What is the vertices f(x)=x^2+10x+6 ?

The vertices f(x)=x^2+10x+6 is Minimum (-5,-19)

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vertices f(x)=x^2+10x+6

vertices

f (x) = x^{2} + 10 x + 6

Minimum (- 5, - 19)

Solution steps

y = x^{2} + 10 x + 6

The parabola parameters are:

a = 1, b = 10, c = 6

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

x_{v} = - \frac{10}{2 \cdot 1}

Simplify

x_{v} = - 5

Plug in

x_{v} = - 5

to find the

y_{v}

value

y_{v} = - 19

Therefore the parabola vertex is

(- 5, - 19)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 5, - 19)

- 3 x^{2} - 12 x - 9

y = x^{2} - 2 x 5

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

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

- 2 x^{2} + 12 x - 14

What is the vertices f(x)=x^2+10x+6 ?
The vertices f(x)=x^2+10x+6 is Minimum (-5,-19)