Sketching Graphs

Core ≈ 40 min Chapter 12 8 exercises
Why this matters

A sketch is a claim your derivative makes about the world. Sign tables turn f′ and f″ from formulas into a map of the curve: where it rises, where it falls, where it cups, where it caps, and where it settles or escapes. Master this and you will never need to plot twenty points again — two small tables will build the picture for you.

1 · Why sign tables matter

A derivative is a rate. Its sign — not its size — tells you the direction of travel. f′(x) > 0 means the function is increasing; f′(x) < 0 means decreasing. Likewise f″ tells the direction of bending: f″(x) > 0 concave up, f″(x) < 0 concave down. A sign table condenses this into one glance: split the number line at every critical point (zeros of f′, zeros of f″, or points where either fails to exist), then test one x per interval.

2 · Table of signs for f′

Recipe: find where f′ vanishes or is undefined, sort them, then test one x per interval and record + or −. Between consecutive sign changes the function is monotonic; a + → − change marks a local max, − → + a local min, and no change means the derivative never actually switched direction.

Worked example 1

Sign table for f′ of f(x) = x³ − 3x

  1. Differentiate: f′(x) = 3x² − 3 = 3(x − 1)(x + 1). Zeros at x = −1 and x = 1.
  2. Test intervals: x = −2 gives f′ = 9 > 0; x = 0 gives f′ = −3 < 0; x = 2 gives f′ = 9 > 0.
  3. Read the shape: increasing on (−∞, −1), decreasing on (−1, 1), increasing on (1, ∞). So x = −1 is a local max (value 2) and x = 1 is a local min (value −2).

3 · Table of signs for f″

Do the same with f″. Its zeros split the line into intervals of fixed concavity. Where f″ changes from + to − (or back), the curve switches from cupping to capping — an inflection point. For our cubic, f″(x) = 6x, which is negative for x < 0 (concave down) and positive for x > 0 (concave up). The sign changes at x = 0, so (0, 0) is the inflection point, exactly halfway between the max and the min.

4 · The Big Method

The Big Method is the full pipeline for any curve. Work through the steps in order — each one removes a whole class of possible surprises:

  1. Domain: where is f defined? Note holes, vertical asymptotes, and endpoints.
  2. Intercepts: set x = 0 and y = 0; find where the curve crosses the axes.
  3. Asymptotes: vertical at denominators that vanish (with nonzero numerator); horizontal from leading-end behaviour — compare degrees, or take the limit at ±∞; oblique when the numerator's degree exceeds the denominator's by one.
  4. Sign table for f′: intercepts of f′ give extrema and monotonicity.
  5. Sign table for f″: concavity and inflection points.
  6. Assemble: plot intercepts, extrema, inflection points, and asymptotes; connect them smoothly; verify against a grapher.
Worked example 2

Sketch f(x) = x³ − 3x with the Big Method

  1. Domain: all reals (polynomial).
  2. Intercepts: f(0) = 0; y = 0 gives x(x² − 3) = 0 → x = 0, ±√3 ≈ ±1.73.
  3. Asymptotes: none (polynomial), but degree 3 with odd leading power: escapes to −∞ left, +∞ right.
  4. f′ table: from Example 1 — increasing (−∞, −1), decreasing (−1, 1), increasing (1, ∞). Local max (−1, 2), local min (1, −2).
  5. f″ table: f″ = 6x; concave down on (−∞, 0), concave up on (0, ∞). Inflection at (0, 0).
  6. Assemble: draw the curve entering from the bottom-left, rising through (−√3, 0) to (−1, 2), falling through (0, 0) to (1, −2), rising through (√3, 0) and exiting top-right. One inflection at the origin, symmetric about it (the cubic is odd).

5 · Rational functions and their asymptotes

For rational functions the asymptotes carry the sketch. f(x) = 1/(x² + 1): domain all reals (x² + 1 never vanishes, so no vertical asymptotes); f(0) = 1 maximum; y = 0 is a horizontal asymptote at both ends; f′(x) = −2x/(x² + 1)² vanishes only at x = 0 (the peak); f″ changes sign at x = ±1/√3 ≈ ±0.58 — two inflection points flanking the peak. The result is a bell curve, every point below y = 1, hugging the x-axis as |x| → ∞.

Contrast with g(x) = x/(x + 1)²: the denominator vanishes at x = −1 with nonzero numerator, so x = −1 is a vertical asymptote; the denominator outgrows the numerator, so y = 0 is the horizontal asymptote. Intercepts: x = 0 (value 0) is the local max; the curve plunges to −∞ as x → −1 from the left and rebounds from +∞ on the right, then settles toward 0 from above as x → ∞. Locate intercepts, asymptote behaviour, and extrema before drawing a single stroke — asymptotes are scaffolding, not decoration.

Watch out

A zero denominator does not automatically mean a vertical asymptote — cancel common factors first. Likewise, if f′(x) = 0 at a point but the sign does not change, it is not an extremum. And infinity behaviour must be checked at both ends: a function can hug y = 0 on the left and blow up on the right if the degrees only mismatch at one side (or if your domain is one-sided).

6 · Verify with a grapher

A grapher is the umpire, not the coach. Sketch first, then plot and check every feature — intercepts, extrema, inflection points, asymptotes — appears where you predicted. Any mismatch means a sign-table slip or a missed asymptote.

Chapter formula sheetf′ > 0 ⇔ f increasing on that interval
f′ < 0 ⇔ f decreasing on that interval
f′ changes + → − ⇒ local max; − → + ⇒ local min
f″ > 0 concave up; f″ < 0 concave down
Inflection point ⇔ f″ changes sign
Vertical asymptote: denominator 0, numerator ≠ 0
Horizontal asymptote: lim x→±∞ f(x)

Common mistakes

  • Sketching before computing f′ and f″ — drawing by eye guarantees missed extrema and wrong concavity.
  • Missing the sign-change check at a zero of f″ and inventing inflection points.
  • Forgetting vertical asymptotes can occur only where the denominator vanishes after cancelling.
  • Assigning a horizontal asymptote by looking at one end only.
  • Letting the curve cross a vertical asymptote — it never does; behaviour on each side must be checked separately.
  • Comparing a sketch to a grapher without asking why each feature appears.

✎ Practice — marked instantly

On the interval (1, 4), f′(x) is negative. What does this tell you about f on that interval?
  • A. f is increasing on (1, 4)
  • B. f is decreasing on (1, 4)
  • C. f is concave up on (1, 4)
  • D. f has an inflection point at x = 2.5

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Find the local maximum value of y = x³ − 3x.

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State the horizontal asymptote of y = 1/(x² + 1). Answer as e.g. y=0.

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Which vertical asymptote does the curve y = x/(x + 1)² have?
  • A. x = 0
  • B. x = 1
  • C. x = −1
  • D. no vertical asymptote

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Give the x-coordinate of the inflection point of y = x³ − 3x.

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Find the local minimum value of y = x³ − 3x.

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f″(x) > 0 on an interval. The graph of f is:
  • A. concave down there
  • B. concave up there
  • C. decreasing there
  • D. flat there

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Find the value c such that y = (3x² + 1)/(x² + 5) has horizontal asymptote y = c.

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