What is the vertices f(x)=x^2+6x+5 ?

The vertices f(x)=x^2+6x+5 is Minimum (-3,-4)

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vertices f(x)=x^2+6x+5

vertices

f (x) = x^{2} + 6 x + 5

Minimum (- 3, - 4)

Solution steps

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

The parabola parameters are:

a = 1, b = 6, c = 5

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} = - 4

Therefore the parabola vertex is

(- 3, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 3, - 4)

4 x^{2} - 12 y^{2} - 16 x - 72 y - 44 = 0

y^{2} = - 2 x

(x - 2)^{2} + (y - 3)^{2} = 64

4 x^{2} - 9 y^{2} = 36

4 x^{2} - 9 y^{2} = 36

What is the vertices f(x)=x^2+6x+5 ?
The vertices f(x)=x^2+6x+5 is Minimum (-3,-4)