What is the vertices f(x)=108-5x^2-7x ?

The vertices f(x)=108-5x^2-7x is Maximum (-7/10 , 2209/20)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices f(x)=108-5x^2-7x

vertices

f (x) = 108 - 5 x^{2} - 7 x

Maximum (- \frac{7}{10}, \frac{2209}{20})

Solution steps

y = 108 - 5 x^{2} - 7 x

Rewrite

y = 108 - 5 x^{2} - 7 x

in the form

y = a x^{2} + b x + c

y = - 5 x^{2} - 7 x + 108

The parabola parameters are:

a = - 5, b = - 7, c = 108

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

x_{v} = - \frac{( - 7 )}{2 ( - 5 )}

Simplify

x_{v} = - \frac{7}{10}

Plug in

x_{v} = - \frac{7}{10}

to find the

y_{v}

value

y_{v} = \frac{2209}{20}

Therefore the parabola vertex is

(- \frac{7}{10}, \frac{2209}{20})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 5

Maximum (- \frac{7}{10}, \frac{2209}{20})

2 x^{2} - x - 1

2 x^{2} - x - 6

6 x^{2} - 13 x + 6

f (x) = - x^{2} + 3 x - 5

y = x^{2} - 10 x - 4

What is the vertices f(x)=108-5x^2-7x ?
The vertices f(x)=108-5x^2-7x is Maximum (-7/10 , 2209/20)