Step 1. Since $f$ is a polynomial, the domain is the set of all real numbers.

Step 2. When $x=0, \, f(x)=2$. Therefore, the $y$-intercept is $(0,2)$. To find the $x$-intercepts, we need to solve the equation $(x-1)^2 (x+2)=0$, which gives us the $x$-intercepts $(1,0)$ and $(-2,0)$

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

Since the leading term of $f$ is $x^3$, we conclude that $f$ behaves like $y=x^3$ as $x\to \pm \infty$.

Step 4. Since $f$ is a polynomial function, it does not have any vertical asymptotes.

Step 5. The first derivative of $f$ is

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

From the table, we see that $f$ has a local maximum at $x=-1$ and a local minimum at $x=1$. Evaluating $f(x)$ at those two points, we find that the local maximum value is $f(-1)=4$ and the local minimum value is $f(1)=0$.

Step 6. The second derivative of $f$ is

The second derivative is zero at $x=0$. Therefore, to determine the concavity of $f$, divide the interval $(−\infty ,\infty)$ into the smaller intervals $(−\infty ,0)$ and $(0,\infty )$, and choose test points $x=-1$ and $x=1$ to determine the concavity of $f$ 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 5, that $f$ has a local maximum at $x=-1$ and a local minimum at $x=1$. In addition, the information found in step 5—namely, $f$ has a local maximum at $x=-1$ and a local minimum at $x=1$, and $f^{\prime}(x)=0$ at those points—combined with the fact that $f^{\prime \prime}$ changes sign only at $x=0$ confirms the results found in step 6 on the concavity of $f$.

Combining this information, we arrive at the graph of $f(x)=(x-1)^2 (x+2)$ shown in the following graph.

Figure 23.

Step 1. The function $f$ is defined as long as the denominator is not zero. Therefore, the domain is the set of all real numbers $x$ except $x=\pm 1$.

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

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

Therefore, $f$ has a horizontal asymptote of $y=-1$ as $x\to \infty$ and $x\to −\infty$.

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

In addition, by looking at each one-sided limit as $x\to −1$, we find that

Step 5. Calculate the first derivative:

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

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

Step 6. Calculate the second derivative:

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

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

Figure 25.

Step 1. The domain of $f$ is the set of all real numbers $x$ except $x=1$.

Step 2. Find the intercepts. We can see that when $x=0$, $f(x)=0$, so $(0,0)$ 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, $f$ must have an oblique asymptote. To find the oblique asymptote, use long division of polynomials to write

Since $1/(x-1)\to 0$ as $x\to \pm \infty$, $f(x)$ approaches the line $y=x+1$ as $x\to \pm \infty$. The line $y=x+1$ is an oblique asymptote for $f$.

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

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

Step 5. Calculate the first derivative:

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

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

Step 6. Calculate the second derivative:

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

From the information gathered, we arrive at the following graph for $f.$

Figure 27.

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

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

Step 3: Since $\underset{x\to \pm \infty }{\lim}(x-1)^{2/3}=\infty$, the function continues to grow without bound as $x\to \infty$ and $x\to −\infty$.

Step 4: The function has no vertical asymptotes.

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

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

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

Therefore, $f$ has a cusp at $x=1$.

Step 6: To determine concavity, we calculate the second derivative of $f$:

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

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

Figure 28.