What is the vertices f(x)=-x^2-14x-42 ?

The vertices f(x)=-x^2-14x-42 is Maximum (-7,7)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

vertices f(x)=-x^2-14x-42

vertices

Maximum (- 7, 7)

Solution steps

y = - x^{2} - 14 x - 42

The parabola parameters are:

a = - 1, b = - 14, c = - 42

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

x_{v} = - \frac{( - 14 )}{2 ( - 1 )}

Simplify

x_{v} = - 7

Plug in

x_{v} = - 7

to find the

y_{v}

value

y_{v} = 7

Therefore the parabola vertex is

(- 7, 7)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 1

Maximum (- 7, 7)

What is the vertices f(x)=-x^2-14x-42 ?
The vertices f(x)=-x^2-14x-42 is Maximum (-7,7)