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

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

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vertices (x+1)^2+1

vertices

(x + 1)^{2} + 1

Minimum (- 1, 1)

Solution steps

y = (x + 1)^{2} + 1

Rewrite

y = (x + 1)^{2} + 1

in the form

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

y = x^{2} + 2 x + 2

The parabola parameters are:

a = 1, b = 2, c = 2

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

x_{v} = - \frac{2}{2 \cdot 1}

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = 1

Therefore the parabola vertex is

(- 1, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 1, 1)

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

- x^{2} - 2 x + 8

y = 4 x^{2} - 8

f (x) = 2 x^{2} - 12 x

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