What is the inverse of 4(x-7)^2+3,x<= 7 ?

The inverse of 4(x-7)^2+3,x<= 7 is (sqrt(x-3))/2+7,-(sqrt(x-3))/2+7

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

inverse of 4(x-7)^2+3,x<= 7

inverse

4 (x - 7)^{2} + 3, x \leq 7

\frac{x - 3}{2} + 7, - \frac{x - 3}{2} + 7

Solution steps

y = 4 (x - 7)^{2} + 3

Replace

x

with

y

x = 4 (y - 7)^{2} + 3

Solve for

y, x = 4 (y - 7)^{2} + 3

\frac{x - 3}{2} + 7, - \frac{x - 3}{2} + 7

What is the inverse of 4(x-7)^2+3,x<= 7 ?
The inverse of 4(x-7)^2+3,x<= 7 is (sqrt(x-3))/2+7,-(sqrt(x-3))/2+7