Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 3 - Section 3.2 - Polynomial Functions and Their Graphs - 3.2 Exercises - Page 266: 34

Answer

$P(x)=-x(x+1)(2x-1)$ Zeros: $-1,\displaystyle \quad 0,\quad \frac{1}{2},$ all single.

Work Step by Step

Factor out $-x$ $P(x)=-x(2x^{2}+x-1)$ $p=\pm 1,\quad q=\pm 1,\pm 2$ $\displaystyle \frac{p}{q}$ are eligible for zeros of P(x). Synthetic division ... \begin{array}{rrrr} -1 \rceil & 2 & 1 & -1 \\ & & -2 & 1 \\ \hline & 2 & -1 & 0 \\ \end{array} $x=-1 $ is a zero $P(x)=-x(x+1)(2x-1)$ Zeros: $-1,\displaystyle \quad 0,\quad \frac{1}{2},$ all single. y-intercept:$ (0,0)$ The leading factor of a third degree polynomial is negative. The end behavior is as for $-x^{3}$ ($+\infty$ at the far left, $-\infty$ at the far right).. Sketch: On the far left, it falls from $+\infty,$ crosses x at $(-1,0)$, falls and turns to cross x at $(0,0)$, rises and turns to cross x again at $(\displaystyle \frac{1}{2},0)$, falling and continues falling to the far right.
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