AMC 10 · 2025 · #9
Grade 8 algebraLet f(x)=100x3−300x2+200x. For how many real numbers a does the graph of y=f(x−a) pass through the point (1,25)?
Pick an answer.
AMC 10 2025 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: Given $f(x) = 100x^3 - 300x^2 + 200x$, count how many real numbers $a$ make the shifted graph $y = f(x - a)$ pass through the point $(1, 25)$.
Givens: $f(x) = 100x^3 - 300x^2 + 200x$; The shifted curve is $y = f(x-a)$ for a real shift $a$; The shifted curve must pass through $(1, 25)$; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) more than $4$
Unknowns: The number of real values of $a$ that work
Understand
Restated: Given $f(x) = 100x^3 - 300x^2 + 200x$, count how many real numbers $a$ make the shifted graph $y = f(x - a)$ pass through the point $(1, 25)$.
Givens: $f(x) = 100x^3 - 300x^2 + 200x$; The shifted curve is $y = f(x-a)$ for a real shift $a$; The shifted curve must pass through $(1, 25)$; Answer choices: (A) $1$, (B) $2$, (C) $3$, (D) $4$, (E) more than $4$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #4 Introduce a Variable, #14 Extreme Principle
The count of valid $a$ is really a count of intersections, so the load-bearing move is to picture the curve. First Tool #4 (Introduce a Variable): setting $t = 1 - a$ turns "passes through $(1,25)$" into the clean question "how many real $t$ satisfy $f(t) = 25$?", and because $a = 1 - t$ is one-to-one the two counts are equal. Then Tool #1 (Draw a Diagram): factor $f$, mark its roots, and read its shape so the crossings with the horizontal line $y = 25$ can literally be seen. Finally Tool #14 (Extreme Principle): comparing the hump's peak height and the curve's end behavior against $25$ pins down exactly how many crossings occur — the whole answer hinges on those boundary heights.
Execute — Answer: C
8.F.A.1 Step 1 Turn the point condition into an equation
- The curve $y = f(x-a)$ passes through $(1, 25)$ exactly when its height at $x = 1$ equals $25$.
- Substituting $x = 1$ gives the single condition $f(1 - a) = 25$.
💡 "Passes through $(1,25)$" is just "the function's output at that input is $25$" — nothing more.
8.EE.C.7 Step 2 Rename the input to count cleanly
- Let $t = 1 - a$.
- Because $a = 1 - t$, every value of $a$ gives exactly one $t$ and every $t$ gives exactly one $a$ — a one-to-one match.
- So the number of valid $a$ equals the number of real $t$ with $f(t) = 25$.
💡 A left-right shift slides the graph but keeps every height it reaches, so counting $a$ is the same as counting where $f$ itself equals $25$.
6.EE.A.3 Step 3 Factor to find the curve's roots
- Factor the cubic to see its shape.
- Pull out $100t$, then factor the quadratic: $f(t) = 100t(t^2 - 3t + 2) = 100t(t-1)(t-2)$.
- So $f$ is zero at $t = 0,\ 1,\ 2$.
💡 Writing the cubic as a product of factors exposes exactly where it touches the $t$-axis.
8.F.B.5 Step 4 Read the sign pattern
- With roots $0, 1, 2$ and a positive leading coefficient, the graph is negative for $t < 0$, forms a positive hump on $(0,1)$, dips negative on $(1,2)$, then rises to $+\infty$ for $t > 2$.
- Since the target height $25$ is positive, crossings can only happen on the positive hump $(0,1)$ and on the final rise $t > 2$.
💡 A positive line like $y = 25$ can only meet the curve where the curve itself is positive.
6.EE.A.2 Step 5 Measure the hump against 25
- Check whether the hump on $(0,1)$ climbs past $25$.
- At the midpoint $t = \tfrac12$: $f\!\left(\tfrac12\right) = 100\cdot\tfrac12\cdot\left(-\tfrac12\right)\left(-\tfrac32\right) = 37.5$.
- Since $f(0)=0<25$, the hump rises to $37.5>25$, then falls back to $f(1)=0<25$.
- Going up past $25$ and back down gives exactly two crossings on $(0,1)$.
💡 If a hill starts and ends below $25$ but its peak is above $25$, the line $y=25$ must be crossed twice.
8.F.B.5 Step 6 Count the final rise, then total
- For $t > 2$, we have $f(2)=0<25$ and $f$ climbs without bound, so it passes $25$ exactly once.
- Adding the two hump crossings to this one gives $2 + 1 = 3$ real solutions $t$, hence $3$ real values of $a$.
- The answer is $\textbf{(C)}$.
💡 A branch that starts below $25$ and grows forever must cross the line once and only once.
8.F.A.1 The curve $y = f(x-a)$ passes through $(1, 25)$ exactly when its height at $x = 8.EE.C.7 Let $t = 1 - a$. Because $a = 1 - t$, every value of $a$ gives exactly one $t$ a 6.EE.A.3 Factor the cubic to see its shape. Pull out $100t$, then factor the quadratic: $ 8.F.B.5 With roots $0, 1, 2$ and a positive leading coefficient, the graph is negative f 6.EE.A.2 Check whether the hump on $(0,1)$ climbs past $25$. At the midpoint $t = \tfrac1 8.F.B.5 For $t > 2$, we have $f(2)=0<25$ and $f$ climbs without bound, so it passes $25$ Review
Reasonableness: A hard cap comes for free: $f(t)=25$ rearranges to the cubic $100t^3 - 300t^2 + 200t - 25 = 0$, and a degree-$3$ polynomial has at most $3$ real roots. That alone kills choices (D) $4$ and (E) more than $4$. We then showed all three roots are real — two on the hump, one past $t=2$ — and the peak margin is comfortable: $37.5$ sits well above $25$, so the two hump crossings are genuine and distinct, not a single tangent point. Everything is consistent with exactly $3$.
Alternative: Use the curve's extreme values instead of a midpoint. The cubic's local maximum is $f\!\left(1-\tfrac{1}{\sqrt3}\right)\approx 38.5$ and its local minimum is $\approx -38.5$. A horizontal line $y=k$ cuts a cubic in exactly three points precisely when $k$ lies strictly between the local minimum and the local maximum. Since $-38.5 < 25 < 38.5$, the line $y=25$ meets the curve three times — matching the count and confirming (C) without testing any specific point.
CCSS standards used (min grade 8)
8.F.A.1Understand that a function is a rule that assigns exactly one output to each input (Reading "the graph of $y=f(x-a)$ passes through $(1,25)$" as the function statement $f(1-a)=25$.)8.EE.C.7Solve linear equations in one variable (Using the linear substitution $t = 1 - a$ as a one-to-one correspondence, so counting valid $a$ becomes counting real solutions of $f(t)=25$.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Factoring $100t^3 - 300t^2 + 200t$ into the equivalent product $100t(t-1)(t-2)$ to expose the roots.)6.EE.A.2Write, read, and evaluate expressions in which letters stand for numbers (Evaluating $f$ at $t = 0, \tfrac12, 1, 2$ to compare the curve's heights with $25$.)8.F.B.5Describe qualitatively the functional relationship between two quantities (Describing the cubic's sign pattern and end behavior to count where it crosses the line $y=25$.)
⭐ A left-right shift never changes how many times a graph reaches a given height, so just count where the cubic $y=f(t)$ crosses $y=25$ — and a cubic can cross a flat line at most three times.
⭐ A left-right shift never changes how many times a graph reaches a given height, so just count where the cubic $y=f(t)$ crosses $y=25$ — and a cubic can cross a flat line at most three times.
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