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

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

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vertices y=(x+1)^2+2

vertices

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

Minimum (- 1, 2)

Solution steps

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

Rewrite

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

in the form

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

y = x^{2} + 2 x + 3

The parabola parameters are:

a = 1, b = 2, c = 3

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} = 2

Therefore the parabola vertex is

(- 1, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 1, 2)

f (x) = - 6 x^{2} - 6 x - 6

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

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

f (x) = - 9 x^{2} + 9 x + 1

+ x^{2} + 4 x - 1