Once you have an equation of the form $(y - 2)^2 = \text{(something in } z\text{)}$, how do you undo the square? Remember that taking a square root usually gives two possibilities.

After taking square roots, you will get two possible expressions for $y$. Think about what $y \ge 2$ tells you about the sign of $y - 2$, and which of the two expressions it allows.

In Desmos, let the horizontal axis represent $z$ and the vertical axis represent $y$. Type x = 3(y - 2)^2 to graph the given relationship, treating $x$ as $z$.

For each option, replace $z$ with $x$ and enter it into Desmos:

• A: y = 2 - sqrt(x/3)
• B: y = sqrt(3x) + 2
• C: y = sqrt(x)/3 + 2
• D: y = 2 + sqrt(x/3) This will plot four candidate curves for $y$ in terms of $x$.

Look at how each candidate graph lines up with the parabola x = 3(y - 2)^2. The correct equation for $y$ will exactly trace the part of the parabola that satisfies $y \ge 2$; the others will either not lie on the parabola or will trace the wrong branch.

Start with the given relationship:

$z = 3(y - 2)^2$.

Divide both sides by 3 to get the squared term by itself:

Now you have an equation where $(y - 2)^2$ is expressed in terms of $z$.

To solve for $y$, undo the square by taking the square root of both sides of

Taking the square root gives

which is the same as

So there are two possible equations for $y$:

• $y = 2 + \sqrt{\dfrac{z}{3}}$
• $y = 2 - \sqrt{\dfrac{z}{3}}$

Next, use the condition on $y$ to decide which one actually works.

Because $y \ge 2$, we know $y - 2 \ge 0$. That means $y - 2$ must be nonnegative.

From

the expression $\sqrt{\dfrac{z}{3}}$ is always nonnegative. So the only way for $y - 2$ to be nonnegative is to take the positive square root:

Add 2 to both sides:

So the correct equation expressing $y$ in terms of $z$ (with $y \ge 2$) is $y = 2 + \sqrt{\dfrac{z}{3}}$.