Why This Matters
Curve sketching is where everything you've learned in calculus comes together—limits, derivatives, and function behavior all unite to help you visualize what a function actually does. On the AP exam, you're being tested on your ability to connect analytical information (derivatives, limits, critical points) to graphical behavior (shape, direction, curvature). Free-response questions frequently ask you to justify why a graph looks a certain way, and multiple-choice questions love to test whether you can match derivative sign charts to curve shapes.
The key insight is that curve sketching isn't about artistic talent—it's about systematic analysis. Each step reveals a different aspect of the function: domain tells you where to look, intercepts anchor the curve, asymptotes set boundaries, derivatives reveal shape. Don't just memorize the steps—understand what information each one provides and how that information constrains what the curve can look like.
Setting the Stage: Domain and Intercepts
Before you can analyze behavior, you need to know where the function exists and where it touches the axes. These foundational steps establish the "playing field" for your sketch.
Find the Domain
- Identify all x-values where the function is defined—look for division by zero, negative values under even roots, or negative arguments in logarithms
- Express in interval notation using unions for multiple intervals; this determines where your curve can actually exist
- Mark excluded values as they often indicate vertical asymptotes or holes in the graph
Determine the Y-Intercept
- Evaluate f(0) to find where the curve crosses the vertical axis—this gives you the point (0,f(0))
- Quick anchor point that's usually easy to calculate and helps orient your entire sketch
- Note if undefined—if x=0 isn't in the domain, there's no y-intercept to plot
Find X-Intercepts
- Solve f(x)=0 by setting the numerator equal to zero (for rational functions) or factoring
- These are your zeros/roots—the points (a,0) where the curve crosses or touches the x-axis
- Multiplicity matters—even multiplicity means the curve touches and bounces; odd multiplicity means it crosses through
Compare: Y-intercept vs. X-intercepts—both anchor your curve to the axes, but y-intercepts require evaluating at a point while x-intercepts require solving an equation. On FRQs, always state intercepts as coordinate pairs, not just numbers.
Boundary Behavior: Asymptotes
Asymptotes tell you what happens at the edges—where the function blows up or where it settles down. These invisible guidelines shape the curve's behavior at extremes.
Identify Vertical Asymptotes
- Find where the denominator equals zero (and the numerator doesn't)—these x-values create infinite discontinuities
- Check the limit from both sides—the function approaches +∞ or −∞ as x→a+ and x→a−
- Draw as dashed vertical lines at x=a; the curve will approach but never cross these boundaries
Identify Horizontal or Slant Asymptotes
- Evaluate limx→±∞f(x) to find horizontal asymptotes—compare degrees of numerator and denominator
- Slant asymptotes occur when the numerator's degree exceeds the denominator's by exactly one; use polynomial long division to find the equation
- End behavior guide—horizontal asymptote y=L means the curve levels off; slant asymptote y=mx+b means it approaches a tilted line
Compare: Vertical vs. Horizontal asymptotes—vertical asymptotes restrict the domain (function is undefined there), while horizontal asymptotes describe end behavior (function can actually cross them in the middle of the graph). This distinction frequently appears in true/false questions.
First Derivative Analysis: Direction and Extrema
The first derivative reveals where the function is climbing or falling. f′(x) is the slope function—positive means rising, negative means falling.
Find Critical Points
- Solve f′(x)=0 and find where f′(x) is undefined—these x-values are your critical points
- Must be in the domain of the original function to count as critical points
- Candidates for extrema—every local max or min occurs at a critical point (but not every critical point is an extremum)
Apply the First Derivative Test
- Create a sign chart for f′(x) using critical points as dividers; test one value in each interval
- Sign change +→− indicates a local maximum; sign change −→+ indicates a local minimum
- No sign change means the critical point is neither—could be a saddle point or point where derivative is undefined
Determine Increasing/Decreasing Intervals
- f′(x)>0 means the function is increasing on that interval; f′(x)<0 means decreasing
- State intervals using domain values and critical points as boundaries
- Justification gold—on FRQs, always cite the derivative's sign when explaining why a function increases or decreases
Compare: Critical points vs. Local extrema—all local extrema occur at critical points, but not all critical points produce extrema. If an FRQ asks you to justify a maximum, you must show the sign change in f′(x), not just that f′(x)=0.
Second Derivative Analysis: Concavity and Inflection
The second derivative tells you how the curve bends. f′′(x) measures the rate of change of the slope—it determines whether the curve cups upward or downward.
Find Inflection Points
- Solve f′′(x)=0 and find where f′′(x) is undefined—these are candidates for inflection points
- Concavity must actually change—verify by checking signs of f′′(x) on either side
- Plot these points—they mark where the curve transitions from "smiling" to "frowning" or vice versa
Determine Concavity Intervals
- f′′(x)>0 means concave up (holds water, like a cup)—the curve bends upward
- f′′(x)<0 means concave down (spills water, like a frown)—the curve bends downward
- Second derivative test shortcut—at a critical point, f′′(c)>0 confirms local min; f′′(c)<0 confirms local max
Compare: First derivative test vs. Second derivative test—both classify critical points, but the first derivative test requires checking intervals while the second derivative test only requires evaluating f′′ at the critical point. Use the second derivative test when f′′ is easy to compute; use the first derivative test when f′′(c)=0 (inconclusive).
Efficiency Shortcuts: Symmetry
Checking for symmetry early can cut your work in half. Symmetric functions behave predictably, so you only need to analyze one side.
Test for Symmetry
- Even function: f(−x)=f(x)—symmetric about the y-axis; sketch the right half and mirror it
- Odd function: f(−x)=−f(x)—symmetric about the origin; sketch one quadrant and rotate 180°
- Neither is most common—but always worth a quick check, especially for polynomials and rational functions
Bringing It Together: The Final Sketch
This is where all your analysis pays off. Every feature you've identified constrains what the curve can look like.
Sketch the Curve
- Plot all key points first—intercepts, critical points, inflection points on a coordinate plane
- Draw asymptotes as dashed lines—vertical, horizontal, and slant; these are boundaries the curve respects
- Connect with appropriate shape—use your increasing/decreasing and concavity information to draw smooth curves between points
Quick Reference Table
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| Domain | Restrictions on x | Where the curve exists |
| Intercepts | f(0) and solutions to f(x)=0 | Anchor points on axes |
| Vertical Asymptotes | Denominator = 0 | Infinite discontinuities |
| Horizontal/Slant Asymptotes | limx→±∞f(x) | End behavior |
| Critical Points | f′(x)=0 or undefined | Candidates for extrema |
| Increasing/Decreasing | Sign of f′(x) | Direction of curve |
| Inflection Points | f′′(x)=0 with sign change | Concavity transitions |
| Concavity | Sign of f′′(x) | Curvature direction |
Self-Check Questions
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If f′(x)>0 and f′′(x)<0 on an interval, what can you conclude about the curve's behavior there? Describe both direction and shape.
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A function has critical points at x=−2 and x=3. The sign chart for f′(x) shows: positive, negative, negative. Classify each critical point and explain your reasoning.
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Compare and contrast how you would find vertical asymptotes versus horizontal asymptotes for the function f(x)=x2−42x2−1.
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Why might a point where f′′(x)=0 not be an inflection point? Give a specific example of a function where this occurs.
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An FRQ gives you f′(x)=x2(x−1) and asks you to find all intervals where f is increasing. Set up your analysis and identify the intervals—what's tricky about the critical point at x=0?