Curve sketching — systematic approach: Worked examples

This page assumes the tools of the parent: First derivative and monotonicity, the Second derivative test, Rational functions and asymptotes, Limits and continuity, and Optimization. We link them where they are used.

The scenario matrix

Every curve-sketching problem is a combination of the following case classes. Each row is a distinct behaviour you must be able to handle; each is covered by at least one worked example below.

#	Case class	What is special about it	Covered by
A	Polynomial, two turning points	$f^{'}$ changes sign twice; ends run to $\pm \infty$	Ex 1
B	Degenerate critical point (flat inflection)	$f^{'} = 0$ but no extremum	Ex 2
C	Rational, horizontal asymptote	denominator degree $\geq$ numerator degree	Ex 3
D	Rational, vertical asymptotes both signs	$f \to + \infty$ one side, $- \infty$ the other	Ex 3
E	Rational, oblique/slant asymptote	numerator degree $=$ denominator degree $+ 1$	Ex 4
F	Even vs odd symmetry	halve the work by reflection	Ex 1 (odd), Ex 3 (even)
G	Crosses its own horizontal asymptote	asymptote governs infinity, not everywhere	Ex 5
H	Zero / degenerate input (limit form $0/0$ )	removable hole vs genuine blow-up	Ex 6
I	Real-world word problem (optimization flavour)	curve shape answers a physical question	Ex 7
J	Exam twist — parameter in the function	behaviour depends on a constant's sign	Ex 8

We now walk each example, labelled with the cell(s) it hits.

Example 1 — Polynomial, two turning points, odd symmetry · cells A, F

Step 1 — Domain & symmetry. $f$ is a polynomial, so it is defined for all real $x$ . Check symmetry: $f (- x) = (- x)^{3} - 3 (- x) = - x^{3} + 3 x = - (x^{3} - 3 x) = - f (x) .$ So $f$ is odd — symmetric about the origin. Why this step? Odd symmetry means once we know the right half, the left half is a $18 0^{\circ}$ rotation — First derivative and monotonicity work is halved.

Step 2 — Intercepts. $f (0) = 0$ ; and $x^{3} - 3 x = x (x^{2} - 3) = 0$ gives $x = 0, \pm 3$ . Why this step? The zeros anchor where the S-curve crosses the axis.

Step 3 — First derivative. $f^{'} (x) = 3 x^{2} - 3 = 3 (x^{2} - 1) .$ Set $f^{'} = 0$ : $x = \pm 1$ . Sign of $f^{'}$ : outside $[- 1, 1]$ we have $x^{2} > 1$ so $f^{'} > 0$ (rising); inside, $f^{'} < 0$ (falling). So $x = - 1$ is a $+ \to -$ change → local max; $x = 1$ is $- \to +$ → local min. Heights: $f (- 1) = (- 1) - 3 (- 1) = 2$ , and by oddness $f (1) = - 2$ . Why this step? The sign of $f^{'}$ is the monotonicity engine; its zeros are the turning candidates.

Step 4 — Second derivative. $f^{''} (x) = 6 x .$ $f^{''} < 0$ for $x < 0$ (concave down, a frown), $f^{''} > 0$ for $x > 0$ (concave up, a smile), switching at $x = 0$ → inflection at $(0, 0)$ . Why this step? The Second derivative test confirms the max/min ( $f^{''} (- 1) = - 6 < 0$ ✓ max, $f^{''} (1) = 6 > 0$ ✓ min) and locates the bend switch.

Figure below — what to observe. The burnt-orange S-curve climbs from the lower left, reaches the teal peak at $(- 1, 2)$ , falls through the ink inflection square at the origin where it briefly straightens, bottoms out at the plum valley $(1, - 2)$ , then climbs again. Notice the picture is a $18 0^{\circ}$ rotation about the origin — that is oddness made visible.

Figure — Curve sketching — systematic approach

Figure 1 — Cubic $f (x) = x^{3} - 3 x$ : local max $(- 1, 2)$ , local min $(1, - 2)$ , inflection at the origin.

Verify: compute directly: $f (- 1) = (- 1)^{3} - 3 (- 1) = - 1 + 3 = 2$ ✓; $f (1) = 1 - 3 = - 2$ ✓; $f^{''} (- 1) = 6 (- 1) = - 6 < 0$ (max) ✓; $f^{''} (1) = 6 > 0$ (min) ✓; $f (0) = 0$ (inflection height) ✓. (Machine-checked below.)

