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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices 2(x+1)^2-3

vertices

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)

2 y^{2} - 8 x = 0

x^{2} + 3 x + 1

x^{2} + 3 x - 1

x^{2} - 7 x - 8

0 = (y - 1)^{2} - 16 (x + 4)