What is the vertices y=x^2+3x+4 ?

The vertices y=x^2+3x+4 is Minimum (-3/2 , 7/4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

vertices y=x^2+3x+4

vertices

Minimum (- \frac{3}{2}, \frac{7}{4})

Solution steps

y = x^{2} + 3 x + 4

The parabola parameters are:

a = 1, b = 3, c = 4

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{7}{4}

Therefore the parabola vertex is

(- \frac{3}{2}, \frac{7}{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{7}{4})

What is the vertices y=x^2+3x+4 ?
The vertices y=x^2+3x+4 is Minimum (-3/2 , 7/4)