Question

In Problems 69– 86, analyze each polynomial function by following Steps 1 through 8 on page 190.

$f\left(x\right)=x^2(x-3)(x+4)$

Short answer

• The graph of the given function has an end behaviour similar to $y=x^4$.
• $x$-intercepts: $0,3,-4$and $y$-intercept: $0$.
• The graph crosses the $x$-axis at $x=3$and$x=-4$and the graph touches the $x$-axis at $x=0$.
• The graph of the given function is shown below:

• Maximum at $(0,0)$, Minimum at $(-2.85,-54.64)$and $(2.10,-24.21)$.
• Apply the information in Steps 2 through 6 to draw a complete graph as shown below:

• Domain:$(0,0)$, Range:$[-54.64,\infty )$.
• Increasing:$(-2.85,0)$and $(2.10,\infty )$, Decreasing: $(-\infty ,-2.85)$and $(0,2.10)$.

Step-by-step solution

Step 1. Given information

The given function is $f\left(x\right)=x^2(x-3)(x+4)$.

Step 2. Determine the end behavior of the graph of the given function.

Solve the given function.

$\begin{array}{rcl}f\left(x\right)& =& x^2(x-3)(x+4)\\ f\left(x\right)& =& x^2(x^2+x-12)\\ & =& x^4+x^3-12x^2\end{array}$

The polynomial function $f$is of degree $4$.

Therefore, the graph of the given function behaves like $y=x^4$.

Step 3. Determine the x- and y-intercepts of the graph of the function.

Substitute $x=0$to find the localid="1646478356901" $y$-intercepts.

$\begin{array}{rcl}f\left(x\right)& =& x^2(x-3)(x+4)\\ f\left(0\right)& =& 0^2(0-3)(0+4)\\ & =& 0\end{array}$

Thus, the localid="1646478330508" $y$-intercept is at the point $(0,0)$.

Now, substitute $f\left(x\right)=0$to find the $x$-intercepts.

$\begin{array}{rcl}f\left(x\right)& =& x^2(x-3)(x+4)\\ 0& =& x^2(x-3)(x+4)\\ & \Rightarrow & x^2=0\\ & & or\\ x-3& =& 0\\ & & or\\ x+4& =& 0\end{array}$

width="71">⇒x=0orx=3orx=-4

Thus, the $x$-intercepts are at the points $x=0,x=3,$and $x=-4$.

Step 4. Determine the zeros of the function and their multiplicity and apply this information to determine whether the graph crosses or touches the x-axis at each x-intercept.

• If the multiplicity is even, then the graph touches the $x$-axis.
• If the multiplicity is odd, then the graph crosses the $x$-axis.

From the given function $f\left(x\right)=x^2(x-3)(x+4)$, conclude that the zeros of the given function are $0,3,$and $-4$.

The zeros localid="1646478582020" $3$and $-4$have multiplicity $1$, so the graph of$f$crosses the $x$-axis at $x=3$and $x=-4$.

The zero $0$has multiplicity $2$, so the graph of $f$touches the $x$-axis at $x=0$.

Step 5. Apply a graphing utility to graph the given function.

The graph of the given function is shown below:

Step 6. Approximate the turning points of the graph.

From the graph of $f$, observe that $f$has two turning points.

Using MAXIMUM, one turning point is at $(0,0)$.

Using MINIIMUM, the other turning point is at $(-2.85,-54.64)$and $(2.10,-24.21)$.

Step 7. Apply the information in Steps 2 through 6 to draw a complete graph of the given function by hand.

The graph is shown below:

Step 8. Find the domain and the range of the given function.

Since the function is a polynomial function.

So, the domain is the set of all real numbers.

Thus, domain$(-\infty ,\infty )$.

From the graph, the range is the set of all possible values for which the function is defined.

Thus, range $[-54.64,\infty )$.

Step 9. Determine where the function is increasing and where it is decreasing by using the graph.

From the graph of$f$, observe the following:

$f$is increasing on the interval$(-2.85,0)$ and $(2.10,\infty )$.

$f$is decreasing on the intervals $(-\infty ,-2.85)$and $(0,2.10)$.