What is the vertices f(x)=-(x-5)^2-4 ?

The vertices f(x)=-(x-5)^2-4 is Maximum (5,-4)

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vertices f(x)=-(x-5)^2-4

vertices

f (x) = - (x - 5)^{2} - 4

Maximum (5, - 4)

Solution steps

y = - (x - 5)^{2} - 4

Rewrite

y = - (x - 5)^{2} - 4

in the form

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

y = - x^{2} + 10 x - 29

The parabola parameters are:

a = - 1, b = 10, c = - 29

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

x_{v} = - \frac{10}{2 ( - 1 )}

Simplify

- \frac{10}{2 ( - 1 )} : 5

x_{v} = 5

Plug in

x_{v} = 5

to find the

y_{v}

value

y_{v} = - 4

Therefore the parabola vertex is

(5, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (5, - 4)

y = \frac{x ^{2} 5 x 8}{x 3}

x^{2} - 4 x

t^{2} + 2 t - 48

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

x = y^{2} - 3

What is the vertices f(x)=-(x-5)^2-4 ?
The vertices f(x)=-(x-5)^2-4 is Maximum (5,-4)