What is the vertices x^2+10x+11 ?

The vertices x^2+10x+11 is Minimum (-5,-14)

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vertices x^2+10x+11

vertices

x^{2} + 10 x + 11

Minimum (- 5, - 14)

Solution steps

y = x^{2} + 10 x + 11

The parabola parameters are:

a = 1, b = 10, c = 11

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} = - 14

Therefore the parabola vertex is

(- 5, - 14)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 5, - 14)

y = x^{2} + 2 x - 35

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

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

y = - x^{2} - 8

x^{2} - 7 x + 12