2025 AMC 10A Problem 9

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 $\boxed{\textbf{(C)}\ 3}$.

A horizontal shift changes where a function attains its values but not which values it attains, so a whole family of shifts asked to pass through one fixed point collapses to a single equation $f(t) = 25$. Counting its solutions needs no formula for them: the roots $0$, $1$, $2$ supply free sample points where $f$ is $0$, the hump between $0$ and $1$ rises to $f\!\left(\tfrac12\right) = 37.5 > 25$, and the degree caps the count at three. Note also that the direction of the shift is irrelevant to the count, so worrying about whether $f(x-a)$ moves the graph left or right is wasted effort here.

Work directly in $a$. Centring the cubic at its middle root, $f(1+u) = 100(1+u)(u)(u-1) = 100u^3 - 100u$, and $t = 1 - a$ means $u = -a$, so the condition becomes
$$-100a^3 + 100a = 25 \iff 4a^3 - 4a + 1 = 0.$$
Testing values, $p(a) = 4a^3 - 4a + 1$ gives $p(-2) = -23$, $p(-1) = 1$, $p\!\left(\tfrac12\right) = -\tfrac12$ and $p(1) = 1$: three sign changes, hence three real values of $a$, one in each of $(-2,-1)$, $\left(0,\tfrac12\right)$ and $\left(\tfrac12,1\right)$.

• Assuming a cubic takes each value once and answering 1, choice (A), when f has a local maximum of about 38 between its roots 0 and 1.
• Substituting into $f$ rather than into the shifted function, for instance solving $f(1) - a = 25$; since $f(1) = 0$ this gives a single value of $a$ and the wrong count.
• Treating $a$ as a free parameter and answering "more than $4$", choice (E), rather than noticing that each $a$ must solve a cubic.