Complete Curve Sketching
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 4.6: “Limits at Infinity and Asymptotes” |
| Direct link | https://openstax.org/books/calculus-volume-1/pages/4-6-limits-at-infinity-and-asymptotes |
| Textbook used in class | Stewart, Calculus, Section 3.5: “Summary of Curve Sketching” |
Opening Scenario
This lesson applies every tool from Chapter 3 in a single extended example. The goal is to carry out all eight steps from the curve-sketching checklist on a function with asymptotes, local extrema, and inflection points, and to produce a sketch that captures every qualitative feature of the graph.
Before computing anything, always predict: look at the formula, identify the degree and sign of leading terms, and form a rough expectation of what the graph should look like. Then the calculation either confirms or corrects that picture.
Worked Example: A Rational Function
Example. Sketch the curve $y = f(x) = \dfrac{2x^2}{x^2 - 1}$.
(This is a rational function with vertical asymptotes. Stewart 3.5 includes a similar analysis.)
Prediction first. The function looks like $2x^2/x^2 = 2$ for large $|x|$ (the $-1$ in the denominator becomes negligible). So the graph should settle near $y = 2$ far to the right and far to the left. There are likely vertical asymptotes where $x^2 - 1 = 0$. The numerator $2x^2 \geq 0$, so the function is non-negative wherever it is defined (except at $x = 0$, where $f(0) = 0$).
Step A: Domain
$x^2 - 1 = 0 \Rightarrow x = \pm 1$. Exclude $x = 1$ and $x = -1$.
Domain: $(-\infty, -1) \cup (-1, 1) \cup (1, \infty)$.
Step B: Intercepts
$y$-intercept: $f(0) = 0/(0-1) = 0$. Point: $(0, 0)$.
$x$-intercepts: $f(x) = 0 \Rightarrow 2x^2 = 0 \Rightarrow x = 0$. Same point: $(0, 0)$.
The graph passes through the origin and that is the only intercept.
Step C: Symmetry
$f(-x) = \dfrac{2(-x)^2}{(-x)^2 - 1} = \dfrac{2x^2}{x^2 - 1} = f(x)$.
Even function: symmetric about the $y$-axis. We only need to analyze $x \geq 0$ and reflect.
Step D: Asymptotes
Vertical. At $x = 1$: $\lim_{x \to 1} f(x)$. Denominator $\to 0$; numerator $\to 2 \neq 0$. So $f(x) \to \pm\infty$.
- $\lim_{x \to 1^+}$: $x^2 - 1 \to 0^+$ (since $(1.1)^2 - 1 = 0.21 > 0$), numerator $\to 2 > 0$: limit is $+\infty$.
- $\lim_{x \to 1^-}$: $x^2 - 1 \to 0^-$ (since $(0.9)^2 - 1 = -0.19 < 0$), numerator $\to 2 > 0$: limit is $-\infty$.
By symmetry, at $x = -1$: $\lim_{x \to -1^-} = +\infty$ and $\lim_{x \to -1^+} = -\infty$.
Horizontal. $\lim_{x \to \pm\infty} \dfrac{2x^2}{x^2 - 1} = \lim_{x\to\infty} \dfrac{2}{1 - 1/x^2} = 2$.
Horizontal asymptote: $y = 2$.
Slant. Degrees are equal ($2$ and $2$); no slant asymptote.
Step E: Intervals of Increase and Decrease
$f(x) = \dfrac{2x^2}{x^2 - 1}$.
Quotient rule: $f'(x) = \dfrac{4x(x^2-1) - 2x^2 \cdot 2x}{(x^2-1)^2} = \dfrac{4x^3 - 4x - 4x^3}{(x^2-1)^2} = \dfrac{-4x}{(x^2-1)^2}$.
$f'(x) = 0$ at $x = 0$. $f'$ is undefined at $x = \pm 1$ (not in the domain).
Sign of $f'(x) = -4x/(x^2-1)^2$: the denominator $(x^2-1)^2 > 0$ except at $\pm 1$. So the sign of $f'$ equals the sign of $-4x$.
| Interval | Sign of $-4x$ | Direction |
|---|---|---|
| $x < -1$ | $+$ (since $x < 0$) | Increasing |
| $-1 < x < 0$ | $+$ | Increasing |
| $0 < x < 1$ | $-$ (since $x > 0$) | Decreasing |
| $x > 1$ | $-$ | Decreasing |
Increasing on $(-\infty, -1)$ and $(-1, 0)$. Decreasing on $(0, 1)$ and $(1, \infty)$.
Step F: Local Extrema
At $x = 0$: $f'$ goes from $+$ to $-$. Local maximum.
$f(0) = 0$.
Local maximum of $0$ at $x = 0$. (The curve dips to its lowest value in the middle piece and the maximum there is $0$.)
No local extremum at $x = \pm 1$ (not in the domain).
Step G: Concavity and Inflection Points
$f'(x) = -4x(x^2-1)^{-2}$. Use the product and chain rules.
$f''(x) = -4(x^2-1)^{-2} + (-4x) \cdot (-2)(x^2-1)^{-3} \cdot 2x$
$= -4(x^2-1)^{-2} + 16x^2(x^2-1)^{-3}$
$= (x^2-1)^{-3}\left[-4(x^2-1) + 16x^2\right]$
$= \frac{-4x^2 + 4 + 16x^2}{(x^2-1)^3}$
$= \frac{12x^2 + 4}{(x^2-1)^3}$
$= \frac{4(3x^2 + 1)}{(x^2-1)^3}$.
