What is the vertices f(x)=x^2+6x-16 ?

The vertices f(x)=x^2+6x-16 is Minimum (-3,-25)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices f(x)=x^2+6x-16

vertices

f (x) = x^{2} + 6 x - 16

Minimum (- 3, - 25)

Solution steps

y = x^{2} + 6 x - 16

The parabola parameters are:

a = 1, b = 6, c = - 16

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

x_{v} = - \frac{6}{2 \cdot 1}

Simplify

x_{v} = - 3

Plug in

x_{v} = - 3

to find the

y_{v}

value

y_{v} = - 25

Therefore the parabola vertex is

(- 3, - 25)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- 3, - 25)

x = 1 - y^{2}

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

- \frac{( x + 3 ) ^{2}}{20} + 1

x^{2} - 2 x + 2

f (x) = - 0.01 x^{2} + 0.7 x + 6

What is the vertices f(x)=x^2+6x-16 ?
The vertices f(x)=x^2+6x-16 is Minimum (-3,-25)