What is the vertices y=(x+1)^2 ?

The vertices y=(x+1)^2 is Minimum (-1,0)

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vertices y=(x+1)^2

vertices

y = (x + 1)^{2}

Minimum (- 1, 0)

Solution steps

y = (x + 1)^{2}

y = (x + 1)^{2}

The parabola parameters are:

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

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

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

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(- 1, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 1, 0)

- x^{2} + 4 x + 5

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

y = 2.6 + 2.2 x - 0.6 x^{2}

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

V (x) = 325 x^{2} - 4600 x + 145000