
# 3.7 Curve skectching

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## Guidelines for Drawing the Graph of a Function

We now have enough analytical tools to draw graphs of a wide variety of algebraic and transcendental functions. Before showing how to graph specific functions, let’s look at a general strategy to use when graphing any function.

Problem-Solving Strategy: Drawing the Graph of a Function

Given a function $$f$$, use the following steps to sketch a graph of $$f$$:

1. Domain: Determine the domain of the function.
2. Intercepts: Locate the $$x$$- and $$y$$-intercepts.
3. Symmetry: If $$f(-x)=f(x)$$, then $$f$$ is symmetric about $$y$$-axis, also called even function. If $$f(-x)=-f(x)$$, then $$f$$ is symmetric about the origin, also called odd function.
4. Asymptote(s): Evaluate $$\displaystyle \lim_{x→∞}f(x)$$ and $$\displaystyle \lim_{x→−∞}f(x)$$ to determine the end behavior. If either of these limits is a finite number $$L$$, then $$y=L$$ is a horizontal asymptote. If either of these limits is $$∞$$ or $$−∞$$, determine whether $$f$$ has an oblique asymptote. If $$f$$ is a rational function such that $$f(x)=\frac{p(x)}{q(x)}$$, where the degree of the numerator is greater than the degree of the denominator, then $$f$$ can be written as $f(x)=\frac{p(x)}{q(x)}=g(x)+\frac{r(x)}{q(x),}$ where the degree of $$r(x)$$ is less than the degree of $$q(x)$$. The values of $$f(x)$$ approach the values of $$g(x)$$ as $$x→±∞$$. If $$g(x)$$ is a linear function, it is known as an oblique asymptote. Determine whether $$f$$ has any vertical asymptotes.
5. Local extrema: Calculate $$f′.$$ Find all critical points and determine the intervals where $$f$$ is increasing and where $$f$$ is decreasing. Determine whether $$f$$ has any local extrema.
6. Concavity: Calculate $$f''.$$ Determine the intervals where $$f$$ is concave up and where $$f$$ is concave down. Use this information to determine whether $$f$$ has any inflection points. The second derivative can also be used as an alternate means to determine or verify that $$f$$ has a local extremum at a critical point.

Now let’s use this strategy to graph several different functions. We start by graphing a polynomial function.

Example $$\PageIndex{1}$$: Sketching a Graph of a Polynomial

Sketch a graph of $$f(x)=(x−1)^2(x+2).$$

Solution

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 $$(x−1)^2(x+2)=0$$, gives us the $$x$$-intercepts $$(1,0)$$ and $$(−2,0)$$.

Step 3: Consider $$f(-x)= ((-x)−1)^2((-x)+2)=(x+1)^2(2-x),$$ which is neither equal to $$-f(x)$$ nor $$f(x)$$. There fore there is no symmetry.

Step 4: We need to evaluate the end behavior of $$f.$$As $$x→∞, (x−1)^2→∞$$ and $$(x+2)→∞$$. Therefore, $$\displaystyle \lim_{x→∞}f(x)=∞$$ As $$x→−∞, (x−1)^2→∞$$ and $$(x+2)→−∞$$. Therefore, $$\displaystyle \lim_{x→∞}f(x)=−∞$$. To get even more information about the end behavior of $$f$$, we can multiply the factors of $$f$$.When doing so, we see that

$f(x)=(x−1)^2(x+2)=x^3−3x+2.$

Since the leading term of $$f$$ is $$x^3$$, we conclude that $$f$$ behaves like $$y=x^3$$ as $$x→±∞.$$

Step 5: Since $$f$$ is a polynomial function, it does not have any vertical asymptotes.

Step 6: The first derivative of $$f$$ is

$f′(x)=3x^2−3.$

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

Interval Test point Sign of Derivative $$f'(x)=3x^2−3=3(x−1)(x+1)$$ Conclusion
$$(−∞,−1)$$ $$x=−2$$ $$(+)(−)(−)=+$$ $$f$$ is increasing
$$(−1,1)$$ $$x=0$$ $$(+)(−)(+)=−$$ $$f$$ decreasing
$$(1,∞)$$ (x=2\) $$(+)(+)(+)=+$$ $$f$$ is increasing

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 7: The second derivative of $$f$$ is

$f''(x)=6x.$

The second derivative is zero at $$x=0.$$ Therefore, to determine the concavity of $$f$$, divide the interval $$(−∞,∞)$$ into the smaller intervals $$(−∞,0)$$ and $$(0,∞)$$, 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.

