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{{ printedBook.courseTrack.name }} {{ printedBook.name }} We want to find the zeros and sketch the graph of the given polynomial function. $f(x)=-x_{3}−x_{2}+9x+9 $ Let's do these things one at a time.

$-x_{3}−x_{2}+9x+9=0$

Factor

MultEqn$LHS⋅(-1)=RHS⋅(-1)$

$x_{3}+x_{2}−9x−9=0$

FactorOutFactor out $x_{2}$

$x_{2}(x+1)−9x−9=0$

FactorOutFactor out $-9$

$x_{2}(x+1)−9(x+1)=0$

FactorOutFactor out $(x+1)$

$(x+1)(x_{2}−9)=0$

FacDiffSquares$a_{2}−b_{2}=(a+b)(a−b)$

$(x+1)(x+3)(x−3)=0$

$(x+1)(x+3)(x−3)=0$

Solve using the Zero Product Property

ZeroProdPropUse the Zero Product Property

$x+1=0x+3=0x−3=0 (I)(II)(III) $

$x=-1x+3=0x−3=0 $

$x=-1x=-3x−3=0 $

$x=-1x=-3x=3 $

To draw the graph of the function, we will find some additional points and consider the end behavior. Let's use a table to find additional points.

$x$ | $-x_{3}−x_{2}+9x+9$ | $f(x)=-x_{3}−x_{2}+9x+9$ |
---|---|---|

$-3.5$ | $-(-3.5)_{3}−(-3.5)_{2}+9(-3.5)+9$ | $8.125$ |

$-2$ | $-(-2)_{3}−(-2)_{2}+9(-2)+9$ | $-5$ |

$1$ | $-1_{3}−1_{2}+9(1)+9$ | $16$ |

The points $(-3.5,8.125),$ $(-2,-5),$ and $(1,16)$ are on the graph of the function. Now, we will determine the leading coefficient and degree of the polynomial function.
$f(x)=-x_{3}−x_{2}+9x+9⇕f(x)=-1x_{3}−x_{2}+9x+9 $
We can see now that the leading coefficient is $-1,$ which is a negative number. Also, the degree is $3,$ which is an odd number. Therefore, the end behavior is **up** and **down**. With this in mind, we will plot the zeros and the obtained points, and draw the graph of the function.