AMC 10 · 2009 · #9

Grade 9 algebra
function-evaluationfunction-compositionquadratic-equations work-backwardsconvert-to-algebra ↑ Prerequisites: function-evaluation
📏 Medium solution 💡 2 insights
Problem
A quadratic is described only through its values at shifted inputs. Find the sum of its three coefficients.

Pick an answer.

(A)
-1
(B)
0
(C)
1
(D)
2
(E)
3
How to solve
Strategy Change Focus / Count the Complement

Chasing a, b, c one at a time is not required. The sum a+b+c is exactly what the standard form returns at x = 1, so change focus from three unknown coefficients to the single number f(1). The given rule reports f at inputs of the form x+3, so work backwards to the x that makes x+3 equal 1. Then rebuild a, b, c by algebra as an independent route that also confirms such an f exists.

1STEP 1

Read a+b+c as one output

The coefficient sum is the value at one.

f(1) = a · 1² + b · 1 + c = a + b + c
2STEP 2

Work backwards to the right input

Working backwards through the shift finds the right input.

x + 3 = 1 ⟹ x = -2
3STEP 3

Evaluate the given quadratic at x = -2

Evaluating there gives 2.

f(1) = f(-2+3) = 3(-2)² + 7(-2) + 4 = 12 - 14 + 4 = 2
4STEP 4

Confirm such an f really exists

Such a quadratic really exists.

a(x+3)² + b(x+3) + c = ax² + (6a+b)x + (9a+3b+c) = 3x² + 7x + 4
5STEP 5

Solve for a, b, c and compare

Solving for the coefficients confirms 2, choice (A).

a = 3, b = 7 - 6a = -11, c = 4 - 9a - 3b = 10 → a + b + c = 3 - 11 + 10 = 2
Answer
2
Feed the reconstructed f back into the original statement: f(x+3) = 3(x+3)² - 11(x+3) + 10 = 3x² + 18x + 27 - 11x - 33 + 10 = 3x² + 7x + 4, exactly the given rule, and f(1) = 3 - 11 + 10 = 2. So the two routes agree, and they lean on different facts: the substitution route uses the identity at the single point x = -2, while the coefficient route uses it at every x and in exchange proves that such an f exists and is unique. Together they show a+b+c = 2 is forced, not merely consistent. The scale is plausible as well — all five choices are small integers and 2 sits among them.
💡Key takeaway

The sum of a polynomial's coefficients is just its value at x = 1, so find the one input that makes the given rule report f(1).

  • Read a+b+c as one output
  • Work backwards to the right input
  • Evaluate the given quadratic at x = -2
  • Confirm such an f really exists
  • Solve for a, b, c and compare