Interval Test Point Sign of $$''(x)=6x$$ Conclusion
$$(−∞,0)$$ $$x=−1$$ $$−$$ $$f$$ is concave down..
$$(0,∞)$$ $$x=1$$ $$+$$ $$f$$ is concave up.

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 $$f′(x)=0$$ at those points—combined with the fact that $$f''$$ 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.

Exercise $$\PageIndex{1}$$

Sketch a graph of $$f(x)=(x−1)^3(x+2).$$

$$f$$ is a fourth-degree polynomial.

Example $$\PageIndex{2}$$: Sketching a Rational Function

Sketch the graph of $$f(x)=\frac{x^2}{(1−x^2)}$$.

Solution

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 $$x=±1.$$

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: Consider $$f(-x) =\frac{(-x)^2}{(1−(-x)^2)}=\frac{x^2}{(1−x^2)}$$=f(x)\). Therefore $$f$$ is symmetric about $$y$$-axis.

Step 4: 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

$$\displaystyle \lim_{x→±∞}\frac{x^2}{1−x^2}=\lim_{x→±∞}\frac{1}{\frac{1}{x^2}−1}=−1.$$

Therefore, $$f$$ has a horizontal asymptote of $$y=−1$$ as $$x→∞$$ and $$x→−∞.$$

Step 5: 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 $$x=±1$$. To determine whether the lines $$x=1$$ or $$x=−1$$ are vertical asymptotes of $$f$$, evaluate $$\displaystyle \lim_{x→1}f(x)$$ and $$\displaystyle \lim_{x→−1}f(x)$$. By looking at each one-sided limit as $$x→1,$$ we see that

$$\displaystyle \lim_{x→1^+}\frac{x^2}{1−x^2}=−∞$$ and $$\displaystyle \lim_{x→1^−}\frac{x^2}{1−x^2}=∞.$$

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

$$\displaystyle \lim_{x→−1^+}\frac{x^2}{1−x^2}=∞$$ and $$\displaystyle \lim_{x→−1^−}\frac{x^2}{1−x^2}=−∞.$$

Step 6: Calculate the first derivative:

$$f′(x)=\frac{(1−x^2)(2x)−x^2(−2x)}{(1−x^2)^2}=\frac{2x}{(1−x^2)^2}$$.

Critical points occur at points $$x$$ where $$f′(x)=0$$ or $$f′(x)$$ is undefined. We see that $$f′(x)=0$$ when $$x=0.$$ The derivative $$f′$$ is not undefined at any point in the domain of $$f$$. However, $$x=±1$$ are not in the domain of $$f$$. Therefore, to determine where $$f$$ is increasing and where $$f$$ is decreasing, divide the interval $$(−∞,∞)$$ into four smaller intervals: $$(−∞,−1), (−1,0), (0,1),$$ and $$(1,∞)$$, and choose a test point in each interval to determine the sign of $$f′(x)$$ 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.

Interval Test point Sign of $$f′(x)=\frac{2x}{(1−x^2)^2}$$ Conclusion
$$(−∞,−1)$$ $$x=−2$$ $$−/+=−$$ $$f$$ is decreasing.
$$(−1,0)$$ $$x=−/2$$ $$−/+=−$$ $$f$$ is decreasing.
$$(0,1)$$ $$x=1/2$$ $$+/+=+$$ $$f$$ is increasing.
$$(1,∞)$$ $$x=2$$ $$+/+=+$$ $$f$$ is increasing.

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

Step 7: Calculate the second derivative:

$$f''(x)=\frac{(1−x^2)^2(2)−2x(2(1−x^2)(−2x))}{(1−x^2)^4}$$

$$=\frac{(1−x^2)[2(1−x^2)+8x^2]}{(1−x^2)^4}$$

$$=\frac{2(1−x^2)+8x^2}{(1−x^2)^3}$$

$$=\frac{6x^2+2}{(1−x^2)^3}.$$

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

Interval Test Point Sign of $$f''(x)=\frac{6x^2+2}{(1−x^2)^3}$$ Conclusion
$$(−∞,−1)$$ $$x=−2$$ $$+/−=−$$ $$f$$ is concave down.
$$(−1,−1)$$ $$x=0$$ $$+/+=+$$ $$f$$ is concave up
$$(1,∞)$$ $$x=2$$ $$+/−=−$$ $$f$$ is concave down.

