What is the vertices f(x)=-1/4 (x+2)^2-2 ?

The vertices f(x)=-1/4 (x+2)^2-2 is Maximum (-2,-2)

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vertices f(x)=-1/4 (x+2)^2-2

vertices

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

Maximum (- 2, - 2)

Solution steps

y = - \frac{1}{4} (x + 2)^{2} - 2

Rewrite

y = - \frac{1}{4} (x + 2)^{2} - 2

in the form

y = a x^{2} + b x + c

y = - \frac{x ^{2}}{4} - x - 3

The parabola parameters are:

a = - \frac{1}{4}, b = - 1, c = - 3

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

x_{v} = - \frac{( - 1 )}{2 ( - \frac{1}{4} )}

Simplify

- \frac{- 1}{2 ( - \frac{1}{4} )} : - 2

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = - 2

Therefore the parabola vertex is

(- 2, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{4}

Maximum (- 2, - 2)

y = (x + 4)^{2}

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

3 x^{2} - 10

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

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

What is the vertices f(x)=-1/4 (x+2)^2-2 ?
The vertices f(x)=-1/4 (x+2)^2-2 is Maximum (-2,-2)