What is the vertices y=2(x+1)^2-8 ?

The vertices y=2(x+1)^2-8 is Minimum (-1,-8)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

vertices y=2(x+1)^2-8

vertices

y = 2 (x + 1)^{2} - 8

Minimum (- 1, - 8)

Solution steps

y = 2 (x + 1)^{2} - 8

Rewrite

y = 2 (x + 1)^{2} - 8

in the form

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

y = 2 x^{2} + 4 x - 6

The parabola parameters are:

a = 2, b = 4, c = - 6

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} = - 8

Therefore the parabola vertex is

(- 1, - 8)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- 1, - 8)

p a r i t y f (x) = sin (x) + cos (x)

in v erse f (x) = 7 x^{3} - 2

cr i t i c a l f (x) = - 16 t^{2} + 60 t + 2

d o main f (x) = x^{4} + 2 x^{3} + 2 x^{2} + x

m o n o t o n e f (x) = \frac{2}{x + 5}

What is the vertices y=2(x+1)^2-8 ?
The vertices y=2(x+1)^2-8 is Minimum (-1,-8)