en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

range of y=x^2+2x+1

Solution

range

y = x^{2} + 2 x + 1

Solution

f (x) \geq 0

+1

Interval Notation

[0, \infty)

Solution steps

Vertex of

x^{2} + 2 x + 1 :

Minimum

(- 1, 0)

For a parabola

a x^{2} + b x + c

with Vertex

(x_{v}, y_{v})

If

a < 0

the range is

f (x) \leq y_{v}

If

a > 0

the range is

f (x) \geq y_{v}

a = 1,

Vertex

(x_{v}, y_{v}) = (- 1, 0)

f (x) \geq 0

Popular Examples

intercepts of f(x)=-(x+6)^2+7 intercepts

f (x) = - (x + 6)^{2} + 7

asymptotes of y=2^{-x}+1 asymptotes

y = 2^{- x} + 1

shift f(x)=2sin(pix-3pi)-4 shift

f (x) = 2 sin (π x - 3 π) - 4

range of y=(x^2)/(x^2-4)range

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

line 4x+3x-12=0 line

4 x + 3 x - 12 = 0

Frequently Asked Questions (FAQ)

What is the range of y=x^2+2x+1 ?
The range of y=x^2+2x+1 is f(x)>= 0