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

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

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vertices-(x+5)^2+4

vertices

- (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 - 21

The parabola parameters are:

a = - 1, b = - 10, c = - 21

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

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

Simplify

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 = 2 x^{2} + 4 x - 8

f (x) = - 2 x^{2} + 8 x - 11

f (x) = - 3 x^{2} + 4 x + 4

f (x) = 2 (x - 5) (x - 1)

- x^{2} + 4 x - 6