What is the vertices 16-(12x+8)^2 ?

The vertices 16-(12x+8)^2 is Maximum (-2/3 ,16)

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vertices 16-(12x+8)^2

vertices

16 - (12 x + 8)^{2}

Maximum (- \frac{2}{3}, 16)

Solution steps

y = 16 - (12 x + 8)^{2}

Rewrite

y = 16 - (12 x + 8)^{2}

in the form

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

y = - 144 x^{2} - 192 x - 48

The parabola parameters are:

a = - 144, b = - 192, c = - 48

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

x_{v} = - \frac{( - 192 )}{2 ( - 144 )}

Simplify

- \frac{- 192}{2 ( - 144 )} : - \frac{2}{3}

x_{v} = - \frac{2}{3}

Plug in

x_{v} = - \frac{2}{3}

to find the

y_{v}

value

y_{v} = 16

Therefore the parabola vertex is

(- \frac{2}{3}, 16)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 144

Maximum (- \frac{2}{3}, 16)

y^{2} - 8 x - 6 y - 23 = 0

g (x) = (x + 3)^{2} + 5

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

f (x) = x^{2} + 8 + 15

y = x^{2} + 4 x - 5

What is the vertices 16-(12x+8)^2 ?
The vertices 16-(12x+8)^2 is Maximum (-2/3 ,16)