Graphing with Technology
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 4.6: “Limits at Infinity and Asymptotes” and 4.5: “Curve Sketching” |
| Textbook used in class | Stewart, Calculus, Section 3.6: “Graphing with Calculus and Calculators” |
Opening Scenario
A graphing calculator or computer can draw any function in seconds. Yet a default window ($-10 \leq x \leq 10$, $-10 \leq y \leq 10$) frequently misses the interesting behavior. A graph that appears flat in the wrong window might be hiding three local extrema and two inflection points outside the frame.
The skill here is not learning to use technology -- it is learning to use calculus to choose the window intelligently, then to verify that every feature the algebra predicts actually appears in the graph.
Quick Reference
Workflow for graphing with technology:
- Find critical numbers ($f' = 0$ or undefined).
- Find inflection points ($f'' = 0$ changes sign).
- Find vertical asymptotes ($f$ undefined) and horizontal asymptotes ($\lim_{x\to\pm\infty} f$).
- Choose a window that contains all critical numbers, inflection points, and representative behavior on each interval.
- Plot. Verify that the graph shows the features calculus predicted.
Key Concepts
1. Why the Default Window Fails
The default square window $[-10,10]\times[-10,10]$ is chosen arbitrarily, not based on the function. Critical points may occur at $x = 50$ or $x = 0.001$. A local minimum at height $-5000$ does not appear in a $[-10,10]$ vertical range. Technology draws exactly what you ask it to; calculus tells you what to ask for.
2. Using Critical Numbers to Set the Horizontal Range
Solve $f'(x) = 0$. Let the critical numbers be $c_1, c_2, \ldots, c_n$. A reasonable horizontal window extends a bit beyond the leftmost and rightmost critical numbers. For example, if critical numbers are $x = -3$ and $x = 5$, use $x \in [-5, 7]$.
If there are no finite critical numbers (e.g., monotone functions), choose a window that shows a clear trend toward asymptotes.
3. Using Function Values to Set the Vertical Range
Evaluate $f$ at each critical number, inflection point, and a few representative $x$-values. The vertical range should cover the minimum and maximum $y$-values found. Include room above and below so extrema do not touch the edge of the graph.
4. Verification: What to Look For
After plotting, confirm that the graph matches the calculus-derived features:
- Local maxima and minima appear where $f'$ changes sign.
- Concavity changes at inflection points.
- Vertical asymptotes appear as the graph shooting off to $\pm\infty$.
- Horizontal behavior matches the computed limits.
If the graph does not show a predicted feature, either the window is wrong or a computation error was made.
Worked Example
Graph $f(x) = x^4 - 4x^3$ on a window that reveals all key features.
Step 1 -- Calculus summary.
$f'(x) = 4x^3 - 12x^2 = 4x^2(x-3)$. Critical numbers: $x = 0$ and $x = 3$.
$f''(x) = 12x^2 - 24x = 12x(x-2)$. Inflection points where $f'' = 0$: $x = 0$ and $x = 2$.
As $x \to \pm\infty$: $f(x) \to +\infty$ (both ends up, since leading term $x^4 > 0$).
Step 2 -- Key values.
$f(0) = 0$, $f(2) = 16 - 32 = -16$, $f(3) = 81 - 108 = -27$.
Step 3 -- Window choice.
Horizontal: $x \in [-1, 5]$ (covers critical numbers at $0$ and $3$, with room).
Vertical: $y \in [-35, 20]$ (covers the minimum $f(3) = -27$ and the behavior near $x = 5$ where $f(5) = 625-500 = 125$; reduce to $[-35, 50]$ to also show the right rise).
Step 4 -- Verification on the graph.
On the graph with window $[-1,5]\times[-35,50]$:
- At $x=0$: inflection point and local minimum-like behavior (actually $f'' = 0$ and $f'=0$ but $f' \geq 0$ on both sides of 0, so $x=0$ is neither a max nor min, only an inflection point).
- At $x=3$: local (and global) minimum visible at $(-27)$.
- Both tails rise upward.
Boxed answer: Window $[-1, 5] \times [-35, 50]$ reveals the inflection at $(0,0)$, the local minimum at $(3,-27)$, and the upward trend of both tails.
