What is the vertices f(x)=-3x^2+18x-8 ?

The vertices f(x)=-3x^2+18x-8 is Maximum (3,19)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices f(x)=-3x^2+18x-8

vertices

f (x) = - 3 x^{2} + 18 x - 8

Maximum (3, 19)

Solution steps

y = - 3 x^{2} + 18 x - 8

The parabola parameters are:

a = - 3, b = 18, c = - 8

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

x_{v} = - \frac{18}{2 ( - 3 )}

Simplify

- \frac{18}{2 ( - 3 )} : 3

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = 19

Therefore the parabola vertex is

(3, 19)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 3

Maximum (3, 19)

y = x^{2} - 6 x + 8

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

y = - 4 x^{2} - 12 x - 8

y = + x^{2} + 8 x + 9

25 x^{2} - 36

What is the vertices f(x)=-3x^2+18x-8 ?
The vertices f(x)=-3x^2+18x-8 is Maximum (3,19)