A family of parabolas have rules of the form \( y=a x^{2}+b x \). For the parabola in this family that passes through the point \( (1,6) \) and has one of its \( x \)-axis intercepts at \( (4,0) \) find the values of \( a \) and \( b \).

To find the values of \( a \) and \( b \) for the parabola given by the equation \( y = ax^2 + bx \), which passes through the point \( (1,6) \) and has an \( x \)-intercept at \( (4,0) \), we can follow these steps: 1. **Substituting the \( x \)-intercept back into the equation**: The parabola has an \( x \)-intercept at \( (4,0) \) so we substitute \( (4,0) \) into the equation: \[ 0 = a(4)^2 + b(4) \] Which simplifies to: \[ 0 = 16a + 4b \quad (1) \] 2. **Substituting the point \( (1,6) \)**: Now, we substitute \( (1,6) \) into the parabola's equation: \[ 6 = a(1)^2 + b(1) \] Simplifying gives us: \[ 6 = a + b \quad (2) \] Now we have a system of equations: 1. From equation (1): \( 16a + 4b = 0 \) 2. From equation (2): \( a + b = 6 \) 3. **Solving the first equation for \( b \)**: We can express \( b \) in terms of \( a \) from equation (1): \[ 4b = -16a \quad \Rightarrow \quad b = -4a \quad (3) \] 4. **Substituting equation (3) into equation (2)**: Substitute \( b = -4a \) into \( a + b = 6 \): \[ a - 4a = 6 \quad \Rightarrow \quad -3a = 6 \quad \Rightarrow \quad a = -2 \] 5. **Finding \( b \) using \( a \)**: Substitute \( a = -2 \) back into equation (3): \[ b = -4(-2) = 8 \] Thus, the values of \( a \) and \( b \) are: \[ \boxed{a = -2, b = 8} \]