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

The vertices f(x)=x^2+x is Minimum (-1/2 ,-1/4)

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+x

vertices

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

Minimum (- \frac{1}{2}, - \frac{1}{4})

Solution steps

y = x^{2} + x

The parabola parameters are:

a = 1, b = 1, c = 0

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

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

Simplify

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

Plug in

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

to find the

y_{v}

value

y_{v} = - \frac{1}{4}

Therefore the parabola vertex is

(- \frac{1}{2}, - \frac{1}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- \frac{1}{2}, - \frac{1}{4})

f (x) = x (115 - 4 x)

x^{2} + 5 x + 1

f (x) = 2 x^{2} - 4 x + 4

f (x) = 2 x^{2} - 4 x + 1

y = - x^{2} + 10 x

What is the vertices f(x)=x^2+x ?
The vertices f(x)=x^2+x is Minimum (-1/2 ,-1/4)