Answer:
See explanation
Step-by-step explanation:
Zeroe of the function is such velue of x at which f(x)=0.
1. Consider the function [tex]f(x)=-x(3x-2)^2 (x+9)^5.[/tex]
Zeros are:
[tex]-x(3x-2)^2(x+9)^5=0\\ \\x=0\text{ or }x=\dfrac{2}{3}\text{ or }x=-9.[/tex]
Zero [tex]x=0[/tex] has multiplicity of 1, zero [tex]x=\dfrac{2}{3}[/tex] has multiplicity of 2, zero [tex]x=-9[/tex] has multiplicity of 5.
At [tex]x=0[/tex] or [tex]x=-9[/tex] the graph of the function crosses the x-axis, at [tex]x=\dfrac{2}{3}[/tex] the graph of the function touches the x-axis.
2. Consider the function [tex]f(x)=x^3+10x^2+25x=x(x^2+10x+25)=x(x+5)^2.[/tex]
Zeros are:
[tex]x(x+5)^2=0\\ \\x=0\text{ or }x=-5.[/tex]
Zero [tex]x=0[/tex] has multiplicity of 1, zero [tex]x=-5[/tex] has multiplicity of 2.
At [tex]x=0[/tex] the graph of the function crosses the x-axis, at [tex]x=-5[/tex] the graph of the function touches the x-axis.