Since $3x^2 + 1 > 0$ always and $4 > 0$, the sign of $f''$ equals the sign of $(x^2 - 1)^3$, which is the sign of $x^2 - 1$.
| Interval | Sign of $x^2 - 1$ | Concavity |
|---|---|---|
| $|x| < 1$ ($-1 < x < 1$) | $-$ | Concave down |
| $|x| > 1$ ($x < -1$ or $x > 1$) | $+$ | Concave up |
$f''$ changes sign at $x = \pm 1$, but those points are not in the domain. There are no inflection points (the concavity transitions happen at the excluded points).
Step H: The Sketch
Features assembled:
- Axes: draw coordinate axes.
- Dashed lines: vertical asymptotes $x = -1$ and $x = 1$; horizontal asymptote $y = 2$.
- Intercept: origin $(0, 0)$.
- Local max: $(0, 0)$ -- the only labeled extremum.
- End behavior: approaches $y = 2$ from below as $|x| \to \infty$ (confirmed: $f(x) = 2x^2/(x^2-1) < 2$ when $x^2 - 1 > 0$, i.e., $|x| > 1$).
Description of the sketch by piece:
- Middle piece ($-1 < x < 1$): Even symmetry. Comes from $-\infty$ as $x \to -1^+$, rises to the local max of $0$ at $x = 0$, then falls back to $-\infty$ as $x \to 1^-$. Concave down throughout this piece.
- Right piece ($x > 1$): Comes from $+\infty$ as $x \to 1^+$, decreases toward the asymptote $y = 2$ from above. Concave up.
- Left piece ($x < -1$): Mirror image of the right piece by even symmetry.
Checks and Verification
- Does $f(x) \to 2$ from above for $x > 1$? For $x > 1$: $x^2 - 1 > 0$, so $f(x) = 2x^2/(x^2-1) = 2 + 2/(x^2-1) > 2$. Yes.
- Is $f(x) < 0$ for $0 < x < 1$? For $0 < x < 1$: $x^2 - 1 < 0$, so $f(x) = 2x^2/(x^2-1) < 0$ (negative over positive... wait: $2x^2 > 0$, denominator $< 0$, so $f(x) < 0$). Yes.
- Is the local max really the maximum on $(-1, 1)$? $f(0) = 0$ and $f(x) < 0$ on $(0,1)$ and by symmetry on $(-1,0)$, so yes, $f(0) = 0$ is the highest value in the middle piece.
Everything is consistent.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Sketching outside the domain | Drawing through $x = 1$ instead of a vertical asymptote | Check domain first; draw vertical asymptotes before the curve |
| Wrong direction at asymptote | Drawing both sides of $x=1$ going upward | Compute both one-sided limits; here right side $\to +\infty$, left side $\to -\infty$ |
| Claiming $y=2$ is a maximum | Seeing the horizontal asymptote and treating it as a value attained | The asymptote is a limit, not a value; $f(x) > 2$ on $|x|>1$ but $f(x) \to 2$ and never equals $2$ |
Leveled Practice
Level 1 -- Full Analysis on a Simpler Function
Problem 1. Carry out all eight steps for $f(x) = x^3 - 3x$.
Show answer
A. Domain: all reals.
B. $y$-intercept: $(0,0)$. $x$-intercepts: $x(x^2-3) = 0 \Rightarrow x = 0, \pm\sqrt{3}$.
C. $f(-x) = -f(x)$: odd function, symmetric about origin.
D. No asymptotes (polynomial).
E. $f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$. Increasing on $(-\infty,-1)$ and $(1,\infty)$; decreasing on $(-1,1)$.
F. Local max at $x=-1$: $f(-1) = -1+3 = 2$. Local min at $x=1$: $f(1) = 1-3=-2$.
G. $f''(x) = 6x$. Concave down on $(-\infty,0)$; concave up on $(0,\infty)$. Inflection point at $(0,0)$.
H. Sketch: rises from $-\infty$, peaks at $(-1,2)$, falls through the origin (inflection), reaches valley $(-1,-2)$... correction: valley at $(1,-2)$, then rises to $+\infty$.
Boxed answer: Local max $2$ at $x=-1$; local min $-2$ at $x=1$; inflection at origin.
Common Misconceptions
a point where $f$ is large must be a local maximum. A local maximum is where $f$ switches from increasing to decreasing. The height of the function at that point is irrelevant to the definition. A function can have a local maximum at $(2, -100)$ (a small peak far below the axis) and no local maximum at a much higher point where the function is monotone. Extrema are identified by sign changes of $f'$, not by the size of $f$.
a horizontal asymptote means the graph never crosses it. A horizontal asymptote $y = L$ describes the long-run behavior: $f(x) \to L$ as $x \to \pm\infty$. The graph can cross the line $y = L$ many times for finite $x$ and still have it as a horizontal asymptote. The asymptote is a statement about limits at infinity, not a barrier for finite inputs.
Mastery Checklist
Mental Model
A complete curve sketch is a translated table of derivative information. The eight steps gather the data; the sketch translates that data into a picture. The picture is correct when it is consistent with every entry in the table: direction agrees with $f'$, bending agrees with $f''$, and the curve approaches each asymptote from the correct side.
Building the sketch is not about art -- it is about making the picture agree with the analysis. Inconsistencies in the sketch reveal computational errors more reliably than any individual check.
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