\(\displaystyle{f{{\left({x}\right)}}}={x}^{{{2}}}{\left({x}+{1}\right)}{\left({x}-{1}\right)}\) The function in factored form

The function has three zeros 0, -1, and 1

So, the graph of f(x) crosses the x-axis at (0,0), (-1,0), and (1,0)

To find the y-intercept, substitute 0 for x in f(x)

\(\displaystyle{f{{\left({x}\right)}}}={x}^{{{2}}}{\left({x}+{1}\right)}{\left({x}-{1}\right)}\)

\(\displaystyle{f{{\left({0}\right)}}}={\left({0}\right)}^{{{2}}}{\left({0}+{1}\right)}{\left({0}-{1}\right)}\) Substitute 0 for x

\(\displaystyle={0}\)

So, the function f(x) crosses the y-axis at (0,0)

\(x^{2} \cdot x \cdot x=x^{4}\)

The leading coefficient is 1

Since the leading coefficient is positive and the function f(x) of degree 4 (even degree)

So, the end behavior is

\(\displaystyle{x}\rightarrow\infty,{f{{\left({x}\right)}}}\rightarrow\infty\)

\(\displaystyle{x}\rightarrow-\infty,{f{{\left({x}\right)}}}\rightarrow\infty\)

See the graph below