\( 3 \quad y = \frac { 3 } { x } \)

**Step 1. Write the equation** We are given the function \[ y = \frac{3}{x}. \] **Step 2. Identify the domain** Since division by zero is undefined, we must have \[ x \neq 0. \] Thus, the domain of the function is all real numbers except \(x = 0\). **Step 3. Determine the vertical asymptote** Because the function is undefined at \(x = 0\), there is a vertical asymptote at \[ x = 0. \] **Step 4. Determine the horizontal asymptote** Rewrite the function as \[ y = \frac{3}{x}. \] For large values (in absolute value) of \(x\), \(y\) approaches 0. Hence, there is a horizontal asymptote at \[ y = 0. \] **Step 5. Find intercepts** - **\(y\)-intercept:** To find the \(y\)-intercept, set \(x = 0\). However, \(x = 0\) is not in the domain, so there is no \(y\)-intercept. - **\(x\)-intercept:** To find the \(x\)-intercept, set \(y = 0\): \[ 0 = \frac{3}{x}. \] Since \(\frac{3}{x}\) can never equal 0 for any finite \(x\), there is no \(x\)-intercept. **Step 6. Analyze the behavior of the function** The function \(y = \frac{3}{x}\) represents a hyperbola. It has two branches: - For \(x > 0\), \(y\) is positive. - For \(x < 0\), \(y\) is negative. **Step 7. Plot some key points** To understand the shape of the graph, consider a few sample values for \(x\): - If \(x = 1\), then \(y = \frac{3}{1} = 3\). - If \(x = 3\), then \(y = \frac{3}{3} = 1\). - If \(x = -1\), then \(y = \frac{3}{-1} = -3\). - If \(x = -3\), then \(y = \frac{3}{-3} = -1\). These points confirm that the branch in the first quadrant (when \(x > 0\) and \(y > 0\)) continuously approaches the horizontal asymptote \(y = 0\) as \(x\) increases and the vertical asymptote \(x = 0\) as \(x\) decreases, and similarly for the branch in the third quadrant. **Summary** The function \[ y = \frac{3}{x} \] has: - Domain: \(x \in \mathbb{R} \setminus \{0\}\), - Vertical asymptote: \(x = 0\), - Horizontal asymptote: \(y = 0\), - No \(x\)- or \(y\)-intercepts, - Two branches (one in the first quadrant and one in the third quadrant) representing a hyperbola.