What is the inverse of y=sqrt((x-1)^2-64)+2 ?

The inverse of y=sqrt((x-1)^2-64)+2 is 1+sqrt(x^2-4x+68),1-sqrt(x^2-4x+68)

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inverse of y=sqrt((x-1)^2-64)+2

inverse

y = (x - 1)^{2} - 64 + 2

1 + x^{2} - 4 x + 68, 1 - x^{2} - 4 x + 68

Solution steps

y = (x - 1)^{2} - 64 + 2

Replace

x

with

y

x = (y - 1)^{2} - 64 + 2

Solve for

y, x = (y - 1)^{2} - 64 + 2

1 + x^{2} - 4 x + 68, 1 - x^{2} - 4 x + 68

What is the inverse of y=sqrt((x-1)^2-64)+2 ?
The inverse of y=sqrt((x-1)^2-64)+2 is 1+sqrt(x^2-4x+68),1-sqrt(x^2-4x+68)