What is the vertices y=(x+5)^2 ?

The vertices y=(x+5)^2 is Minimum (-5,0)

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vertices y=(x+5)^2

vertices

y = (x + 5)^{2}

Minimum (- 5, 0)

Solution steps

y = (x + 5)^{2}

y = (x + 5)^{2}

The parabola parameters are:

a = 1, m = - 5, n = - 5

x_{v} = \frac{m + n}{2}

x_{v} = \frac{( - 5 ) + ( - 5 )}{2}

Simplify

x_{v} = - 5

Plug in

x_{v} = - 5

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(- 5, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 5, 0)

y = x^{2} + 5 x - 14

f (x) = - 2 (x - 1)^{2} - 5

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

+ \frac{x ^{2} + x}{2}

f (x) = 4 x^{2} - 32 x - 7