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 $$x=1.$$

Exercise $$\PageIndex{2}$$

Sketch a graph of $$f(x)=\frac{(3x+5)}{(8+4x).}$$

Hint

A line $$y=L$$ is a horizontal asymptote of $$f$$ if the limit as $$x→∞$$ or the limit as $$x→−∞$$ of $$f(x)$$ is $$L$$. A line $$x=a$$ is a vertical asymptote if at least one of the one-sided limits of $$f$$ as $$x→a$$ is $$∞$$ or $$−∞.$$

Example $$\PageIndex{3}$$: Sketching a Rational Function with an Oblique Asymptote

Sketch the graph of $$f(x)=\frac{x^2}{(x−1)}$$

Solution

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: Consider $$f(-x)=\frac{(-x)^2}{((-x)−1)}= \frac{x^2}{(-x−1)}$$, which is neither equal to $$-f(x)$$ nor $$f(x)$$. There fore there is no symmetry.

Step 4: 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

$$f(x)=\frac{x^2}{x−1}=x+1+\frac{1}{x−1}$$.

Since $$1/(x−1)→0$$ as $$x→±∞, f(x)$$ approaches the line $$y=x+1$$ as $$x→±∞$$. The line $$y=x+1$$ is an oblique asymptote for $$f$$.

Step 5: 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 $$x→1,$$ we find

$$\displaystyle \lim_{x→1^+}\frac{x^2}{x−1}=∞$$ and $$\displaystyle \lim_{x→1^−}\frac{x^2}{x−1}=−∞.$$

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 6: Calculate the first derivative:

$$f′(x)=\frac{(x−1)(2x)−x^2(1)}{(x−1)^2}=\frac{x^2−2x}{(x−1)^2}.$$

We have $$f′(x)=0$$ 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 $$(−∞,∞)$$ into the smaller intervals $$(−∞,0), (0,1), (1,2),$$ and $$(2,∞)$$, and choose a test point from each interval to evaluate the sign of $$f′(x)$$ 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.

Interval Test Point Sign of $$f''(x)=\frac{2}{(x−1)^3}$$ Conclusion
$$(−∞,1)$$ $$x=0$$ $$+/−=−$$ $$f$$ is concave down.
$$(1,∞)$$ (x=2\) $$+/+=+$$ $$f$$ is concave up

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

Exercise $$\PageIndex{3}$$

Find the oblique asymptote for $$f(x)=\frac{(3x^3−2x+1)}{(2x^2−4)}$$.

Hint

Use long division of polynomials.

$$y=\frac{3}{2}x$$

Example $$\PageIndex{4}$$: Sketching the Graph of a Function with a Cusp

Sketch a graph of $$f(x)=(x−1)^{2/3}$$

Solution

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: Consider $$f(-x)=((-x)−1)^{2/3}=(x+1)^{2/3}$$, which is neither equal to $$-f(x)$$ nor $$f(x)$$. There fore there is no symmetry.

Step 4: Since $$\displaystyle \lim_{x→±∞}(x−1)^{2/3}=∞,$$ the function continues to grow without bound as $$x→∞$$ and $$x→−∞.$$

Step 5: The function has no vertical asymptotes.

Step 6: To determine where $$f$$ is increasing or decreasing, calculate $$f′.$$ We find

$f′(x)=\frac{2}{3}(x−1)^{−1/3}=\frac{2}{3(x−1)^{1/3}}$

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

Interval Test Point Sign of $$f′(x)=\frac{2}{3(x−1)^{1/3}}$$ Conclusion
$$(−∞,1)$$ $$x=0$$ $$+/−=−$$ $$f$$ is decreasing
$$(1,∞)$$ $$x=2$$ $$+/+=+$$ $$f$$ is increasing

We conclude that $$h$$as 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 $$f′(1)$$ is undefined, so to determine the behavior of the function at this critical point, we need to examine $$\displaystyle \lim_{x→1}f′(x).$$ Looking at the one-sided limits, we have

