Step 1. Since [latex]f[/latex] is a polynomial, the domain is the set of all real numbers.

Step 2. When [latex]x=0,f\left(x\right)=2.[/latex] Therefore, the [latex]y[/latex]-intercept is [latex]\left(0,2\right).[/latex] To find the [latex]x[/latex]-intercepts, we need to solve the equation [latex]{\left(x-1\right)}^{2}\left(x+2\right)=0,[/latex] gives us the [latex]x[/latex]-intercepts [latex]\left(1,0\right)[/latex] and [latex]\left(-2,0\right)[/latex]

Step 3. We need to evaluate the end behavior of [latex]f.[/latex] As [latex]x\to \infty ,[/latex] [latex]{\left(x-1\right)}^{2}\to \infty [/latex] and [latex]\left(x+2\right)\to \infty .[/latex] Therefore, [latex]\underset{x\to \infty }{\text{lim}}f\left(x\right)=\infty .[/latex] As [latex]x\to \text{−}\infty ,[/latex] [latex]{\left(x-1\right)}^{2}\to \infty [/latex] and [latex]\left(x+2\right)\to \text{−}\infty .[/latex] Therefore, [latex]\underset{x\to \infty }{\text{lim}}f\left(x\right)=\text{−}\infty .[/latex] To get even more information about the end behavior of [latex]f,[/latex] we can multiply the factors of [latex]f.[/latex] When doing so, we see that

Since the leading term of [latex]f[/latex] is [latex]{x}^{3},[/latex] we conclude that [latex]f[/latex] behaves like [latex]y={x}^{3}[/latex] as [latex]x\to \text{±}\infty .[/latex]

Step 4. Since [latex]f[/latex] is a polynomial function, it does not have any vertical asymptotes.

Step 5. The first derivative of [latex]f[/latex] is

Therefore, [latex]f[/latex] has two critical points: [latex]x=1,-1.[/latex] Divide the interval [latex]\left(\text{−}\infty ,\infty \right)[/latex] into the three smaller intervals: [latex]\left(\text{−}\infty ,-1\right),[/latex] [latex]\left(-1,1\right),[/latex] and [latex]\left(1,\infty \right).[/latex] Then, choose test points [latex]x=-2,[/latex] [latex]x=0,[/latex] and [latex]x=2[/latex] from these intervals and evaluate the sign of [latex]{f}^{\prime }\left(x\right)[/latex] at each of these test points, as shown in the following table.

From the table, we see that [latex]f[/latex] has a local maximum at [latex]x=-1[/latex] and a local minimum at [latex]x=1.[/latex] Evaluating [latex]f\left(x\right)[/latex] at those two points, we find that the local maximum value is [latex]f\left(-1\right)=4[/latex] and the local minimum value is [latex]f\left(1\right)=0.[/latex]

Step 6. The second derivative of [latex]f[/latex] is

The second derivative is zero at [latex]x=0.[/latex] Therefore, to determine the concavity of [latex]f,[/latex] divide the interval [latex]\left(\text{−}\infty ,\infty \right)[/latex] into the smaller intervals [latex]\left(\text{−}\infty ,0\right)[/latex] and [latex]\left(0,\infty \right),[/latex] and choose test points [latex]x=-1[/latex] and [latex]x=1[/latex] to determine the concavity of [latex]f[/latex] on each of these smaller intervals as shown in the following table.

We note that the information in the preceding table confirms the fact, found in step [latex]5,[/latex] that [latex]f[/latex] has a local maximum at [latex]x=-1[/latex] and a local minimum at [latex]x=1.[/latex] In addition, the information found in step [latex]5[/latex]—namely, [latex]f[/latex] has a local maximum at [latex]x=-1[/latex] and a local minimum at [latex]x=1,[/latex] and [latex]{f}^{\prime }\left(x\right)=0[/latex] at those points—combined with the fact that [latex]f\text{″}[/latex] changes sign only at [latex]x=0[/latex] confirms the results found in step [latex]6[/latex] on the concavity of [latex]f.[/latex]

Combining this information, we arrive at the graph of [latex]f\left(x\right)={\left(x-1\right)}^{2}\left(x+2\right)[/latex] shown in the following graph.

