot 1}

Simplify

x_{v} = - 3

Plug in

x_{v} = - 3

to find the

y_{v}

value

y_{v} = 2

Therefore the parabola vertex is

(- 3, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 3, 2)

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

f (x) = 5 x^{2} + 1

\frac{5 x ^{2}}{2}

x^{2} - 2 x

- x^{2} - 2 x + 3

What is the vertices f(x)=x^2+6x+11 ?
The vertices f(x)=x^2+6x+11 is Minimum (-3,2)