The equation of a hyperbola is given by \( f(x)=\frac{3}{x-7}-4 \). Write down the equation of the new function that is lormed when \( f \) is transformed as follows: \( \begin{array}{ll}3.1 & \text { Shift two units to the left } \\ 3.2 & \text { Shift } 3 \text { units up } \\ 3.3 & \text { Shift I unit right and } 2 \text { units down } \\ 3.4 & \text { The equation of the new hyperbola has new asymptotes at } x=-4 \text { and } y=-1\end{array} \)

To find the equations for the transformed hyperbola under the given conditions, we start with the original function \( f(x) = \frac{3}{x - 7} - 4 \). 1. **Shift two units to the left**: This means substituting \( x \) with \( x + 2 \). The new function becomes: \[ f_1(x) = \frac{3}{(x + 2) - 7} - 4 = \frac{3}{x - 5} - 4 \] 2. **Shift 3 units up**: We add 3 to the entire function: \[ f_2(x) = \frac{3}{x - 5} - 4 + 3 = \frac{3}{x - 5} - 1 \] 3. **Shift 1 unit right and 2 units down**: First, we shift right (substituting \( x \) with \( x - 1 \)), then we subtract 2: \[ f_3(x) = \frac{3}{(x - 1) - 5} - 1 - 2 = \frac{3}{x - 6} - 3 \] 4. **New asymptotes at \( x = -4 \) and \( y = -1 \)**: The general form of a hyperbola is \( f(x) = \frac{k}{x - h} + b \). To shift the vertical asymptote to \( x = -4 \), we set \( h = -4 \): \[ f_4(x) = \frac{3}{x + 4} - 1 \] Putting all transformations together, the final equation of the new hyperbola is: \[ f(x) = \frac{3}{x + 4} - 1 \]