en

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

range of y=-2x^2+8x-1

Solution

range

y = - 2 x^{2} + 8 x - 1

Solution

f (x) \leq 7

+1

Interval Notation

(- \infty, 7]

Solution steps

Vertex of

- 2 x^{2} + 8 x - 1 :

Maximum

(2, 7)

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}) = (2, 7)

f (x) \leq 7

Popular Examples

shift f(x)=sin(x+5)-3 shift

f (x) = sin (x + 5) - 3

inflection 2x^4+8x^3 inflection points

2 x^{4} + 8 x^{3}

domain of y=x^2-7 domain

y = x^{2} - 7

inflection x^3-7x^2+36 inflection points

x^{3} - 7 x^{2} + 36

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

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

Frequently Asked Questions (FAQ)

What is the range of y=-2x^2+8x-1 ?
The range of y=-2x^2+8x-1 is f(x)<= 7