What is the vertices f(x)=2(x+1)^2-3 ?

The vertices f(x)=2(x+1)^2-3 is Minimum (-1,-3)

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vertices f(x)=2(x+1)^2-3

vertices

f (x) = 2 (x + 1)^{2} - 3

Minimum (- 1, - 3)

Solution steps

y = 2 (x + 1)^{2} - 3

Rewrite

y = 2 (x + 1)^{2} - 3

in the form

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

y = 2 x^{2} + 4 x - 1

The parabola parameters are:

a = 2, b = 4, c = - 1

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

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

Simplify

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = - 3

Therefore the parabola vertex is

(- 1, - 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- 1, - 3)

y = - 2 x^{2} - 8 x + 3

x^{2} - 2 x + 1

y^{2} = 5 x + 3

f (x) = - 3 x^{2} + 1

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

What is the vertices f(x)=2(x+1)^2-3 ?
The vertices f(x)=2(x+1)^2-3 is Minimum (-1,-3)