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 $b = -an = -2$ .

2003 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2003 AMC 12B Problems/Problem 20

Problem

Contents

Solution

Solution 1

Solution 2

See also