Solution: Assume $ h(x) = ax^2 + bx + c $. Using $ h(1) = a + b + c = 4 $, $ h(2) = 4a + 2b + c = 11 $, $ h(3) = 9a + 3b + c = 22 $. Subtract first equation from second: $ 3a + b = 7 $. Subtract second from third: $ 5a + b = 11 $. Subtract these: $ 2a = 4 \Rightarrow a = 2 $. Then $ 3(2) + b = 7 \Rightarrow b = 1 $. From $ 2 + 1 + c = 4 \Rightarrow c = 1 $. Thus, $ h(x) = 2x^2 + x + 1 $, and $ h(4) = 32 + 4 + 1 = 37 $. \boxed{37}

["How to Solve Quadratic Equations Using System of Equations: A Step-by-Step Example", "Solving quadratic equations is a fundamental skill in algebra, but when presented with multiple values of a quadratic function at different points, a system of equations offers an efficient and powerful solution method. This article explores how to determine a quadratic function $ h(x) = ax^2 + bx + c $ using known outputs, using subtraction and elimination techniques—providing both clarity and a practical example sure to boost your problem-solving confidence.", "---", "### Understanding the Problem", "We begin with a quadratic function of the form:\n$$ h(x) = ax^2 + bx + c $$\nWe’re given three key values:\n- $ h(1) = 4 $\n- $ h(2) = 11 $\n- $ h(3) = 22 $", "Our goal is to find the coefficients $ a $, $ b $, and $ c $, then use the function to compute $ h(4) = 37 $. This approach leverages the concept that a quadratic is uniquely determined by three distinct points.", "---", "### Step-by-Step Solution Using Subtraction", "To eliminate variables and isolate one coefficient at a time, start by subtracting equations:", "1. Use $ h(2) - h(1) $:\n$$\n(4a + 2b + c) - (a + b + c) = 11 - 4\n\Rightarrow 3a + b = 7 \quad \ ext{(Equation A)}\n$$", "2. Use $ h(3) - h(2) $:\n$$\n(9a + 3b + c) - (4a + 2b + c) = 22 - 11\n\Rightarrow 5a + b = 11 \quad \ ext{(Equation B)}\n$$", "Now subtract Equation A from Equation B:\n$$\n(5a + b) - (3a + b) = 11 - 7\n\Rightarrow 2a = 4 \Rightarrow a = 2\n$$", "With $ a = 2 $, substitute back into Equation A:\n$$\n3(2) + b = 7 \Rightarrow 6 + b = 7 \Rightarrow b = 1\n$$", "Now use $ h(1) = a + b + c = 4 $:\n$$\n2 + 1 + c = 4 \Rightarrow c = 1\n$$", "---", "### Final Expression and Evaluation", "Having determined all coefficients:\n$$\nh(x) = 2x^2 + x + 1\n$$", "Evaluate at $ x = 4 $:\n$$\nh(4) = 2(4)^2 + 4 + 1 = 2(16) + 4 + 1 = 32 + 4 + 1 = 37\n$$", "So, $ \boxed{h(4) = 37} $", "---", "### Why This Method Works", "Quadratic functions are second-degree polynomials with three unknowns ($ a, b, c $). Knowing $ h(x) $ at three distinct values creates a system with three independent equations. Subtracting equations removes the constant $ c $, reducing the system step-by-step until a single unknown is isolated. This method highlights logical algebra and computational strategy—essential tools for mastering functional equations.", "---", "### Bringing It to Practice", "Understanding how to derive and evaluate quadratic functions from sample points empowers students and problem-solvers alike. This structured approach transforms abstract algebra into a clear, actionable process—proving that even complex equations can be untangled with patience and precision.", "Try it yourself: pick any quadratic function, generate three outputs, and solve for its coefficients using system elimination. The satisfaction of arriving at $ h(4) = 37 $ (or any other value) will reinforce these core algebra skills."]