Step 1. The function [latex]f[/latex] is defined as long as the denominator is not zero. Therefore, the domain is the set of all real numbers [latex]x[/latex] except [latex]x=\text{±}1.[/latex]

Step 2. Find the intercepts. If [latex]x=0,[/latex] then [latex]f\left(x\right)=0,[/latex] so [latex]0[/latex] is an intercept. If [latex]y=0,[/latex] then [latex]\frac{{x}^{2}}{\left(1-{x}^{2}\right)}=0,[/latex] which implies [latex]x=0.[/latex] Therefore, [latex]\left(0,0\right)[/latex] is the only intercept.

Step 3. Evaluate the limits at infinity. Since [latex]f[/latex] is a rational function, divide the numerator and denominator by the highest power in the denominator: [latex]{x}^{2}.[/latex] We obtain

Therefore, [latex]f[/latex] has a horizontal asymptote of [latex]y=-1[/latex] as [latex]x\to \infty [/latex] and [latex]x\to \text{−}\infty .[/latex]

Step 4. To determine whether [latex]f[/latex] has any vertical asymptotes, first check to see whether the denominator has any zeroes. We find the denominator is zero when [latex]x=\text{±}1.[/latex] To determine whether the lines [latex]x=1[/latex] or [latex]x=-1[/latex] are vertical asymptotes of [latex]f,[/latex] evaluate [latex]\underset{x\to 1}{\text{lim}}f\left(x\right)[/latex] and [latex]\underset{x\to \text{−}1}{\text{lim}}f\left(x\right).[/latex] By looking at each one-sided limit as [latex]x\to 1,[/latex] we see that

In addition, by looking at each one-sided limit as [latex]x\to \text{−}1,[/latex] we find that

Step 5. Calculate the first derivative:

Critical points occur at points [latex]x[/latex] where [latex]{f}^{\prime }\left(x\right)=0[/latex] or [latex]{f}^{\prime }\left(x\right)[/latex] is undefined. We see that [latex]{f}^{\prime }\left(x\right)=0[/latex] when [latex]x=0.[/latex] The derivative [latex]{f}^{\prime }[/latex] is not undefined at any point in the domain of [latex]f.[/latex] However, [latex]x=\text{±}1[/latex] are not in the domain of [latex]f.[/latex] Therefore, to determine where [latex]f[/latex] is increasing and where [latex]f[/latex] is decreasing, divide the interval [latex]\left(\text{−}\infty ,\infty \right)[/latex] into four smaller intervals: [latex]\left(\text{−}\infty ,-1\right),[/latex] [latex]\left(-1,0\right),[/latex] [latex]\left(0,1\right),[/latex] and [latex]\left(1,\infty \right),[/latex] and choose a test point in each interval to determine the sign of [latex]{f}^{\prime }\left(x\right)[/latex] in each of these intervals. The values [latex]x=-2,[/latex] [latex]x=-\frac{1}{2},[/latex] [latex]x=\frac{1}{2},[/latex] and [latex]x=2[/latex] are good choices for test points as shown in the following table.

From this analysis, we conclude that [latex]f[/latex] has a local minimum at [latex]x=0[/latex] but no local maximum.

Step 6. Calculate the second derivative:

