What is the vertices+1+x^2 ?

The vertices+1+x^2 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+1+x^2

vertices

+ 1 + x^{2}

Minimum (0, 1)

Solution steps

y = 1 + x^{2}

Rewrite

y = + 1 + x^{2}

in the form

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

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)

y = (x - 3)^{2} - 2

x^{2} - 5 x - 36

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

100 (1 - \frac{x}{40})^{2}

y = x^{2} + 12 x + 16