1.) Evaluating that function, to find the points of interception of the graph i.e. when y=0. So
[tex]\begin{gathered} f(x)=\frac{(x^2-1)^3}{(1+2x)^5}\Rightarrow0=\frac{(x^2-1)^3}{(1+2x)^5} \\ (x^2-1)^3=0 \\ \sqrt[3]{(x^2-1)^3}=0 \\ x^2-1=0 \\ x^2=1 \\ x=1,-1 \end{gathered}[/tex]As we can see, when we set y=0, and cross multiply, we can set the points that intercept the x-axis (-1,0) and (1,0)