Example 2 — Degenerate critical point (flat inflection) · cell B

Step 1 — First derivative and its sign. $f^{'} (x) = 3 x^{2} .$ This is $\geq 0$ everywhere, equal to $0$ only at $x = 0$ . So $f^{'}$ does not change sign at $x = 0$ — it is $+$ on both sides. Why this step? The First Derivative Test says an extremum needs a sign change. No change ⇒ no max, no min.

Step 2 — Second derivative. $f^{''} (x) = 6 x,$ which changes sign $- \to +$ at $x = 0$ . So $x = 0$ is an inflection point — but with a horizontal tangent. This is a flat (stationary) inflection. Why this step? It shows the parent's mistake-callout in action: $f^{'} = 0$ alone does not guarantee an extremum.

Verify: $f^{'} (- 1) = 3 (- 1)^{2} = 3 > 0$ and $f^{'} (1) = 3 > 0$ (same sign both sides ⇒ no extremum); $f^{''} (- 1) = - 6 < 0$ and $f^{''} (1) = 6 > 0$ (concavity flips ⇒ inflection). (Machine-checked below.)

Example 3 — Rational with horizontal wall & twin vertical walls, even · cells C, D, F

Step 1 — Domain & symmetry. $x^{2} - 1 = 0$ at $x = \pm 1$ , so domain is $R ∖ {- 1, 1}$ . And $f (- x) = \frac{x ^{2}}{x ^{2} - 1} = f (x)$ → even, mirror over the $y$ -axis. Why this step? Even symmetry lets us study $x \geq 0$ and reflect.

Step 2 — Asymptotes (the limits made explicit). Vertical: as $x \to 1^{-}$ , denominator $\to 0^{-}$ , numerator $\to 1$ , so $f \to - \infty$ ; as $x \to 1^{+}$ , denominator $\to 0^{+}$ , so $f \to + \infty$ . (By evenness the walls at $- 1$ mirror this.) Horizontal: divide top and bottom by $x^{2}$ : $lim_{x \to \pm \infty} \frac{x ^{2}}{x ^{2} - 1} = lim \frac{1}{1 - 1/ x ^{2}} = 1, so y = 1.$ Why this step? Asymptotes are the skeleton the curve drapes over; the sided limits tell us which way the curve escapes each wall.

Step 3 — First derivative (quotient rule). $f^{'} (x) = \frac{2 x ( x ^{2} - 1 ) - x ^{2} ( 2 x )}{( x ^{2} - 1 ) ^{2}} = \frac{- 2 x}{( x ^{2} - 1 ) ^{2}} .$ Denominator is a square, always $> 0$ . So $f^{'}$ has the sign of $- 2 x$ : positive for $x < 0$ , negative for $x > 0$ → local max at $(0, 0)$ . Why this step? Sign of $f^{'}$ gives climb/fall; only critical point in the domain is $x = 0$ .

Step 4 — Value regions. For $∣ x ∣ < 1$ : $x^{2} - 1 < 0$ so $f < 0$ (curve sits below the $x$ -axis, peaking at the origin). For $∣ x ∣ > 1$ : $f > 1$ (above the ceiling). Second derivative $f^{''} (x) = \frac{6 x ^{2} + 2}{( x ^{2} - 1 ) ^{3}}$ is $> 0$ when $∣ x ∣ > 1$ (concave up) and $< 0$ when $∣ x ∣ < 1$ (concave down); no inflection since the sign only flips across the forbidden $x = \pm 1$ . Why this step? Locates the curve relative to its asymptotes in each region.

Figure below — what to observe. The two plum dashed verticals at $x = \pm 1$ are the walls; the teal dashed horizontal at $y = 1$ is the ceiling. In the middle strip the orange curve makes a low hump peaking at the ink dot $(0, 0)$ and dives to $- \infty$ as it nears each wall. Outside the walls the two orange branches sit above $y = 1$ and sweep down toward it far out. The whole picture is mirror-symmetric across the $y$ -axis — that is evenness.

Figure 2 — $f (x) = x^{2} / (x^{2} - 1)$ : walls at $x = \pm 1$ , ceiling $y = 1$ , even hump peaking at the origin.

