What is the vertices-((x+3)^2)/(20)+1 ?

The vertices-((x+3)^2)/(20)+1 is Maximum (-3,1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices-((x+3)^2)/(20)+1

vertices

- \frac{( x + 3 ) ^{2}}{20} + 1

Maximum (- 3, 1)

Solution steps

y = - \frac{( x + 3 ) ^{2}}{20} + 1

y = - \frac{( x + 3 ) ^{2}}{20} + 1

The parabola parameters are:

a = - \frac{1}{20}, m = 25 - 3, n = - 25 - 3

x_{v} = \frac{m + n}{2}

x_{v} = \frac{( 2 5 - 3 ) + ( - 2 5 - 3 )}{2}

Simplify

\frac{( 2 5 - 3 ) + ( - 2 5 - 3 )}{2} : - 3

x_{v} = - 3

Plug in

x_{v} = - 3

to find the

y_{v}

value

y_{v} = 1

Therefore the parabola vertex is

(- 3, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{20}

Maximum (- 3, 1)

x^{2} - 2 x + 2

f (x) = - 0.01 x^{2} + 0.7 x + 6

y = + x^{2} - 3

f (x) = (x - 1)^{2} - 4

x^{2} - x - 20

What is the vertices-((x+3)^2)/(20)+1 ?
The vertices-((x+3)^2)/(20)+1 is Maximum (-3,1)