# 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

## See also

 2003 AMC 12B (Problems • Answer Key • Resources) Preceded byProblem 19 Followed byProblem 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

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