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

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

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

vertices

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

Minimum (- 1, - 3)

Solution steps

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

Rewrite

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

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} = - 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 = 1

Minimum (- 1, - 3)

3 x^{2} - 10

f (x) = - x^{2} + 2 x - 5

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

f (x) = 2 x^{2} + 4 x - 6

f (x) = 10 x^{2} - 17

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