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

The vertices f(x)=-x^2-6x-10 is Maximum (-3,-1)

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

vertices

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

Maximum (- 3, - 1)

Solution steps

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

The parabola parameters are:

a = - 1, b = - 6, c = - 10

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

x_{v} = - \frac{( - 6 )}{2 ( - 1 )}

Simplify

x_{v} = - 3

Plug in

x_{v} = - 3

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(- 3, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (- 3, - 1)

y = 2 x (x - 4) + 7

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

f (x) = x^{2} + 6 x - 8

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

f (x) = x^{2} + 6 x + 4

What is the vertices f(x)=-x^2-6x-10 ?
The vertices f(x)=-x^2-6x-10 is Maximum (-3,-1)