$\lim_{x→1^+}\frac{2}{3(x−1)^{1/3}}=∞\) and $$\displaystyle \lim_{x→1^−}\frac{2}{3(x−1)^{1/3}}=−∞.$ Therefore, \(f$$ has a cusp at $$x=1.$$

Step 7: To determine concavity, we calculate the second derivative of $$f:$$

$f''(x)=−\dfrac{2}{9}(x−1)^{−4/3}=\dfrac{−2}{9(x−1)^{4/3}}.$

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

Interval Test Point Sign of $$f''(x)=\frac{−2}{3(x−1)^{4/3}}$$ Conclusion
$$(−∞,1)$$ $$x=0$$ $$−/+=−$$ $$f$$ is concave down
$$(1,∞)$$ $$x=2$$ \−(+/+=−\) $$f$$ is concave down

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

Exercise $$\PageIndex{4}$$

Consider the function $$f(x)=5−x^{2/3}$$. Determine the point on the graph where a cusp is located. Determine the end behavior of $$f$$.

Hint

A function $$f$$ has a cusp at a point a if $$f(a)$$ exists, $$f'(a)$$ is undefined, one of the one-sided limits as $$x→a$$ of $$f'(x) is +∞$$, and the other one-sided limit is $$−∞.$$

The function $$f$$ has a cusp at $$(0,5) \lim_{x→0^−}f′(x)=∞$$, $$\displaystyle \lim_{x→0^+}f′(x)=−∞$$. For end behavior, $$\displaystyle \lim_{x→±∞}f(x)=−∞.$$

Example $$\PageIndex{5}$$: Exponential function

Sketch a graph of $$f(x)=e^{\frac{-x^2}{2}}$$.

## Key Concepts

• The limit of $$f(x)$$ is $$L$$ as $$x→∞$$ (or as $$x→−∞)$$ if the values $$f(x)$$ become arbitrarily close to $$L$$ as $$x$$becomes sufficiently large.
• The limit of $$f(x)$$ is $$∞$$ as $$x→∞$$ if $$f(x)$$ becomes arbitrarily large as $$x$$ becomes sufficiently large. The limit of $$f(x)$$ is $$−∞$$ as $$x→∞$$ if $$f(x)<0$$ and $$|f(x)|$$ becomes arbitrarily large as $$x$$ becomes sufficiently large. We can define the limit of $$f(x)$$ as $$x$$ approaches $$−∞$$ similarly.
• For a polynomial function $$p(x)=a_nx^n+a_{n−1}x^{n−1}+…+a_1x+a_0,$$ where $$a_n≠0$$, the end behavior is determined by the leading term $$a_nx^n$$. If $$n≠0, p(x)$$ approaches $$∞$$ or $$−∞$$at each end.
• For a rational function $$f(x)=\frac{p(x)}{q(x),}$$ the end behavior is determined by the relationship between the degree of $$p$$ and the degree of $$q$$. If the degree of $$p$$ is less than the degree of $$q$$, the line $$y=0$$ is a horizontal asymptote for $$f$$. If the degree of $$p$$ is equal to the degree of $$q$$, then the line $$y=\frac{a_n}{b_n}$$ is a horizontal asymptote, where $$a_n$$ and $$b_n$$ are the leading coefficients of $$p$$ and $$q$$, respectively. If the degree of $$p$$ is greater than the degree of $$q$$, then $$f$$ approaches $$∞$$ or $$−∞$$ at each end.

## Glossary

end behavior
the behavior of a function as $$x→∞$$ and $$x→−∞$$
horizontal asymptote
if $$\displaystyle \lim_{x→∞}f(x)=L$$ or $$\displaystyle \lim_{x→−∞}f(x)=L$$, then $$y=L$$ is a horizontal asymptote of $$f$$
infinite limit at infinity
a function that becomes arbitrarily large as x becomes large
limit at infinity
a function that becomes arbitrarily large as $$x$$ becomes large
oblique asymptote
the line $$y=mx+b$$ if $$f(x)$$ approaches it as $$x→∞$$ or$$x→−∞$$

### Contributors

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• Pamini Thangarajah (Mount Royal University, Calgary, Alberta, Canada)