To determine the intervals where [latex]f[/latex] is concave up and where [latex]f[/latex] is concave down, we first need to find all points [latex]x[/latex] where [latex]f\text{″}\left(x\right)=0[/latex] or [latex]f\text{″}\left(x\right)[/latex] is undefined. Since the numerator [latex]6{x}^{2}+2\ne 0[/latex] for any [latex]x,[/latex] [latex]f\text{″}\left(x\right)[/latex] is never zero. Furthermore, [latex]f\text{″}[/latex] is not undefined for any [latex]x[/latex] in the domain of [latex]f.[/latex] However, as discussed earlier, [latex]x=\text{±}1[/latex] are not in the domain of [latex]f.[/latex] Therefore, to determine the concavity of [latex]f,[/latex] we divide the interval [latex]\left(\text{−}\infty ,\infty \right)[/latex] into the three smaller intervals [latex]\left(\text{−}\infty ,-1\right),[/latex] [latex]\left(-1,-1\right),[/latex] and [latex]\left(1,\infty \right),[/latex] and choose a test point in each of these intervals to evaluate the sign of [latex]f\text{″}\left(x\right).[/latex] in each of these intervals. The values [latex]x=-2,[/latex] [latex]x=0,[/latex] and [latex]x=2[/latex] are possible test points as shown in the following table.

Combining all this information, we arrive at the graph of [latex]f[/latex] shown below. Note that, although [latex]f[/latex] changes concavity at [latex]x=-1[/latex] and [latex]x=1,[/latex] there are no inflection points at either of these places because [latex]f[/latex] is not continuous at [latex]x=-1[/latex] or [latex]x=1.[/latex]

Step 1. The domain of [latex]f[/latex] is the set of all real numbers [latex]x[/latex] except [latex]x=1.[/latex]

Step 2. Find the intercepts. We can see that when [latex]x=0,[/latex] [latex]f\left(x\right)=0,[/latex] so [latex]\left(0,0\right)[/latex] is the only intercept.

Step 3. Evaluate the limits at infinity. Since the degree of the numerator is one more than the degree of the denominator, [latex]f[/latex] must have an oblique asymptote. To find the oblique asymptote, use long division of polynomials to write

Since [latex]1\text{/}\left(x-1\right)\to 0[/latex] as [latex]x\to \text{±}\infty ,[/latex] [latex]f\left(x\right)[/latex] approaches the line [latex]y=x+1[/latex] as [latex]x\to \text{±}\infty .[/latex] The line [latex]y=x+1[/latex] is an oblique asymptote for [latex]f.[/latex]

Step 4. To check for vertical asymptotes, look at where the denominator is zero. Here the denominator is zero at [latex]x=1.[/latex] Looking at both one-sided limits as [latex]x\to 1,[/latex] we find

Therefore, [latex]x=1[/latex] is a vertical asymptote, and we have determined the behavior of [latex]f[/latex] as [latex]x[/latex] approaches [latex]1[/latex] from the right and the left.

Step 5. Calculate the first derivative:

We have [latex]{f}^{\prime }\left(x\right)=0[/latex] when [latex]{x}^{2}-2x=x\left(x-2\right)=0.[/latex] Therefore, [latex]x=0[/latex] and [latex]x=2[/latex] are critical points. Since [latex]f[/latex] is undefined at [latex]x=1,[/latex] we need to divide the interval [latex]\left(\text{−}\infty ,\infty \right)[/latex] into the smaller intervals [latex]\left(\text{−}\infty ,0\right),[/latex] [latex]\left(0,1\right),[/latex] [latex]\left(1,2\right),[/latex] and [latex]\left(2,\infty \right),[/latex] and choose a test point from each interval to evaluate the sign of [latex]{f}^{\prime }\left(x\right)[/latex] in each of these smaller intervals. For example, let [latex]x=-1,[/latex] [latex]x=\frac{1}{2},[/latex] [latex]x=\frac{3}{2},[/latex] and [latex]x=3[/latex] be the test points as shown in the following table.

From this table, we see that [latex]f[/latex] has a local maximum at [latex]x=0[/latex] and a local minimum at [latex]x=2.[/latex] The value of [latex]f[/latex] at the local maximum is [latex]f\left(0\right)=0[/latex] and the value of [latex]f[/latex] at the local minimum is [latex]f\left(2\right)=4.[/latex] Therefore, [latex]\left(0,0\right)[/latex] and [latex]\left(2,4\right)[/latex] are important points on the graph.

