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

The vertices y=(x-8)^2 is Minimum (8,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices y=(x-8)^2

vertices

y = (x - 8)^{2}

Minimum (8, 0)

Solution steps

y = (x - 8)^{2}

y = (x - 8)^{2}

The parabola parameters are:

a = 1, m = 8, n = 8

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

x_{v} = \frac{8 + 8}{2}

Simplify

x_{v} = 8

Plug in

x_{v} = 8

to find the

y_{v}

value

y_{v} = 0

Therefore the parabola vertex is

(8, 0)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (8, 0)

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

f (x) = (x + 2) (x + 3)

y^{2} = 6 x - 4 y - 12

y = + x^{2} + 8

- 3 x^{2} + 30 x - 8