What is the vertices f(x)=0.75x^2-5 ?

The vertices f(x)=0.75x^2-5 is Minimum (0,-5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices f(x)=0.75x^2-5

vertices

f (x) = 0.75 x^{2} - 5

Minimum (0, - 5)

Solution steps

y = 0.75 x^{2} - 5

The parabola parameters are:

a = 0.75, b = 0, c = - 5

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

x_{v} = - \frac{0}{2 \cdot 0.75}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = - 5

Therefore the parabola vertex is

(0, - 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 0.75

Minimum (0, - 5)

x^{2} + 9 x + 4

5 x^{2} + 10 x + 6

+ x^{2} - 5

x^{2} + 8 x + 19

y = x^{2} + 7 x + 12

What is the vertices f(x)=0.75x^2-5 ?
The vertices f(x)=0.75x^2-5 is Minimum (0,-5)