Step 1

\(\displaystyle{f{{\left({x}\right)}}}={2}{x}{\left({x}-{3}\right)}{\left({x}+{2}\right)}\) The function in factored form

The function has three zeros 0,3, and -2

So, the graph of f(x) crosses the x-axis at (0,0),(3,0), and (-2,0)

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

\(\displaystyle{f{{\left({x}\right)}}}={2}{x}{\left({x}-{3}\right)}{\left({x}+{2}\right)}\)

\(\displaystyle{f{{\left({0}\right)}}}={2}{\left({0}\right)}{\left({0}-{3}\right)}{\left({0}+{2}\right)}\) Substitute 0 for x

\(\displaystyle={0}\)

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

Step 2

\(\displaystyle{2}{x}\cdot{x}\cdot{x}={2}{x}^{{{3}}}\)

The leading coefficient is 2

Since the leading coefficient is positive and the function f(x) of degree 3 (odd 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

\(\displaystyle{f{{\left({x}\right)}}}={2}{x}{\left({x}-{3}\right)}{\left({x}+{2}\right)}\) The function in factored form

The function has three zeros 0,3, and -2

So, the graph of f(x) crosses the x-axis at (0,0),(3,0), and (-2,0)

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

\(\displaystyle{f{{\left({x}\right)}}}={2}{x}{\left({x}-{3}\right)}{\left({x}+{2}\right)}\)

\(\displaystyle{f{{\left({0}\right)}}}={2}{\left({0}\right)}{\left({0}-{3}\right)}{\left({0}+{2}\right)}\) Substitute 0 for x

\(\displaystyle={0}\)

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

Step 2

\(\displaystyle{2}{x}\cdot{x}\cdot{x}={2}{x}^{{{3}}}\)

The leading coefficient is 2

Since the leading coefficient is positive and the function f(x) of degree 3 (odd 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