Given:
[tex]f\mleft(x\mright)=-2x^3\left(x-2\right)^2\left(x+3\right)[/tex]
Required:
We need to find the zeros of the function, the end behavior, and the maximum number of turning points and draw the graph.
Explanation:
a)
Set f(x)=0 to find the zeros of the given function.
[tex]-2x^3(x-2)^2(x+3)=0[/tex]
[tex]x^3(x-2)^2(x+3)=0[/tex][tex]x^3=0,(x-2)^2=0,(x+3)=0[/tex]
[tex]x=0,x-2=0,x+3=0[/tex]
[tex]x=0,x=2,x=-3.[/tex]
The zeros are 0,2 and -3.
Recall that The number of times a given factor appears in the factored form of the equation of a polynomial is called the multiplicity.
The multiplicity 0 is 3.
The multiplicity 2 is 2.
The multiplicity -3 is 1.
b)
The end behavior:
Taking the limit of negative infinity to the given function.
[tex]\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(-2x^3(x-2)^2(x+3))[/tex]
[tex]\lim_{x\to-\infty}f(x)=-2(-\infty)^3(-\infty-2)^2(-\infty+3)[/tex]
[tex]\lim_{x\to-\infty}f(x)=-(-\infty)(\infty)(-\infty)[/tex]
[tex]\lim_{x\to-\infty}f(x)=-\infty[/tex][tex]As\text{ }x\rightarrow-\infty\text{ then }f(x)\rightarrow-\infty[/tex]
Taking the limit of infinity to the given function.
[tex]\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(-2x^3(x-2)^2(x+3))[/tex]
[tex]\lim_{x\to-\infty}f(x)=-2(\infty)^3(\infty-2)^2(\infty+3)[/tex]
[tex]\lim_{x\to-\infty}f(x)=-(\infty)(\infty)(\infty)[/tex]
[tex]\lim_{x\to-\infty}f(x)=-\infty[/tex]
[tex]As\text{ }x\rightarrow\infty\text{ then }f(x)\rightarrow-\infty[/tex]
Recall that the maximum number of turning points of a polynomial function is always one less than the degree of the function.
The degree of the given function is 3+2+1=6.
The maximum number of turning points =6-1=5.
Final answer:
a)
Zeros of the function
[tex]x=0,x=2,x=-3.[/tex]
The multiplicity 0 is 3.
The multiplicity 2 is 2.
The multiplicity -3 is 1.
b)
End behavior:
[tex]As\text{ }x\rightarrow-\infty\text{ then }f(x)\rightarrow-\infty[/tex]
[tex]As\text{ }x\rightarrow\infty\text{ then }f(x)\rightarrow-\infty[/tex]
The maximum number of turning points is 5.
c)
The graph of the given function
