What is the vertices f(x)=8x-13-x^2 ?

The vertices f(x)=8x-13-x^2 is Maximum (4,3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices f(x)=8x-13-x^2

vertices

f (x) = 8 x - 13 - x^{2}

Maximum (4, 3)

Solution steps

y = 8 x - 13 - x^{2}

Rewrite

y = 8 x - 13 - x^{2}

in the form

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

y = - x^{2} + 8 x - 13

The parabola parameters are:

a = - 1, b = 8, c = - 13

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

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

Simplify

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

x_{v} = 4

Plug in

x_{v} = 4

to find the

y_{v}

value

y_{v} = 3

Therefore the parabola vertex is

(4, 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (4, 3)

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

What is the vertices f(x)=8x-13-x^2 ?
The vertices f(x)=8x-13-x^2 is Maximum (4,3)