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

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

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

vertices

f (x) = (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 + 3

The parabola parameters are:

a = 1, b = 4, c = 3

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)

- x^{2} + 10 x

x^{2} + 4 x - 8 y + 12 = 0

y = - \frac{1}{4} x^{2} + 8 x - 4

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

- \frac{x ^{2}}{2} - 2 x

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