What is the vertices y=x^2+5x-14 ?

The vertices y=x^2+5x-14 is Minimum (-5/2 ,-81/4)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices y=x^2+5x-14

vertices

y = x^{2} + 5 x - 14

Minimum (- \frac{5}{2}, - \frac{81}{4})

Solution steps

y = x^{2} + 5 x - 14

The parabola parameters are:

a = 1, b = 5, c = - 14

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

x_{v} = - \frac{5}{2 \cdot 1}

Simplify

x_{v} = - \frac{5}{2}

Plug in

x_{v} = - \frac{5}{2}

to find the

y_{v}

value

y_{v} = - \frac{81}{4}

Therefore the parabola vertex is

(- \frac{5}{2}, - \frac{81}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- \frac{5}{2}, - \frac{81}{4})

f (x) = - 2 (x - 1)^{2} - 5

f (x) = y^{2} + 3 y + 4

+ \frac{x ^{2} + x}{2}

f (x) = 4 x^{2} - 32 x - 7

2 (x + 1)^{2}

What is the vertices y=x^2+5x-14 ?
The vertices y=x^2+5x-14 is Minimum (-5/2 ,-81/4)