What is the vertices f(x)=x^2+10x+9 ?

The vertices f(x)=x^2+10x+9 is Minimum (-5,-16)

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vertices f(x)=x^2+10x+9

vertices

f (x) = x^{2} + 10 x + 9

Minimum (- 5, - 16)

Solution steps

y = x^{2} + 10 x + 9

The parabola parameters are:

a = 1, b = 10, c = 9

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} = - 16

Therefore the parabola vertex is

(- 5, - 16)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 5, - 16)

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

x^{2} = 8 (y + 4)

y = 3 x^{2} - 12 x + 8

f (x) = - x^{2} - 4 x + 1

y = x^{2} - 4 x + 7

What is the vertices f(x)=x^2+10x+9 ?
The vertices f(x)=x^2+10x+9 is Minimum (-5,-16)