Step 6. Calculate the second derivative:

We see that [latex]f\text{″}\left(x\right)[/latex] is never zero or undefined for [latex]x[/latex] in the domain of [latex]f.[/latex] Since [latex]f[/latex] is undefined at [latex]x=1,[/latex] to check concavity we just divide the interval [latex]\left(\text{−}\infty ,\infty \right)[/latex] into the two smaller intervals [latex]\left(\text{−}\infty ,1\right)[/latex] and [latex]\left(1,\infty \right),[/latex] and choose a test point from each interval to evaluate the sign of [latex]f\text{″}\left(x\right)[/latex] in each of these intervals. The values [latex]x=0[/latex] and [latex]x=2[/latex] are possible test points as shown in the following table.

From the information gathered, we arrive at the following graph for [latex]f.[/latex]

Step 1. Since the cube-root function is defined for all real numbers [latex]x[/latex] and [latex]{\left(x-1\right)}^{2\text{/}3}={\left(\sqrt[3]{x-1}\right)}^{2},[/latex] the domain of [latex]f[/latex] is all real numbers.

Step 2: To find the [latex]y[/latex]-intercept, evaluate [latex]f\left(0\right).[/latex] Since [latex]f\left(0\right)=1,[/latex] the [latex]y[/latex]-intercept is [latex]\left(0,1\right).[/latex] To find the [latex]x[/latex]-intercept, solve [latex]{\left(x-1\right)}^{2\text{/}3}=0.[/latex] The solution of this equation is [latex]x=1,[/latex] so the [latex]x[/latex]-intercept is [latex]\left(1,0\right).[/latex]

Step 3: Since [latex]\underset{x\to \text{±}\infty }{\text{lim}}{\left(x-1\right)}^{2\text{/}3}=\infty ,[/latex] the function continues to grow without bound as [latex]x\to \infty [/latex] and [latex]x\to \text{−}\infty .[/latex]

Step 4: The function has no vertical asymptotes.

Step 5: To determine where [latex]f[/latex] is increasing or decreasing, calculate [latex]{f}^{\prime }.[/latex] We find

This function is not zero anywhere, but it is undefined when [latex]x=1.[/latex] Therefore, the only critical point is [latex]x=1.[/latex] Divide the interval [latex]\left(\text{−}\infty ,\infty \right)[/latex] into the smaller intervals [latex]\left(\text{−}\infty ,1\right)[/latex] and [latex]\left(1,\infty \right),[/latex] and choose test points in each of these intervals to determine the sign of [latex]{f}^{\prime }\left(x\right)[/latex] in each of these smaller intervals. Let [latex]x=0[/latex] and [latex]x=2[/latex] be the test points as shown in the following table.

We conclude that [latex]f[/latex] has a local minimum at [latex]x=1.[/latex] Evaluating [latex]f[/latex] at [latex]x=1,[/latex] we find that the value of [latex]f[/latex] at the local minimum is zero. Note that [latex]{f}^{\prime }\left(1\right)[/latex] is undefined, so to determine the behavior of the function at this critical point, we need to examine [latex]\underset{x\to 1}{\text{lim}}{f}^{\prime }\left(x\right).[/latex] Looking at the one-sided limits, we have

Therefore, [latex]f[/latex] has a cusp at [latex]x=1.[/latex]

Step 6: To determine concavity, we calculate the second derivative of [latex]f\text{:}[/latex]

We find that [latex]f\text{″}\left(x\right)[/latex] is defined for all [latex]x,[/latex] but is undefined when [latex]x=1.[/latex] Therefore, divide the interval [latex]\left(\text{−}\infty ,\infty \right)[/latex] into the smaller intervals [latex]\left(\text{−}\infty ,1\right)[/latex] and [latex]\left(1,\infty \right),[/latex] and choose test points to evaluate the sign of [latex]f\text{″}\left(x\right)[/latex] in each of these intervals. As we did earlier, let [latex]x=0[/latex] and [latex]x=2[/latex] be test points as shown in the following table.

From this table, we conclude that [latex]f[/latex] is concave down everywhere. Combining all of this information, we arrive at the following graph for [latex]f.[/latex]