To find the inverse of the relation given by:

$x = y^2 - 2y$

we need to solve for $y$ in terms of $x$ and then interchange $x$ and $y$ to find the inverse.

Step 1: Rewrite the Equation in Terms of $y$

The original equation is: $x = y^2 - 2y$

Rearrange it to form a quadratic equation in terms of $y$: $y^2 - 2y - x = 0$

Step 2: Solve for $y$ Using the Quadratic Formula

The quadratic formula is: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For our equation $y^2 - 2y - x = 0$:

• $a = 1$
• $b = -2$
• $c = -x$

Substitute these values: $y = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-x)}}{2 \cdot 1}$ $y = \frac{2 \pm \sqrt{4 + 4x}}{2}$ $y = \frac{2 \pm 2\sqrt{1 + x}}{2}$ $y = 1 \pm \sqrt{1 + x}$

Step 3: Determine the Branch of the Solution

Since we are solving for the inverse, we select only one branch of the solution. Let’s assume $y = 1 + \sqrt{1 + x}$ (the other branch would represent another solution that might not satisfy the inverse relation contextually).

Step 4: Interchange $x$ and $y$ for the Inverse

Now, let $y = x$ and $x = 1 + \sqrt{1 + y}$, so: $y = 1 + \sqrt{1 + x}$

Answer Selection

The correct answer is: None of the choices match this exact equation. Please confirm if the given answers need adjustment.

Would you like further clarification on any steps?

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1. How do you determine the correct branch in finding the inverse?
2. Why did we interchange $x$ and $y$ in the final answer?
3. What is the significance of the quadratic formula in finding inverses?
4. How do we interpret extraneous solutions in inverse functions?
5. What would happen if we chose the other branch?

Tip: Always check solutions against the context given in inverse functions, especially when square roots are involved.