Question: Solve the exercise by solving a system of equations. Find the quadratic equation of the form \( y…

Solve the exercise by solving a system of equations.

Find the quadratic equation of the form \( y = ax^2 + bx + c \) whose graph passes through the points \( (2, 3) \), \( (-2, 7) \), and \( (1, -5) \).

Solution

To find the quadratic equation \( y = ax^2 + bx + c \) that passes through the given points \((2, 3)\), \((-2, 7)\), and \((1, -5)\), we need to create and solve a system of equations. Start by substituting each point into the equation \( y = ax^2 + bx + c \): For the point \((2, 3)\), \[ 3 = 4a + 2b + c \] For the point \((-2, 7)\), \[ 7 = 4a - 2b + c \] For the point \((1, -5)\), \[ -5 = a + b + c \] Now we have the system of equations: \[ \begin{cases} 4a + 2b + c = 3 & \\ 4a - 2b + c = 7 & \\ a + b + c = -5 & \end{cases} \] Subtract the second equation from the first: \[ (4a + 2b + c) - (4a - 2b + c) = 3 - 7 \] \[ 4b = -4 \implies b = -1 \] Subtract the third equation from the second: \[ (4a - 2b + c) - (a + b + c) = 7 + 5 \] \[ 3a - 3b = 12 \] \[ a - b = 4 \] Substitute \( b = -1 \) into \( a - b = 4 \): \[ a + 1 = 4 \implies a = 3 \] Substitute \( a = 3 \) and \( b = -1 \) into the third equation: \[ 3 + (-1) + c = -5 \] \[ 2 + c = -5 \implies c = -7 \] Therefore, the quadratic equation is: \[ y = 3x^2 - x - 7 \]