What is the vertices f(x)=x^2-1 ?

The vertices f(x)=x^2-1 is Minimum (0,-1)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

vertices f(x)=x^2-1

vertices

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

Minimum (0, - 1)

Solution steps

y = x^{2} - 1

The parabola parameters are:

a = 1, b = 0, c = - 1

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

x_{v} = - \frac{0}{2 \cdot 1}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(0, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (0, - 1)

36 x^{2} + 36 y^{2} - 12 x - 27 y - 8 = 0

y^{2} = 2 x

y^{2} - 3 y - x + 3 = 0

\frac{x ^{2}}{4} - \frac{y ^{2}}{9} = 1

y^{2} + x + y = 0