#10 · Quadratic table? Look for twins.
The question. The table shows three values of the quadratic function f. What is the value of f(4)?
| x | f(x) |
|---|---|
| 8 | −12 |
| 20 | −12 |
| 24 | 52 |
The slow way. Every quadratic can be written f(x) = ax² + bx + c. Plug in each row of the table and you get three equations:
Solve them, then plug in 4. It works, but there's a faster way.
Look for twins. Two inputs with the same output are twins. Here, 8 and 20 both give −12. A parabola is symmetric, so they sit equally far from its axis of symmetry, which is halfway between them: (8 + 20) ÷ 2 = 14.
Find the other twins. 4 is 10 to the left of 14, and 24 is 10 to the right, so 4 and 24 mirror each other too. That means f(4) = f(24), and the table already gives f(24) = 52.
So f(4) = 52. You never needed a, b, or c.
Or use Desmos
Enter the table (its columns are x₁ and y₁), then a regression on a new line: y₁ ~ ax₁² + bx₁ + c. Desmos finds a = 1, b = −28, c = 148. Then type a·4² + b·4 + c and Desmos shows 52. It works, but here spotting the twins is faster.
Go further: the same table, different questions
- What is the vertex's x-coordinate? Average the twins: (8 + 20) ÷ 2 = 14.
- What is the minimum value? Now you need a, which sets how stretched the parabola is. The twins give f(x) = a(x − 8)(x − 20) − 12 (slide the graph up 12 and the twins become zeros). The third row finds a: 64a − 12 = 52, so a = 1. Since a is positive the parabola opens upward, and f(14) = (6)(−6) − 12 = −48.
The halfway idea is a standard fact about parabolas: UWorld notes that the axis of symmetry is halfway between any two inputs with the same output, and Albert's quadratics review says the same.
The takeaway: quadratic table? Look for twins, two inputs with the same output. Average them to find the symmetry line, then look for the input the same distance away on the other side. The answer may already be in the table.