Verify: $f (0) = 0$ ; $f^{'} (0) = \frac{- 2 \cdot 0}{( 0 - 1 ) ^{2}} = 0$ ; $f^{''} (0) = \frac{6 \cdot 0 + 2}{( 0 - 1 ) ^{3}} = \frac{2}{- 1} = - 2 < 0$ (concave down ⇒ max); $lim_{x \to \infty} f = 1$ ; $lim_{x \to 1^{+}} f = + \infty$ . (Machine-checked below.)

Example 4 — Rational with an oblique (slant) asymptote · cell E

Step 1 — Long division to reveal the line. $\frac{x ^{2} + 1}{x} = x + \frac{1}{x} .$ As $x \to \pm \infty$ , the term $\frac{1}{x} \to 0$ , so the curve hugs the line $y = x$ . That line is the oblique asymptote. Why this step? Polynomial division separates the "line part" from a remainder that vanishes at infinity — exactly the parent's oblique-asymptote recipe. Cross-check with the formula, being careful to state which direction of infinity each limit uses: $m = lim_{x \to + \infty} \frac{f ( x )}{x} = lim_{x \to + \infty} \frac{x ^{2} + 1}{x ^{2}} = 1, c = lim_{x \to + \infty} (f (x) - x) = lim_{x \to + \infty} \frac{1}{x} = 0.$ The same two limits taken as $x \to - \infty$ give the identical $m = 1, c = 0$ , so the single line $y = x$ is the asymptote on both ends.

Step 2 — Vertical asymptote & domain. Domain $R ∖ {0}$ . As $x \to 0^{+}$ , $f = x + \frac{1}{x} \to + \infty$ ; as $x \to 0^{-}$ , $f \to - \infty$ . So $x = 0$ is a two-signed vertical wall. Why this step? Denominator zero with non-zero numerator ⇒ genuine blow-up.

Step 3 — Symmetry & first derivative (full monotonicity). $f (- x) = - x - \frac{1}{x} = - f (x)$ → odd. $f^{'} (x) = 1 - \frac{1}{x ^{2}} = \frac{x ^{2} - 1}{x ^{2}} .$ The denominator $x^{2} > 0$ always, so the sign of $f^{'}$ is the sign of $x^{2} - 1$ . Walk all four intervals split by the critical points $x = \pm 1$ and the wall $x = 0$ :

$(- \infty, - 1)$ : $x^{2} > 1$ ⇒ $f^{'} > 0$ → increasing.
$(- 1, 0)$ : $x^{2} < 1$ ⇒ $f^{'} < 0$ → decreasing.
$(0, 1)$ : $x^{2} < 1$ ⇒ $f^{'} < 0$ → decreasing.
$(1, \infty)$ : $x^{2} > 1$ ⇒ $f^{'} > 0$ → increasing.

So on the left branch the curve rises to $x = - 1$ then falls into the wall, giving a local max at $(- 1, - 2)$ ; on the right branch it falls from the wall to $x = 1$ then rises, giving a local min at $(1, 2)$ . Why this step? The interval-by-interval sign of $f^{'}$ is the full monotonicity picture, and the $+ \to -$ / $- \to +$ changes pin the extrema.

Step 4 — Second derivative (concavity, all cases). $f^{''} (x) = \frac{2}{x ^{3}} .$ The sign of $f^{''}$ is just the sign of $x^{3}$ :

$x > 0$ : $f^{''} > 0$ → concave up (the right branch cups upward, sitting above the slant line $y = x$ ).
$x < 0$ : $f^{''} < 0$ → concave down (the left branch caps downward, sitting below $y = x$ ).

There is no inflection point: $f^{''}$ only changes sign across the forbidden $x = 0$ , not at any point of the domain. Concavity confirms the extrema too — $f^{''} (1) = 2 > 0$ (min ✓), $f^{''} (- 1) = - 2 < 0$ (max ✓). Why this step? The Second derivative test rounds out the bending and rules out a hidden inflection.

Figure below — what to observe. The teal dashed line $y = x$ runs corner to corner; the two burnt-orange branches lie one on each side of the plum vertical wall $x = 0$ . The right branch dips to the plum minimum $(1, 2)$ just above the slant line (concave up), the left branch rises to the teal maximum $(- 1, - 2)$ just below it (concave down). Each branch flattens toward $y = x$ far from the origin — that is the curve "hugging" its slant asymptote.

Figure 3 — $f (x) = (x^{2} + 1) / x$ : slant asymptote $y = x$ , wall $x = 0$ , min $(1, 2)$ , max $(- 1, - 2)$ .

