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)=-5x^2+2x+8

Solution

range

f (x) = - 5 x^{2} + 2 x + 8

Solution

f (x) \leq \frac{41}{5}

+1

Interval Notation

(- \infty, \frac{41}{5}]

Solution steps

Vertex of

- 5 x^{2} + 2 x + 8 :

Maximum

(\frac{1}{5}, \frac{41}{5})

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 = - 5,

Vertex

(x_{v}, y_{v}) = (\frac{1}{5}, \frac{41}{5})

f (x) \leq \frac{41}{5}

Popular Examples

domain of sqrt(4-6a)domain

4 - 6 a

inverse of f(x)=(-2x-5)/(6x-7)inverse

f (x) = \frac{- 2 x - 5}{6 x - 7}

critical f(x)=e^{-2.5x^2}critical points

f (x) = e^{- 2.5 x^{2}}

domain of f(x)=(9x-6)/(sqrt(x+9))domain

f (x) = \frac{9 x - 6}{x + 9}

domain of f(x)=x^7 domain

f (x) = x^{7}

Frequently Asked Questions (FAQ)

What is the range of f(x)=-5x^2+2x+8 ?
The range of f(x)=-5x^2+2x+8 is f(x)<= 41/5