en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

range of 2x^2+x

Solution

range

2 x^{2} + x

Solution

f (x) \geq - \frac{1}{8}

+1

Interval Notation

[- \frac{1}{8}, \infty)

Solution steps

Vertex of

2 x^{2} + x :

Minimum

(- \frac{1}{4}, - \frac{1}{8})

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 = 2,

Vertex

(x_{v}, y_{v}) = (- \frac{1}{4}, - \frac{1}{8})

f (x) \geq - \frac{1}{8}

Popular Examples

domain of f(x)= 4/(x-1)+2 domain

f (x) = \frac{4}{x - 1} + 2

domain of f(x)=((2-x^2))/((x^2+2x-8))domain

f (x) = \frac{( 2 - x ^{2} )}{( x ^{2} + 2 x - 8 )}

asymptotes of f(x)= 6/((x-7)^3)asymptotes

f (x) = \frac{6}{( x - 7 ) ^{3}}

perpendicular y=(-1)/3 x+2 perpendicular

y = \frac{- 1}{3} x + 2

domain of f(x)= 6/(x+4)+1/(2-x)domain

f (x) = \frac{6}{x + 4} + \frac{1}{2 - x}

Frequently Asked Questions (FAQ)

What is the range of 2x^2+x ?
The range of 2x^2+x is f(x)>=-1/8