6 Let \( f \) be the function defined by \( f(x)=\frac{a x+b}{c x+d} \) and its representative curve. Calculate the real numbers \( a, b, c \) and \( d \) so that (H) passes through the points \( A(4 ; 1) \) and \( B(-2 ; 4) \) and admits the line of equation \( y=3 \) as asymptote.

To find the values of \( a, b, c, \) and \( d \), we know that the function \( f(x) = \frac{ax + b}{cx + d} \) needs to satisfy the following conditions: 1. The function passes through the points \( A(4, 1) \) and \( B(-2, 4) \). 2. The horizontal asymptote is \( y = 3 \). From the asymptote condition, we know that as \( x \to \infty \), \( f(x) \approx \frac{a}{c} \). For this to equal 3, we have the equation: \[ \frac{a}{c} = 3 \quad \Rightarrow \quad a = 3c. \] Now we'll substitute this relationship into the equations derived from the points A and B. **Using Point A (4, 1):** \[ f(4) = \frac{3c(4) + b}{c(4) + d} = 1, \] which leads to: \[ 12c + b = 4c + d. \] Rearranging gives us: \[ 8c + b - d = 0 \quad \text{(1)}. \] **Using Point B (-2, 4):** \[ f(-2) = \frac{3c(-2) + b}{c(-2) + d} = 4, \] so we have: \[ -6c + b = -2c + d. \] Rearranging gives us: \[ -4c + b - d = 0 \quad \text{(2)}. \] Now we can solve the system of equations formed by (1) and (2): From (1): \[ b = d - 8c, \] Substituting into (2): \[ -4c + (d - 8c) - d = 0, \] which simplifies to: \[ -4c - 8c = 0 \quad \Rightarrow \quad -12c = 0. \] This implies \( c = 0 \), which isn’t viable since it would make \( f(x) \) undefined. Next, let's directly encode \( b \) and \( d \) in terms of \( c \) from the asymptote condition. From the above solutions, we can express both equations in \( d \): Using (1) and solving for \( d \): \[ d = 8c + b, \] and substituting \( d \) into (2): \[ -4c + b - (8c + b) = 0 \Rightarrow -4c - 8c = 0 \implies b = 12c. \] Now with \( b = 12c \): Substituting \( b = 12c \) back into the equation for \( d \): \[ d = 8c + 12c = 20c. \] So we have \( a = 3c, b = 12c, d = 20c \). Now we just need arbitrary values for \( c \) to find specific values. If we choose \( c = 1 \): - \( a = 3(1) = 3 \) - \( b = 12(1) = 12 \) - \( d = 20(1) = 20 \) Thus, our coefficients are: \[ a = 3, \quad b = 12, \quad c = 1, \quad d = 20. \] The final function is: \[ f(x) = \frac{3x + 12}{x + 20}. \]