en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

range of f(x)=x^2+2x+5

Solution

range

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

Solution

f (x) \geq 4

+1

Interval Notation

[4, \infty)

Solution steps

Vertex of

x^{2} + 2 x + 5 :

Minimum

(- 1, 4)

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, 4)

f (x) \geq 4

Popular Examples

asymptotes of f(x)=(2x^2+x-1)/(x^2+x-2)asymptotes

f (x) = \frac{2 x ^{2} + x - 1}{x ^{2} + x - 2}

slope ofintercept y-1=-2x slope intercept

y - 1 = - 2 x

critical f(x)=x^2e^{-3x}critical points

f (x) = x^{2} e^{- 3 x}

asymptotes of f(x)=3^{x-4}asymptotes

f (x) = 3^{x - 4}

midpoint (1,-1),(3,5)midpoint

(1, - 1), (3, 5)

Frequently Asked Questions (FAQ)

What is the range of f(x)=x^2+2x+5 ?
The range of f(x)=x^2+2x+5 is f(x)>= 4