What is the vertices y=-1/16 (x+2)^2-6 ?

The vertices y=-1/16 (x+2)^2-6 is Maximum (-2,-6)

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vertices y=-1/16 (x+2)^2-6

vertices

y = - \frac{1}{16} (x + 2)^{2} - 6

Maximum (- 2, - 6)

Solution steps

y = - \frac{1}{16} (x + 2)^{2} - 6

Rewrite

y = - \frac{1}{16} (x + 2)^{2} - 6

in the form

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

y = - \frac{x ^{2}}{16} - \frac{x}{4} - \frac{25}{4}

The parabola parameters are:

a = - \frac{1}{16}, b = - \frac{1}{4}, c = - \frac{25}{4}

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

x_{v} = - \frac{( - \frac{1}{4} )}{2 ( - \frac{1}{16} )}

Simplify

- \frac{- \frac{1}{4}}{2 ( - \frac{1}{16} )} : - 2

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = - 6

Therefore the parabola vertex is

(- 2, - 6)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{16}

Maximum (- 2, - 6)

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

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

y = 4 x^{2} - 12 x + 10

x^{2} - 12 x + 16 y + 68 = 0

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

What is the vertices y=-1/16 (x+2)^2-6 ?
The vertices y=-1/16 (x+2)^2-6 is Maximum (-2,-6)