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

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

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

vertices

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

Minimum (- 2, 1)

Solution steps

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

Rewrite

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

in the form

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

y = x^{2} + 4 x + 5

The parabola parameters are:

a = 1, b = 4, c = 5

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

x_{v} = - \frac{4}{2 \cdot 1}

Simplify

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = 1

Therefore the parabola vertex is

(- 2, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 2, 1)

3 x^{2} + 6 x + 2

2 x^{2} - 3 x

f (x) = 4 (x - 2)^{2} + 11

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

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