What is the vertices y=2.6+2.2x-0.6x^2 ?

The vertices y=2.6+2.2x-0.6x^2 is Maximum (1.83333…,4.61666…)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices y=2.6+2.2x-0.6x^2

vertices

y = 2.6 + 2.2 x - 0.6 x^{2}

Maximum (1.83333 \dots, 4.61666 \dots)

Solution steps

y = 2.6 + 2.2 x - 0.6 x^{2}

Rewrite

y = 2.6 + 2.2 x - 0.6 x^{2}

in the form

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

y = - 0.6 x^{2} + 2.2 x + 2.6

The parabola parameters are:

a = - 0.6, b = 2.2, c = 2.6

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

x_{v} = - \frac{2.2}{2 ( - 0.6 )}

Simplify

- \frac{2.2}{2 ( - 0.6 )} : 1.83333 \dots

x_{v} = 1.83333 \dots

Plug in

x_{v} = 1.83333 \dots

to find the

y_{v}

value

y_{v} = 4.61666 \dots

Therefore the parabola vertex is

(1.83333 \dots, 4.61666 \dots)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 0.6

Maximum (1.83333 \dots, 4.61666 \dots)

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

V (x) = 325 x^{2} - 4600 x + 145000

f (x) = 3 + x - 2 x^{2}

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

y = x^{2} - 16

What is the vertices y=2.6+2.2x-0.6x^2 ?
The vertices y=2.6+2.2x-0.6x^2 is Maximum (1.83333…,4.61666…)