en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular Functions >

range of f(x)=-x^2+8x

Solution

range

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

Solution

f (x) \leq 16

+1

Interval Notation

(- \infty, 16]

Solution steps

Vertex of

- x^{2} + 8 x :

Maximum

(4, 16)

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}) = (4, 16)

f (x) \leq 16

Graph

Popular Examples

intercepts of f(x)=6(x+7)-5

range of f(x)=6-(x+2)^2

inverse of (-(3))/((x-1)-1)

domain of f(x)=sqrt(40-4x)

parallel y=-5/3 x-3

Frequently Asked Questions (FAQ)

What is the range of f(x)=-x^2+8x ?
The range of f(x)=-x^2+8x is f(x)<= 16