Verify: slant slope $m = lim_{x \to \infty} \frac{x ^{2} + 1}{x ^{2}} = 1$ , intercept $c = lim_{x \to \infty} \frac{1}{x} = 0$ ; $f (1) = 1 + 1 = 2$ , $f (- 1) = - 1 - 1 = - 2$ ; $f^{''} (1) = 2 > 0$ (min), $f^{''} (- 1) = - 2 < 0$ (max); $f^{'}$ changes $+ \to -$ at $x = - 1$ and $- \to +$ at $x = 1$ . (Machine-checked below.)

Example 5 — A curve that crosses its horizontal asymptote · cell G

Step 1 — Horizontal asymptote. $lim_{x \to \pm \infty} \frac{x}{x ^{2} + 1} = lim \frac{1/ x}{1 + 1/ x ^{2}} = 0 \Rightarrow y = 0.$ Why this step? Degree of top $<$ degree of bottom ⇒ the curve flattens to the $x$ -axis at infinity.

Step 2 — Crossing point. Set $f (x) = 0$ : $\frac{x}{x ^{2} + 1} = 0 \Rightarrow x = 0$ . The curve passes through the origin — which lies on $y = 0$ . So it crosses its own asymptote at $x = 0$ . Why this step? The asymptote is an end-behaviour statement, not a barrier; solving $f = 0$ locates any crossing.

Step 3 — Turning points (bounded bump). $f^{'} (x) = \frac{( x ^{2} + 1 ) - x ( 2 x )}{( x ^{2} + 1 ) ^{2}} = \frac{1 - x ^{2}}{( x ^{2} + 1 ) ^{2}} = 0 \Rightarrow x = \pm 1.$ $f (1) = \frac{1}{2}$ (max), $f (- 1) = - \frac{1}{2}$ (min). The curve rises from the origin to a gentle peak at $(1, \frac{1}{2})$ , then falls back toward $y = 0$ . Why this step? Confirms the shape: two humps straddling a crossing at the origin.

Figure below — what to observe. The teal dashed line $y = 0$ is the horizontal asymptote, and the burnt-orange curve crosses right through it at the ink square on the origin — an asymptote is not a fence. The curve rises to the plum maximum $(1, \frac{1}{2})$ , then eases back down toward the axis; by oddness it mirrors to the plum minimum $(- 1, - \frac{1}{2})$ on the left. Far out in either direction it settles onto $y = 0$ from above (right) and below (left).

Figure 4 — $f (x) = x / (x^{2} + 1)$ : crosses its horizontal asymptote $y = 0$ at the origin, max $(1, \frac{1}{2})$ , min $(- 1, - \frac{1}{2})$ .

Verify: $lim_{x \to \infty} f = 0$ ; $f (0) = 0$ (on the asymptote); $f (1) = \frac{1}{1 + 1} = \frac{1}{2}$ , $f (- 1) = - \frac{1}{2}$ . (Machine-checked below.)

Example 6 — Degenerate input: removable hole vs blow-up · cell H

Step 1 — Test $g$ for a removable hole. Plugging $x = 1$ gives $\frac{0}{0}$ — an indeterminate form, not automatically $\infty$ . Factor: $g (x) = \frac{( x - 1 ) ( x + 1 )}{x - 1} = x + 1 (x \neq = 1) .$ The $(x - 1)$ cancels, so $g$ is just the line $y = x + 1$ with a hole punched out at $x = 1$ (the value would have been $2$ ). Why this step? Limits distinguish a $0/0$ that simplifies (removable) from a $\frac{nonzero}{0}$ (true pole).

Step 2 — Test $h$ for a pole (both one-sided limits). At $x = 1$ , $h = \frac{1}{0}$ with non-zero numerator, so it cannot cancel. Take each side separately:

As $x \to 1^{+}$ , the denominator $x - 1 \to 0^{+}$ (small positive), so $h = \frac{1}{0 ^{+}} \to + \infty$ .
As $x \to 1^{-}$ , the denominator $x - 1 \to 0^{-}$ (small negative), so $h = \frac{1}{0 ^{-}} \to - \infty$ .

Because the two one-sided limits run off to opposite infinities, $x = 1$ is a genuine two-sided vertical asymptote — the curve shoots up on the right of the wall and down on the left. Why this step? Non-zero over zero can never be removed; the sign of the tiny denominator on each side fixes which infinity the branch chases.

