en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Functions >

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}

Graph

Popular Examples

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

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

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

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

domain of f(x)= 6/(x+4)+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