Equation: x² + y 2 = 1.
Solve the given equation using the “assumption method” let x = 2.  Sketch it after solving.Then determine whether y is a function of x.

Let's go through the problem step by step.

### Step 1: Given Equation
The equation provided is:

\[
x^2 + y^2 = 1
\]

### Step 2: Apply the Assumption Method (Let \( x = 2 \))
We are asked to assume \( x = 2 \) and then solve for \( y \).

Substituting \( x = 2 \) into the equation:

\[
2^2 + y^2 = 1
\]

Simplifying further:

\[
4 + y^2 = 1
\]

### Step 3: Solve for \( y \)
Now, solve for \( y \):

\[
y^2 = 1 - 4
\]
\[
y^2 = -3
\]

Taking the square root of both sides:

\[
y = \pm\sqrt{-3}
\]

Since the square root of a negative number is not real, \( y \) would be an imaginary number in this case:

\[
y = \pm i\sqrt{3}
\]

### Step 4: Sketch the Graph
The original equation \( x^2 + y^2 = 1 \) represents a circle with a radius of 1 centered at the origin (0, 0).

However, substituting \( x = 2 \) into the equation leads to a value of \( y \) that is not real, indicating that there are no real points on this circle corresponding to \( x = 2 \).

### Step 5: Determine if \( y \) is a Function of \( x \)
For \( y \) to be a function of \( x \), each \( x \) value must correspond to exactly one \( y \) value. However, the equation \( x^2 + y^2 = 1 \) typically corresponds to two values of \( y \) for each \( x \) within the domain \([-1, 1]\) (except at \( x = \pm 1 \)).

Since \( y = \pm \sqrt{1 - x^2} \), \( y \) is not a function of \( x \) because it does not pass the vertical line test—there are two values of \( y \) for most values of \( x \) in the domain.

### Summary
- **No real solution** exists for \( y \) when \( x = 2 \).
- The graph of \( x^2 + y^2 = 1 \) is a circle centered at the origin with a radius of 1.
- \( y \) is not a function of \( x \) because each \( x \) does not map to a single \( y \) value.

Would you like to see the detailed graph or have any other questions?

Here are 5 related questions to further explore the topic:
1. What would the graph of \( x^2 + y^2 = 1 \) look like if the radius were increased to 2?
2. How would the solution change if the assumption was \( x = 0.5 \)?
3. What is the general form of a circle's equation in the coordinate plane?
4. Can you find the points where the circle intersects the x-axis and y-axis?
5. What does the vertical line test mean in terms of determining if a relation is a function?

**Tip:** When determining if a relation is a function, always check if each input (x-value) corresponds to only one output (y-value).

Learn how to solve the equation x^2 + y^2 = 1 using the assumption method with detailed step-by-step solution. This problem covers quadratic equations, circle equations, and the concept of imaginary numbers. Suitable for high school students.

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