Common Errors Summary
| Error | Example | Correction |
|---|---|---|
| Using default window without calculus | Graphing on $[-10,10]\times[-10,10]$ without checking critical numbers | Compute $f'(x) = 0$ first; let critical numbers guide the window |
| Setting vertical range too narrow | Minimum at $y = -27$ but using $[-10,10]$ vertically | Evaluate $f$ at critical numbers; set vertical range to cover those values |
| Trusting the picture without verification | “The graph looks like it has one max” but calculus predicts two | Cross-check: every sign change of $f'$ should appear as a turning point in the graph |
Common Misconceptions
a graph produced by technology is automatically complete and accurate.
This is the concept-image-conflicts-definition error about what a plotted curve represents. A graphing tool draws only the portion of the curve inside the chosen window. If the window does not contain a local minimum at $y = -27$, the minimum is simply absent from the picture, not absent from the function. For $f(x) = x^5 - 5x^4 + 5x^3 + 1$, the default $[-10,10] \times [-10,10]$ window crops the local minimum at $(3, -26)$ entirely, making the graph appear monotonically increasing when it is not.
finding critical numbers is optional when technology is available.
This is the height-vs-slope error applied to graphing workflow. Critical numbers are where $f' = 0$, a slope condition, not a height condition. The graph’s height at those points must be evaluated separately to set an appropriate vertical window. Skipping the derivative analysis means the window is chosen arbitrarily, and features at heights outside the default range remain invisible no matter how carefully the technology is operated.
Leveled Practice
Level 1 -- Plan the Window
Problem 1. For $f(x) = x^3 - 6x^2 + 9x + 1$, find the critical numbers and inflection point, then choose a viewing window.
Show answer
$f'(x) = 3x^2 - 12x + 9 = 3(x-1)(x-3)$. Critical numbers: $x = 1$ and $x = 3$.
$f''(x) = 6x - 12$. Inflection at $x = 2$.
Values: $f(1) = 1 - 6 + 9 + 1 = 5$; $f(2) = 8-24+18+1 = 3$; $f(3) = 27-54+27+1 = 1$.
Window: $[-1, 5] \times [-2, 8]$ shows all key features.
Level 2 -- Diagnose a Bad Window
Problem 2. A student graphs $f(x) = x^5 - 5x^4 + 5x^3 + 1$ on the default window $[-10,10]\times[-10,10]$ and reports “it looks like a monotone increasing curve with no turning points.” Where did the student go wrong, and what window should be used?
Show answer
$f'(x) = 5x^4 - 20x^3 + 15x^2 = 5x^2(x^2-4x+3) = 5x^2(x-1)(x-3)$.
Critical numbers: $x = 0, 1, 3$. All lie in $[-10,10]$, so the $x$-range is adequate.
Values: $f(0) = 1$, $f(1) = 1-5+5+1 = 2$, $f(3) = 243-405+135+1 = -26$.
The minimum $f(3) = -26$ lies outside the default vertical range $[-10,10]$. The graph was cut off at the bottom, hiding the local minimum at $x = 3$.
Better window: $[-1, 5] \times [-35, 10]$. This reveals the local max at $(1,2)$, the local min at $(3,-26)$, and the sign of the leading term ($x^5 \to -\infty$ as $x \to -\infty$, $+\infty$ as $x \to +\infty$).
Mastery Checklist
Mental Model
Technology is a verification tool, not a discovery tool. Calculus tells you what the graph must look like; technology shows you that it does. The workflow is: use calculus to predict all key features (where the extrema are, how concavity changes, what happens at infinity), then choose a window that would show those features, then plot to confirm. If the picture does not match the calculus, something is wrong -- either the window or the calculation.
Connections
Looking back
- Critical numbers (Section 3.1): Used to locate extrema for window selection.
- Concavity and inflection points (Section 3.3): Inflection points are additional landmarks for window selection.
- Curve sketching (Section 3.5): Graphing with technology extends the pencil-and-paper process with more precision.
Looking ahead
- Analyzing function families (Section 3.6): Apply the same calculus-guided analysis to families of functions with a parameter.