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

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

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vertices y=x^2+6x+5

vertices

y = 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)

f (x) = 3 x^{2} + 6 x - 1

t^{2} + 2

y = - x^{2} - 4 x + 1

x^{2} - 3 x + 5

3 x^{2} - 3 x