{{ item.displayTitle }}

No history yet!

Student

Teacher

{{ item.displayTitle }}

{{ item.subject.displayTitle }}

{{ searchError }}

{{ courseTrack.displayTitle }} {{ statistics.percent }}% Sign in to view progress

{{ printedBook.courseTrack.name }} {{ printedBook.name }} We want to find the zeros and sketch the graph of the given polynomial function. $f(x)=x_{4}+x_{3}−6x_{2} $ Let's do these things one at a time.

$x_{4}+x_{3}−6x_{2}=0$

$x_{2}(x+3)(x−2)=0$

$x_{2}(x+3)(x−2)=0$

Solve using the Zero Product Property

ZeroProdPropUse the Zero Product Property

$x_{2}=0x+3=0x−2=0 (I)(II)(III) $

$x=0x+3=0x−2=0 $

$x=0x=-3x−2=0 $

$x=0x=-3x=2 $

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.

$x$ | $x_{4}+x_{3}−6x_{2}$ | $f(x)=x_{4}+x_{3}−6x_{2}$ |
---|---|---|

$-2$ | $(-2)_{4}+(-2)_{3}−6(-2)_{2}$ | $-16$ |

$-1$ | $(-1)_{4}+(-1)_{3}−6(-1)_{2}$ | $-6$ |

$1$ | $1_{4}+1_{3}−6(1)_{2}$ | $-4$ |

The points $(-2,-16),$ $(-1,-6),$ and $(1,-4)$ are on the graph of the function. Now, we will determine the leading coefficient and degree of the polynomial function.
$f(x)=x_{4}+x_{3}−6x_{2}⇕f(x)=1x_{4}+x_{3}−6x_{2} $
We can see now that the leading coefficient is $1,$ which is a positive number. Also, the degree is $4,$ which is an even number. Therefore, the end behavior is **up** and **up**. With this in mind, we will plot and connect the zeros and the obtained points to draw the graph.