Key Concepts and Formulas

• Quadratic Function: A function of the form $f(x) = ax^2 + bx + c$, where $a, b, c$ are constants and $a \neq 0$.
• Evaluating a Function: Substituting a specific value for the variable (e.g., $x$) into the function's expression to find the corresponding output.
• System of Linear Equations: A set of two or more linear equations that share the same variables. Solving such a system involves finding values for the variables that satisfy all equations simultaneously. Methods include substitution and elimination.

Step-by-Step Solution

Step 1: Express $f(x)$ at the given points and set up initial equations.

We are given the quadratic function $f(x) = ax^2 + bx + c$. We need to evaluate this function at the points $x=0, 1, -2,$ and $3$.

• At $x=0$: $f(0) = a(0)^2 + b(0) + c = c$.
• At $x=1$: $f(1) = a(1)^2 + b(1) + c = a+b+c$.
• At $x=-2$: $f(-2) = a(-2)^2 + b(-2) + c = 4a-2b+c$.
• At $x=3$: $f(3) = a(3)^2 + b(3) + c = 9a+3b+c$.

We are given specific values for some of these evaluations:

• $f(1) = 3 \implies a+b+c = 3$ (Equation 1)
• $f(3) = 4 \implies 9a+3b+c = 4$ (Equation 2)
• $f(-2) = \lambda \implies 4a-2b+c = \lambda$ (Equation 3)

Step 2: Utilize the sum condition to establish a relationship between $c$ and $\lambda$.

We are given that $f(0)+f(1)+f(-2)+f(3)=14$. Substitute the known values and expressions: $c + 3 + \lambda + 4 = 14$ Combine the constant terms: $c + \lambda + 7 = 14$ Isolate $c + \lambda$: $c + \lambda = 14 - 7$ $c + \lambda = 7$ (Equation 4)

This equation is key because it directly relates $c$ and $\lambda$, the value we want to find. We can express $\lambda$ as $\lambda = 7-c$.

Step 3: Form a system of linear equations in $a, b, c$ and solve it.

From Equation 3, substitute $\lambda = 7-c$: $4a-2b+c = 7-c$ Rearrange this equation to get a linear equation in $a, b, c$: $4a-2b+2c = 7$ (Equation 5)

Now we have a system of three linear equations with three unknowns ($a, b, c$):

1. $a+b+c = 3$
2. $9a+3b+c = 4$
3. $4a-2b+2c = 7$

We will use elimination to solve this system.

• Eliminate $c$ from Equations 1 and 2: Subtract Equation 1 from Equation 2: $(9a+3b+c) - (a+b+c) = 4 - 3$ $8a+2b = 1$ (Equation 6)

• Eliminate $c$ from Equations 1 and 5: Multiply Equation 1 by 2: $2(a+b+c) = 2(3) \implies 2a+2b+2c = 6$ (Equation 1') Subtract Equation 1' from Equation 5: $(4a-2b+2c) - (2a+2b+2c) = 7 - 6$ $2a-4b = 1$ (Equation 7)

Now we have a system of two linear equations with two unknowns ($a, b$): 6. $8a+2b = 1$ 7. $2a-4b = 1$

• Solve for $a$ and $b$: Multiply Equation 6 by 2 to make the coefficient of $b$ equal and opposite to that in Equation 7: $2(8a+2b) = 2(1) \implies 16a+4b = 2$ (Equation 6') Add Equation 7 and Equation 6': $(2a-4b) + (16a+4b) = 1 + 2$ $18a = 3$ $a = \frac{3}{18} = \frac{1}{6}$

  Substitute $a = \frac{1}{6}$ into Equation 6: $8\left(\frac{1}{6}\right) + 2b = 1$ $\frac{4}{3} + 2b = 1$ $2b = 1 - \frac{4}{3}$ $2b = -\frac{1}{3}$ $b = -\frac{1}{6}$

• Solve for $c$: Substitute $a = \frac{1}{6}$ and $b = -\frac{1}{6}$ into Equation 1: $\frac{1}{6} + \left(-\frac{1}{6}\right) + c = 3$ $0 + c = 3$ $c = 3$

Step 4: Calculate $\lambda$ using the relationship found in Step 2.

From Equation 4, we have $c + \lambda = 7$. We found that $c=3$. Substitute this value: $3 + \lambda = 7$ $\lambda = 7 - 3$ $\lambda = 4$

Common Mistakes & Tips

• Algebraic Errors: Carefully check each step of algebraic manipulation, especially when dealing with signs and fractions. A single error can propagate and lead to an incorrect final answer.
• Misinterpreting Conditions: Ensure all given conditions ($f(1)=3$, $f(-2)=\lambda$, $f(3)=4$, and the sum equation) are translated correctly into equations involving $a, b, c$, and $\lambda$.
• Using Equation 4 Early: Equation 4, $c+\lambda=7$, is a direct consequence of the sum condition. It's often more efficient to use this relationship to simplify the system of equations involving $a, b, c$ rather than trying to solve for $a, b, c$ first and then using the sum condition at the very end.

Summary

The problem required us to find the value of $\lambda$, which represents $f(-2)$ for a quadratic function $f(x) = ax^2 + bx + c$. We used the given function values at $x=1$ and $x=3$ to form two equations. The condition $f(-2)=\lambda$ provided a third equation. Crucially, the sum of function values at four points ($0, 1, -2, 3$) yielded a direct relationship between $c$ and $\lambda$ ($c+\lambda=7$). By substituting this relationship into the equation for $f(-2)$, we obtained a system of three linear equations in $a, b, c$. Solving this system gave us the values of $a, b,$ and $c$. Finally, using the value of $c$ and the relationship $c+\lambda=7$, we determined $\lambda$ to be 4.

The final answer is $\boxed{4}$.