What is the vertices x^2+6x+5 ?

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

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vertices x^2+6x+5

vertices

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)

y = 3 x^{2} - 5 x 7

y^{2} = - 8 x

y = - x^{2} + 4 x - 3

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

- x^{2} - 4 x - 2