QUESTION 12 The graph of a hyperbola with equation \( y=f(x) \) bas the following properties: - Domain: \( x \in R, x \neq 5 \) - Renge: \( y \in R, y \neq 1 \) - Passes through the point (2:0) Determine \( f(x) \)

To determine the function \( f(x) \) for the hyperbola given the properties, we can start by considering the general form of a hyperbola. The standard form of a hyperbola can be expressed as: \[ \frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 \] or \[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \] However, since we have specific domain and range restrictions, we can assume a more suitable form for our hyperbola. ### Step 1: Identify the form of the hyperbola Given that the domain excludes \( x = 5 \) and the range excludes \( y = 1 \), we can express the function in the form: \[ f(x) = \frac{A}{x - 5} + 1 \] This form ensures that: - The function is undefined at \( x = 5 \) (which matches the domain condition). - The function approaches \( y = 1 \) as \( x \) approaches \( 5 \) from either side, but never actually reaches \( y = 1 \) (which matches the range condition). ### Step 2: Use the point (2, 0) to find \( A \) Now, we will use the point (2, 0) to find the value of \( A \): \[ f(2) = \frac{A}{2 - 5} + 1 = 0 \] This simplifies to: \[ \frac{A}{-3} + 1 = 0 \] Solving for \( A \): \[ \frac{A}{-3} = -1 \] \[ A = 3 \] ### Step 3: Write the final function Now that we have determined \( A \), we can write the function: \[ f(x) = \frac{3}{x - 5} + 1 \] ### Conclusion Thus, the function \( f(x) \) that satisfies all the given conditions is: \[ f(x) = \frac{3}{x - 5} + 1 \]