What is the vertices f(x)=x^2+10x+7 ?

The vertices f(x)=x^2+10x+7 is Minimum (-5,-18)

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vertices f(x)=x^2+10x+7

vertices

f (x) = x^{2} + 10 x + 7

Minimum (- 5, - 18)

Solution steps

y = x^{2} + 10 x + 7

The parabola parameters are:

a = 1, b = 10, c = 7

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} = - 18

Therefore the parabola vertex is

(- 5, - 18)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 5, - 18)

f (x) = x^{2} - 3 x - 18

- 2 x + x (x - 2) - 2 x (x - 1) - (x - 3)

(5 x - 8)^{2}

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

- x^{2} + 7

What is the vertices f(x)=x^2+10x+7 ?
The vertices f(x)=x^2+10x+7 is Minimum (-5,-18)