2025 AMC 10A Problem 9

1. Turn the condition into an equation. The graph of $y = f(x - a)$ passes through $(1, 25)$ when putting $x = 1$ gives $y = 25$, that is, when
$$f(1 - a) = 25.$$

1. Count heights instead of shifts. Write $t = 1 - a$. Each value of $t$ matches exactly one value of $a$ (namely $a = 1 - t$), and the other way around. So the question becomes: how many numbers $t$ make $f(t) = 25$? In picture form: how many times does the graph of $f$ reach height $25$?

3. Factor $f$ to understand its shape. Pull out $100x$:
$$f(x) = 100x(x^2 - 3x + 2) = 100x(x - 1)(x - 2).$$
So the graph crosses zero at $x = 0$, $1$ and $2$.

1. Make a small table of values. Use the zeros plus a few easy points in between:

For example, $f\!\left(\tfrac12\right) = 100 \cdot \tfrac12 \cdot \left(-\tfrac12\right) \cdot \left(-\tfrac32\right) = 37.5$.

5. Find where the graph passes height $25$.
- From $t = 0$ to $t = \tfrac12$ it goes from $0$ up to $37.5$, so it passes $25$ once.
- From $t = \tfrac12$ to $t = 1$ it goes from $37.5$ down to $0$, so it passes $25$ again.
- From $t = 2$ to $t = 3$ it goes from $0$ up to $600$, so it passes $25$ a third time.

1. Show there are no more. The equation $f(t) = 25$ is a cubic equation, and a cubic equation has at most $3$ real solutions. We found $3$, so that is all of them.

Each of the three values of $t$ gives one value of $a$, so there are exactly $3$ values of $a$.

The answer is $\boxed{\textbf{(C)}\ 3}$.

Shifting a graph left or right moves its points sideways but keeps every height. So "which shifts pass through a point at height $25$" is the same question as "where does the original graph reach height $25$". To count those spots you do not need to solve the cubic: a table of values shows where the graph goes above and below $25$, and the degree tells you the most solutions there can be.

Write the equation directly in $a$. Expanding $f(1 - a)$ gives $100a - 100a^3$, so the condition is
$$100a - 100a^3 = 25, \qquad\text{or}\qquad 4a^3 - 4a + 1 = 0.$$
Let $p(a) = 4a^3 - 4a + 1$ and test a few values: $p(-2) = -23$, $p(-1) = 1$, $p\!\left(\tfrac12\right) = -\tfrac12$, $p(1) = 1$. The sign flips three times, so there is a solution between $-2$ and $-1$, one between $-1$ and $\tfrac12$, and one between $\tfrac12$ and $1$. A cubic has at most three, so the count is $3$.

First factor:
$$f(x) = 100x(x^2 - 3x + 2) = 100x(x-1)(x-2),$$
so $f$ has roots $0$, $1$ and $2$.

The graph of $y = f(x-a)$ passes through $(1,25)$ precisely when $f(1 - a) = 25$. Put $t = 1 - a$. As $a$ ranges over the reals so does $t$, and different $a$ give different $t$, so the number of valid $a$ equals the number of real solutions of $f(t) = 25$. In other words, the question is how many times the fixed cubic reaches the height $25$.

Since $f(t) = 25$ means $100t(t-1)(t-2) = 25$, set
$$g(t) = t(t-1)(t-2) - \tfrac14 .$$
Evaluate at four convenient points:
$$g(0) = -\tfrac14 < 0, \qquad g\!\left(\tfrac12\right) = \tfrac12 \cdot \left(-\tfrac12\right)\cdot\left(-\tfrac32\right) - \tfrac14 = \tfrac38 - \tfrac14 > 0,$$
$$g(1) = -\tfrac14 < 0, \qquad g\!\left(\tfrac52\right) = \tfrac52\cdot\tfrac32\cdot\tfrac12 - \tfrac14 = \tfrac{15}{8} - \tfrac14 > 0.$$
The sign changes three times, so $g$ has a root in each of $\left(0,\tfrac12\right)$, $\left(\tfrac12,1\right)$ and $\left(1,\tfrac52\right)$. A cubic has at most three real roots, so there are exactly three, and therefore exactly three values of $a$.

The answer is $\boxed{\textbf{(C)}\ 3}$.