2003 AMC 12B Problems/Problem 20

Problem

Part of the graph of $f(x) = ax^3 + bx^2 + cx + d$ is shown. What is $b$ ?

$\mathrm{(A)}\ -4 \qquad\mathrm{(B)}\ -2 \qquad\mathrm{(C)}\ 0 \qquad\mathrm{(D)}\ 2 \qquad\mathrm{(E)}\ 4$

Solution

Solution 1

Since $\begin{align*} -f(-1) = a - b + c - d = 0 = f(1) = a + b + c + d \end{align*}$

It follows that $b + d = 0$ . Also, $d = f(0) = 2$ , so $b = -2 \Rightarrow \mathrm{(B)}$ .

Solution 2

Two of the roots of $f(x) = 0$ are $\pm 1$ , and we let the third one be $n$ . Then $a(x-1)(x+1)(x-n) = ax^3-anx^2-ax+an = ax^3 + bx^2 + cx + d = 0$ Notice that $f(0) = d = an = 2$ , so $b = -an = -2 \Rightarrow \mathrm{(B)}$ .

Solution 3

Notice that if $g(x) = 2 - 2x^2$ , then $f - g$ vanishes at $x = -1, 0, 1$ and so $f(x) - g(x) = ax(x-1)(x+1) = ax^3 - ax$ implies by $x^2$ coefficient, $b + 2 = 0, b = -2 \Rightarrow \mathrm{(B)}$ .

Solution 4

The roots of this equation are $-1, 1, \text{ and } x$ , letting $x$ be the root not shown in the graph. By Vieta, we know that $-1+1+x=x=-\frac{b}{a}$ and $-1\cdot 1\cdot x=-x=-\frac{d}{a}$ . Therefore, $x=\frac{d}{a}$ . Setting the two equations for $x$ equal to each other, $\frac{d}{a}=-\frac{b}{a}$ . We know that the y-intercept of the polynomial is $d$ , so $d=2$ . Plugging in for $d$ , $\frac{2}{a}=-\frac{b}{a}$ .

Therefore, $b=-2 \Rightarrow \boxed{B}$

Solution 5

From the graph, we have $f(0)=2$ so $d=2$ . Also from the graph, we have $f(1)=a+b+c+2=0$ . But we also have from the graph $f(-1)=-a+b-c+2=0$ . Summing $f(1)+f(2)$ we get $2b+4=0$ so $b = -2 \Rightarrow \mathrm{(B)}$ .

Solution by franzliszt

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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