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

The vertices f(x)=x^2+3x is Minimum (-3/2 ,-9/4)

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+3x

vertices

f (x) = x^{2} + 3 x

Minimum (- \frac{3}{2}, - \frac{9}{4})

Solution steps

y = x^{2} + 3 x

The parabola parameters are:

a = 1, b = 3, c = 0

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

x_{v} = - \frac{3}{2 \cdot 1}

Simplify

x_{v} = - \frac{3}{2}

Plug in

x_{v} = - \frac{3}{2}

to find the

y_{v}

value

y_{v} = - \frac{9}{4}

Therefore the parabola vertex is

(- \frac{3}{2}, - \frac{9}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- \frac{3}{2}, - \frac{9}{4})

x^{2} + 8 y + 2 x - 23 = 0

y = x^{2} - 10 x + 24

2 x^{2} - 5 x + 5

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

y^{2} = 20 x

What is the vertices f(x)=x^2+3x ?
The vertices f(x)=x^2+3x is Minimum (-3/2 ,-9/4)