What are the domain and range for 2(y-1)= - the square root of 1/3x-6

The domain is: [18,∞),{x|x≥18}

The range is: [0,∞),{y|y≥0}

Hmmm.

y = 1 - 1/(2√(3x-6))

I get a horizontal asymptote at y=1, with the range (-∞,1)

To determine the domain and range of the given equation, we need to solve for both x and y. Let's start with solving for y.

We have the equation: 2(y-1) = -√(1/3x-6)

First, eliminate the coefficient of 2 by dividing both sides of the equation by 2:
(y-1) = -√(1/3x-6) / 2

Next, isolate y by adding 1 to both sides of the equation:
y = 1 - √(1/3x-6) / 2

Now that we have solved for y, we can determine its range. The range represents all the possible values that y can take. In this case, the square root function (√) can only produce non-negative values, so the expression inside the square root (1/3x-6) must be non-negative.

1/3x-6 ≥ 0

To solve for x, multiply both sides of the inequality by 3:
1 ≥ 3(1/3x-6)

Now, distribute the 3 on the right side:
1 ≥ x - 18

Add 18 to both sides:
19 ≥ x

Therefore, the domain for this equation is x ≤ 19.

Lastly, let's consider the range. Since we've determined that the expression inside the square root (1/3x-6) must be non-negative, it means that the square root (√) is always positive or zero. Therefore, the range for y is y ≥ 1.

In summary, the domain is x ≤ 19, and the range is y ≥ 1.