The points \( (-2,2) \) and \( (3,-23) \) lie on the curve given by the equation \( y=a x^{2}+b x+4 \). Use an algebraic method to find the values of \( a \) and \( b \).

To find the values of \( a \) and \( b \) in the equation \( y = ax^2 + bx + 4 \) using the given points \((-2,2)\) and \((3,-23)\), we can substitute the coordinates of these points into the equation and then solve the resulting system of equations. Substitute the coordinates of the first point \((-2,2)\) into the equation: \[ 2 = a(-2)^2 + b(-2) + 4 \] \[ 2 = 4a - 2b + 4 \] \[ 4a - 2b = -2 \] Substitute the coordinates of the second point \((3,-23)\) into the equation: \[ -23 = a(3)^2 + b(3) + 4 \] \[ -23 = 9a + 3b + 4 \] \[ 9a + 3b = -27 \] Now, we have a system of two equations: \[ 4a - 2b = -2 \] \[ 9a + 3b = -27 \] We can solve this system of equations to find the values of \( a \) and \( b \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}4a-2b=-2\\9a+3b=-27\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}a=\frac{-1+b}{2}\\9a+3b=-27\end{array}\right.\) - step2: Substitute the value of \(a:\) \(9\times \frac{-1+b}{2}+3b=-27\) - step3: Simplify: \(\frac{9\left(-1+b\right)}{2}+3b=-27\) - step4: Multiply both sides of the equation by LCD: \(\left(\frac{9\left(-1+b\right)}{2}+3b\right)\times 2=-27\times 2\) - step5: Simplify the equation: \(-9+15b=-54\) - step6: Move the constant to the right side: \(15b=-54+9\) - step7: Add the numbers: \(15b=-45\) - step8: Divide both sides: \(\frac{15b}{15}=\frac{-45}{15}\) - step9: Divide the numbers: \(b=-3\) - step10: Substitute the value of \(b:\) \(a=\frac{-1-3}{2}\) - step11: Simplify: \(a=-2\) - step12: Calculate: \(\left\{ \begin{array}{l}a=-2\\b=-3\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}a=-2\\b=-3\end{array}\right.\) - step14: Rewrite: \(\left(a,b\right) = \left(-2,-3\right)\) The solution to the system of equations is \( a = -2 \) and \( b = -3 \). Therefore, the values of \( a \) and \( b \) in the equation \( y = ax^2 + bx + 4 \) are \( a = -2 \) and \( b = -3 \).