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)=x^2-8x+12

Solution

range

f (x) = x^{2} - 8 x + 12

Solution

f (x) \geq - 4

+1

Interval Notation

[- 4, \infty)

Solution steps

Vertex of

x^{2} - 8 x + 12 :

Minimum

(4, - 4)

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, - 4)

f (x) \geq - 4

Popular Examples

inverse of f(x)= x/(x-20)inverse

f (x) = \frac{x}{x - 20}

parity f(x)=2x-tan(x)parity

f (x) = 2 x - tan (x)

inverse of f(x)= 5/x+4 inverse

f (x) = \frac{5}{x} + 4

inverse of y=(5x)/(2x+3)inverse

y = \frac{5 x}{2 x + 3}

extreme 2x^3-9x^2-24x+30 extreme points

2 x^{3} - 9 x^{2} - 24 x + 30

Frequently Asked Questions (FAQ)

What is the range of f(x)=x^2-8x+12 ?
The range of f(x)=x^2-8x+12 is f(x)>=-4