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

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

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

vertices

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

Minimum (- \frac{1}{2}, - 1)

Solution steps

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

Rewrite

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

in the form

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

y = 4 x^{2} + 4 x

The parabola parameters are:

a = 4, b = 4, c = 0

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

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

Simplify

- \frac{4}{2 \cdot 4} : - \frac{1}{2}

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

Plug in

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

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(- \frac{1}{2}, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 4

Minimum (- \frac{1}{2}, - 1)

What is the vertices y=+(2x+1)^2-1 ?
The vertices y=+(2x+1)^2-1 is Minimum (-1/2 ,-1)