Let \( f(x)=a x^{2}+b x+c \) be the quadratic function whose graph goes through the points \( (3,28),(-4,70) \), and \( (-3,40) \). Find \( f(x) \) by solving the system for \( a, b \), and \( c \). The parabola fitting these three points is \( f(x)=\square \)

To find the coefficients \( a \), \( b \), and \( c \) for the quadratic function \( f(x) = ax^2 + bx + c \) that passes through the points \( (3, 28) \), \( (-4, 70) \), and \( (-3, 40) \), we can set up a system of equations using these points. 1. **Using the point \( (3, 28) \)**: \( 28 = a(3^2) + b(3) + c \) This simplifies to: \( 9a + 3b + c = 28 \) --- (1) 2. **Using the point \( (-4, 70) \)**: \( 70 = a(-4^2) + b(-4) + c \) This simplifies to: \( 16a - 4b + c = 70 \) --- (2) 3. **Using the point \( (-3, 40) \)**: \( 40 = a(-3^2) + b(-3) + c \) This simplifies to: \( 9a - 3b + c = 40 \) --- (3) Now we have the following system of equations: \[ \begin{align*} 9a + 3b + c &= 28 \quad (1) \\ 16a - 4b + c &= 70 \quad (2) \\ 9a - 3b + c &= 40 \quad (3) \\ \end{align*} \] Next, we can eliminate \( c \) from these equations. Subtract equation (1) from equations (2) and (3): From (2) - (1): \[ (16a - 4b + c) - (9a + 3b + c) = 70 - 28 \] Which simplifies to: \[ 7a - 7b = 42 \implies a - b = 6 \quad (4) \] From (3) - (1): \[ (9a - 3b + c) - (9a + 3b + c) = 40 - 28 \] Which simplifies to: \[ -6b = 12 \implies b = -2 \quad (5) \] Substituting \( b = -2 \) into equation (4): \[ a - (-2) = 6 \implies a + 2 = 6 \implies a = 4 \] Now substituting \( a = 4 \) and \( b = -2 \) back into one of the original equations to find \( c \). We can use equation (1): \[ 9(4) + 3(-2) + c = 28 \] \[ 36 - 6 + c = 28 \] \[ c = 28 - 30 = -2 \] Now we have \( a = 4 \), \( b = -2 \), and \( c = -2 \). Therefore, the quadratic function is: \[ f(x) = 4x^2 - 2x - 2 \] The parabola fitting these three points is \( f(x)=4x^{2}-2x-2 \).