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2025 AMC 10A Problems/Problem 9: Difference between revisions

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


Dividing all terms by 25, we get <imath>4x^3-12x^2+8x-1=0.</imath> 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 <imath>x</imath> approaches infinity, <imath>f(x)</imath> also approaches infinity. When <imath>x</imath> approaches negative infinity, <imath>f(x)</imath> also approaches negative infinity. Substituting <imath>x=0</imath> gives us <imath>f(x)=-1.</imath> Substituting <imath>x=1</imath> gives us <imath>f(x)=-1</imath> 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 <imath>1/2.</imath> Substituting <imath>x=1/2,</imath> we get <imath>1/2-3+4-1=1/2.</imath> Aha! So connecting the dots, we see there is two solutions from <imath>0\leq x \leq 1.</imath> Also, <imath>f(x)</imath> must continue increasing after <imath>x>1,</imath> which tells us there is a third root when <imath>x>1.</imath> 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}.
Dividing all terms by 25, we get <imath>4x^3-12x^2+8x-1=0.</imath> 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 <imath>x</imath> approaches infinity, <imath>f(x)</imath> also approaches infinity. When <imath>x</imath> approaches negative infinity, <imath>f(x)</imath> also approaches negative infinity. Substituting <imath>x=0</imath> gives us <imath>f(x)=-1.</imath> Substituting <imath>x=1</imath> gives us <imath>f(x)=-1</imath> 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 <imath>1/2.</imath> Substituting <imath>x=1/2,</imath> we get <imath>1/2-3+4-1=1/2.</imath> Aha! So connecting the dots, we see there is two solutions from <imath>0\leq x \leq 1.</imath> Also, <imath>f(x)</imath> must continue increasing after <imath>x>1,</imath> which tells us there is a third root when <imath>x>1.</imath> Now we can conclude that the problem has 3 real roots, meaning that 3 values of a satisfy the problem. The answer is <imath>\boxed {C}.</imath>


~ eqb5000
~ eqb5000

Revision as of 13:55, 6 November 2025

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 $x$ approaches infinity, $f(x)$ also approaches infinity. When $x$ approaches negative infinity, $f(x)$ also approaches negative infinity. 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

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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