What is the vertices (x-3)^2=8(y-5) ?

The vertices (x-3)^2=8(y-5) is Minimum (3,5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices (x-3)^2=8(y-5)

vertices

(x - 3)^{2} = 8 (y - 5)

Minimum (3, 5)

Solution steps

(x - 3)^{2} = 8 (y - 5)

Rewrite

(x - 3)^{2} = 8 (y - 5)

in the form

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

y = \frac{1 \cdot x ^{2}}{8} - \frac{3 x}{4} + \frac{9}{8} + 5

The parabola parameters are:

a = \frac{1}{8}, b = - \frac{3}{4}, c = \frac{9}{8} + 5

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

x_{v} = - \frac{( - \frac{3}{4} )}{2 ( \frac{1}{8} )}

Simplify

- \frac{- \frac{3}{4}}{2 ( \frac{1}{8} )} : 3

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = 5

Therefore the parabola vertex is

(3, 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{1}{8}

Minimum (3, 5)

(x - 2) = 3 (y - 5)^{2}

y = \frac{x ^{2} x ^{1}}{x}

X^{2}

3 - x^{2}

y = x^{2} - 5 x + 6

What is the vertices (x-3)^2=8(y-5) ?
The vertices (x-3)^2=8(y-5) is Minimum (3,5)