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

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

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vertices (x+1)^2=2(y+3)

vertices

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

Minimum (- 1, - 3)

Solution steps

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

Rewrite

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

in the form

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

y = \frac{1 \cdot x ^{2}}{2} + x + \frac{1}{2} - 3

The parabola parameters are:

a = \frac{1}{2}, b = 1, c = \frac{1}{2} - 3

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

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

Simplify

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

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 = \frac{1}{2}

Minimum (- 1, - 3)

y = x (2 x + 3)

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

f (x) = x^{2} - 4 x + 5

y^{2} = 8 x

x \leq y^{2}

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