2025 AMC 10A Problems/Problem 9

Let $f(x) = 100x^3 - 300x^2 + 200x$ . For how many real numbers $a$ does the graph of $y = f(x - a)$ pass through the point $(1, 25)$ ?

(A) $1$ (B) $2$ (C) $3$ (D) $4$ (E) more than $4$

Solution 1

This problem is essentially asking how many values of $x$ satisfy $f(x)=25,$ since the graph $f(x-a)$ is a shift $a$ units right from the original graph of $f(x).$ Now we just need to determine the amount of real solutions to $100x^3-300x^2+200x=25.$

Dividing all terms by 25, we get $4x^3-12x^2+8x-1=0.$ Using the rational root theorem and testing, we find no rational roots. Our next method to determine the amount of roots is graphing. Notice how when $x$ $\to$ $+\infty,$ $f(x)$ $\to$ $+\infty.$ When $x$ $\to$ $-\infty,$ $f(x)$ $\to$ $-\infty.$ Substituting $x=0$ gives us $f(x)=-1.$ Substituting $x=1$ gives us $f(x)=-1$ also. This makes us think to see if there was any sudden bump or curve in the graph, so we test the midpoint which is $1/2.$ Substituting $x=1/2,$ we get $1/2-3+4-1=1/2.$ Aha! So connecting the dots, we see there is two solutions from $0\leq x \leq 1.$ Also, $f(x)$ must continue increasing after $x>1,$ which tells us there is a third root when $x>1.$ Now we can conclude that the problem has 3 real roots, meaning that 3 values of a satisfy the problem. The answer is $\boxed {C}.$

~ eqb5000/Esteban Q.

Solution

The problem boils down to how many real roots does the equation $100(1-a)^3 - 300(1-a)^2 + 200(1-a) = 25$ have? Using Descarte's Rule of Signs we find there shouldn't be any imaginary roots, so the answer is $\boxed{3}$ real roots.

~grogg007

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2025 AMC 10A Problems/Problem 9

Solution 1

Solution