What is the vertices f(x)=-(x+3)^2+5 ?

The vertices f(x)=-(x+3)^2+5 is Maximum (-3,5)

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vertices f(x)=-(x+3)^2+5

vertices

f (x) = - (x + 3)^{2} + 5

Maximum (- 3, 5)

Solution steps

y = - (x + 3)^{2} + 5

Rewrite

y = - (x + 3)^{2} + 5

in the form

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

y = - x^{2} - 6 x - 4

The parabola parameters are:

a = - 1, b = - 6, c = - 4

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} = 5

Therefore the parabola vertex is

(- 3, 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (- 3, 5)

4 t^{2}

x^{2} - 6 x - 4 y + 5 = 0

(x + 1)^{2} + 1

y = x^{2} - 5 x - 6

- x^{2} - 2 x + 8

What is the vertices f(x)=-(x+3)^2+5 ?
The vertices f(x)=-(x+3)^2+5 is Maximum (-3,5)