We want to find the and sketch the graph of the given .
f(x)=x6−11x5+30x4
Let's do these things one at a time.
Zeros of the Function
To find the zeros, we need to find the values of
x for which
f(x)=0.
f(x)=0⇔x6−11x5+30x4=0
Since the function is not written in , we will begin by the .
x6−11x5+30x4=0 x4(x2−11x+30)=0 x4(x2−5x−6x+30)=0 x4(x(x−5)−6x+30)=0 x4(x(x−5)−6(x−5))=0
x4(x−5)(x−6)=0
Now, we can apply the .
x4(x−5)(x−6)=0
Solve using the Zero Product Property
x4=0x−5=0x−6=0(I)(II)(III) x=0x−5=0x−6=0 x=0x=5x−6=0
x=0x=5x=6
We found that the zeros of the function are
x=0, x=5, and
x=6. Graph
To draw the graph of the function, we will find some additional points and consider the . Let's use a table to find additional points.
x
|
x6−11x5+30x4
|
f(x)=x6−11x5+30x4
|
-1
|
(-1)6−11(-1)5+30(-1)4
|
42
|
2
|
26−11(2)5+30(2)4
|
192
|
4
|
46−11(4)5+30(4)4
|
512
|
5.5
|
5.56−11(5.5)5+30(5.5)4
|
≈-228.8
|
The points (-1,42), (2,192), (4,512), and (5.5,-228.8) are on the graph of the function. Now, we will determine the and of the polynomial function.
f(x)=x6−11x5+30x4⇕f(x)=1x6−11x5+30x4
We can see now that the leading coefficient is 1, which is a positive number. Also, the degree is 6, which is an . Therefore, the end behavior is up and up. With this in mind, we will plot the zeros, obtained points, and connect them with a smooth curve.