AMC 10 · 2006 · #8
Grade 8 algebraA parabola with equation y=x2+bx+c passes through the points (2,3) and (4,3). What is c?
Pick an answer.
AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: A parabola with equation $y = x^2 + bx + c$ passes through the two points $(2,3)$ and $(4,3)$. The leading coefficient is fixed at $1$, but $b$ and $c$ are unknown constants. Find the value of the constant term $c$.
Givens: The parabola has equation $y = x^2 + bx + c$ (leading coefficient fixed at $1$); It passes through the point $(2,3)$; It passes through the point $(4,3)$; Answer choices: (A) $2$, (B) $5$, (C) $7$, (D) $10$, (E) $11$
Unknowns: The constant term $c$ (with $b$ found along the way)
Understand
Restated: A parabola with equation $y = x^2 + bx + c$ passes through the two points $(2,3)$ and $(4,3)$. The leading coefficient is fixed at $1$, but $b$ and $c$ are unknown constants. Find the value of the constant term $c$.
Givens: The parabola has equation $y = x^2 + bx + c$ (leading coefficient fixed at $1$); It passes through the point $(2,3)$; It passes through the point $(4,3)$; Answer choices: (A) $2$, (B) $5$, (C) $7$, (D) $10$, (E) $11$
Plan
Primary tool: #4 Introduce a Variable
Secondary: #7 Identify Subproblems, #3 Eliminate Possibilities
The equation already carries two unknown constants, $b$ and $c$, so Tool #4 (Introduce a Variable) says: treat them as the quantities to solve for and turn each passing-through condition into an equation. That gives two equations in two unknowns. Tool #7 (Identify Subproblems) breaks the job into stages — first eliminate $c$ to pin down $b$, then use $b$ to recover $c$ — instead of attacking both at once. Tool #3 (Eliminate Possibilities) is a safety net: $c$ must be one of the five listed values, so any answer outside the choices signals an arithmetic slip.
Execute — Answer: E
6.EE.A.2 Step 1 Turn each point into an equation
- A point is on the parabola exactly when its coordinates satisfy the equation, so substitute each point's $x$ and $y$.
- For $(2,3)$: $3 = 2^2 + 2b + c = 4 + 2b + c$, which tidies to $2b + c = -1$.
- For $(4,3)$: $3 = 4^2 + 4b + c = 16 + 4b + c$, which tidies to $4b + c = -13$.
- Now there are two equations in the two unknowns $b$ and $c$.
💡 A point sits on a curve precisely when plugging its coordinates into the equation makes both sides equal.
8.EE.C.8 Step 2 Subtract to cancel c and find b
- Both equations contain a lone $+c$, so subtracting the first from the second makes the $c$ terms disappear and leaves only $b$: $(4b + c) - (2b + c) = -13 - (-1)$ gives $2b = -12$, so $b = -6$.
- Eliminating one unknown this way turns the pair of equations into a single easy one.
💡 When two equations share the same $+c$, subtracting them wipes $c$ out and isolates the other unknown.
8.EE.C.8 Step 3 Back-substitute to get c
- Put $b = -6$ back into the first equation $2b + c = -1$: $2(-6) + c = -1$, so $-12 + c = -1$ and therefore $c = 11$.
- That value is on the list, matching choice (E).
💡 Once one unknown is pinned down, the original equation hands you the other in a single step.
6.EE.A.2 A point is on the parabola exactly when its coordinates satisfy the equation, so 8.EE.C.8 Both equations contain a lone $+c$, so subtracting the first from the second mak 8.EE.C.8 Put $b = -6$ back into the first equation $2b + c = -1$: $2(-6) + c = -1$, so $- Review
Reasonableness: Rebuild the parabola with $b = -6$ and $c = 11$: $y = x^2 - 6x + 11$. Test the two given points — at $x = 2$: $4 - 12 + 11 = 3$, and at $x = 4$: $16 - 24 + 11 = 3$ — both land on $y = 3$, exactly as required. The constant term $c = 11$ is the $y$-intercept, and since the vertex sits at $x = 3$ with height $9 - 18 + 11 = 2$, the curve rises from $2$ up to $11$ as $x$ moves from $3$ out to $0$, which is consistent. Value $11$ is choice (E).
Alternative: Skip finding $b$ by using the roots directly. The two points both have $y = 3$, so $x = 2$ and $x = 4$ are exactly the solutions of $x^2 + bx + c = 3$, i.e. of $x^2 + bx + (c - 3) = 0$. A monic quadratic with roots $2$ and $4$ is $(x - 2)(x - 4) = x^2 - 6x + 8$, so its constant term equals the product of the roots: $c - 3 = 2 \cdot 4 = 8$, giving $c = 11$ (and $b = -6$). Same answer (E).
CCSS standards used (min grade 8)
6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Evaluating the squares $2^2 = 4$ and $4^2 = 16$ when each point is substituted into $x^2 + bx + c$.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Substituting the coordinates of $(2,3)$ and $(4,3)$ into $y = x^2 + bx + c$ to produce an equation from each condition.)8.EE.C.8Analyze and solve pairs of simultaneous linear equations (Solving the system $2b + c = -1$ and $4b + c = -13$ by elimination and back-substitution to find $b$ and then $c$.)
⭐ Plug each point into the equation to get two equations, then subtract them so the shared $c$ cancels — that hands you $b$, and one more substitution gives $c$.
⭐ Plug each point into the equation to get two equations, then subtract them so the shared $c$ cancels — that hands you $b$, and one more substitution gives $c$.
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