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)

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

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

f (x) = - x^{2} + 6 x - 5

f (x) = x^{2} - 5 x + 6

- x^{2} + 4 x - 3