QUESTION 3 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 formed when \( f \) is transformed as follows: \( 3.1 \quad \) Shift two units to the left \( 3.2 \quad \) Shift 3 units up \( 3.3 \quad \) Shift 1 unit right and 2 units down \( 3.4 \quad \) The equation of the new hyperbola has new asymptotes at \( x=-4 \) and \( y=-1 \)

To transform the function \( f(x)=\frac{3}{x-7}-4 \) according to the specified shifts, we can break it down step by step. \( 3.1 \quad \) Shifting two units to the left means we replace \( x \) with \( x + 2 \): \[ f_1(x) = \frac{3}{(x + 2) - 7} - 4 = \frac{3}{x - 5} - 4. \] \( 3.2 \quad \) Shifting three units up involves adding 3 to the entire function: \[ f_2(x) = \frac{3}{x - 5} - 4 + 3 = \frac{3}{x - 5} - 1. \] \( 3.3 \quad \) Shifting 1 unit right and 2 units down requires us to replace \( x \) with \( x - 1 \) and subtracting 2: \[ f_3(x) = \frac{3}{(x - 1) - 5} - 1 - 2 = \frac{3}{x - 6} - 3. \] \( 3.4 \quad \) To adjust the asymptotes to \( x = -4 \) and \( y = -1 \), we rewrite the hyperbola: The standard form is \( f(x) = \frac{a}{x-h} + k \), where \( h \) is the x-coordinate of the vertical asymptote and \( k \) is the y-coordinate of the horizontal asymptote. Thus, the equation becomes: \[ f(x) = \frac{3}{x + 4} - 1. \] So the final answers are as follows: - \( 3.1: \quad f_1(x) = \frac{3}{x - 5} - 4 \) - \( 3.2: \quad f_2(x) = \frac{3}{x - 5} - 1 \) - \( 3.3: \quad f_3(x) = \frac{3}{x - 6} - 3 \) - \( 3.4: \quad f_4(x) = \frac{3}{x + 4} - 1 \)