AMC 10 · 2006 · #12

Grade 9 algebra
Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity
completing-the-squarequadratic-equationscoordinate-geometry convert-to-algebra ↑ Prerequisites: quadratic-equations
📏 Medium solution 💡 3 insights
Problem
A parabola has its vertex at a point whose coordinates are equal, and crosses the vertical axis at the negative of that value. Find the middle coefficient.

Pick an answer.

(A)
-p
(B)
0
(C)
2
(D)
4
(E)
p
How to solve
Strategy Organize Information in More Ways

Both givens are statements about the graph — where the vertex sits and where the curve meets the y-axis — but the form y=ax²+bx+c hides the vertex completely. Tool #15 (Organize Information in More Ways) rewrites the same parabola as y=a(x-p)²+p, where the vertex is written into the formula and only one unknown, a, is left. Tool #4 (Introduce a Variable) then keeps the bookkeeping honest: expanding shows b=-2ap, so b depends on a and p only through their product, and finding that single product is the real goal. Tool #11 (Work Backwards) supplies it — the y-intercept is a fact about the finished curve, and running it backwards through the vertex form gives ap. Finally tool #6 (Guess and Check) closes the gap that this kind of argument usually leaves open: the algebra only shows what b must be if such a parabola exists, so at the end the candidate curve is built and tested against both conditions.

1STEP 1

Rewrite with the vertex visible

Rewriting makes the vertex visible in the formula.

y=ax²+bx+c=a(x+b/2a)²+c-b²/4a=a(x-p)²+p
2STEP 2

See what b is made of

Expanding shows the middle coefficient is a single product.

a(x-p)²+p=ax²-2apx+(ap²+p) → b=-2ap
3STEP 3

Cash in the y-intercept

The crossing point delivers exactly that product.

ap²+p=-p → ap²=-2p p ≠ 0{⟹} ap=-2
4STEP 4

Put the two facts together

Combining gives the coefficient 4.

b=-2(ap)=-2(-2)=4
5STEP 5

Check the parabola really exists

An explicit parabola shows this really happens, so the answer is 4, choice (D).

y=-2/px²+4x-p: y(0)=-p, -b/2a=p, y(p)=-2p+4p-p=p → (D)
Answer
4
Every number in the problem is written with p, so an answer with no p in it deserves a test. Take p=1: then a=-2 and y=-2x²+4x-1, whose vertex is (1,1) and whose y-intercept is -1, with b=4. Now take p=-3: then a=2/3 and y=2/3x²+4x+3, whose turning point is at x=-4/2(2/3)=-3 with height 2/3(9)-12+3=-3, and whose y-intercept is 3=-p; again b=4. Two very different parabolas, same b, which is what a constant answer predicts. That comparison also disposes of the p-flavoured choices: (A) -p would have to be -1 in the first case and 3 in the second, and (E) p would have to be 1 then -3, while the actual b never moves. Choices (B) 0 and (C) 2 already fail at p=1. A last sanity check on the shape: for p ＞ 0 the vertex (p,p) sits above the intercept (0,-p), so the curve must open downward, and indeed a=-2/p ＜ 0.
💡Key takeaway

Rewrite the parabola as y=a(x-p)²+p and b turns out to be -2ap, so a and p never matter on their own — the y-intercept forces the product ap=-2, and b=4 no matter which p you started with.

  • Rewrite with the vertex visible
  • See what b is made of
  • Cash in the y-intercept
  • Put the two facts together
  • Check the parabola really exists

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