Verify: $lim_{x \to 1} g (x) = 1 + 1 = 2$ (finite ⇒ removable hole; $g$ undefined at $1$ but the limit exists). For $h$ : $lim_{x \to 1^{+}} h = + \infty$ , $lim_{x \to 1^{-}} h = - \infty$ (two-signed pole). (Machine-checked below.)

Example 7 — Word problem: the shape of a cost-per-unit curve · cell I

Step 1 — Simplify (spot the slant asymptote in disguise). $A (x) = x + \frac{400}{x}, x > 0.$ For large $x$ , $A \approx x$ (rising line); for small $x$ , the $\frac{400}{x}$ term dominates and $A \to + \infty$ . A minimum must sit between. Why this step? This is Example 4's structure applied to a real curve; the slant $y = x$ tells the large-scale trend. This is exactly the kind of "find the lowest point" question studied in Optimization.

Step 2 — Minimise with the first derivative. $A^{'} (x) = 1 - \frac{400}{x ^{2}} = 0 \Rightarrow x^{2} = 400 \Rightarrow x = 20 (take + root since x > 0) .$ Check it is a minimum: $A^{''} (x) = \frac{800}{x ^{3}} > 0$ for $x > 0$ → concave up → minimum. Why this step? Optimization: the lowest point of a smooth positive curve is where slope $= 0$ and $A^{''} > 0$ .

Step 3 — Minimum value. $A (20) = 20 + \frac{400}{20} = 20 + 20 = 40 dollars .$ Why this step? Answers the physical question: making $20$ (hundred) items gives the cheapest average cost, $40 each.

Verify: critical $x = 20$ (positive root of $x^{2} = 400$ ); $A (20) = 20 + 20 = 40$ dollars/item; $A^{''} (20) = \frac{800}{2 0 ^{3}} = \frac{800}{8000} = \frac{1}{10} > 0$ (concave up ⇒ minimum). Units: dollars per item, consistent throughout. (Machine-checked below.)

Example 8 — Exam twist: behaviour depends on a parameter · cell J

Step 1 — Differentiate and locate critical points. $f^{'} (x) = 3 x^{2} + a .$ Solve $3 x^{2} + a = 0 \Rightarrow x^{2} = - \frac{a}{3}$ . Why this step? Turning points exist only where $f^{'} = 0$ has real solutions. Whether $x^{2} = - \frac{a}{3}$ has real roots depends entirely on the sign of the right-hand side.

Step 2 — Case split on the sign of $a$ (all three regimes).

$a < 0$ : then $- \frac{a}{3} > 0$ , so $x = \pm - a /3$ are two distinct real critical points — a local max and a local min: an S-curve with a genuine wiggle. Example $a = - 3$ recovers $x^{3} - 3 x$ from Example 1, with roots at $x = \pm 1$ . Two turning points.
$a = 0$ : then $f^{'} = 3 x^{2} \geq 0$ , zero only at $x = 0$ and with no sign change — this is the flat (stationary) inflection $x^{3}$ of Example 2. Zero turning points.
$a > 0$ : then $- \frac{a}{3} < 0$ , so $x^{2} = - \frac{a}{3}$ has no real solution; in fact $f^{'} = 3 x^{2} + a \geq a > 0$ everywhere, so the curve is strictly increasing with no flat spot at all. Zero turning points.

Why this step? The single quantity $- \frac{a}{3}$ (positive, zero, or negative) is the hinge; the three bullets exhaust every possible value of the parameter.

Step 3 — Concluding summary.

Sign of $a$	Real roots of $f^{'} = 0$	Turning points	Shape
$a < 0$	two ( $x = \pm - a /3$ )	2 (max & min)	wiggling S
$a = 0$	one ( $x = 0$ ), no sign change	0	flat inflection cubic
$a > 0$	none	0	strictly increasing

So the cubic $x^{3} + a x$ has a genuine "hill and valley" only when $a < 0$ ; at $a = 0$ and for all $a > 0$ it climbs monotonically. The transition value is $a = 0$ . Why this step? Gives the reader the one-line rule to quote in an exam: "two turning points iff $a < 0$ ."

Verify: $a = - 3$ gives critical points $x = \pm 1$ (two real); $a = 0$ gives only $x = 0$ with $f^{'} = 3 x^{2} \geq 0$ (no sign change, zero turning points); $a = 3$ gives $f^{'} = 3 x^{2} + 3 \geq 3 > 0$ everywhere (no real critical points). (Machine-checked below.)

Recall Self-test across the matrix

Match each curve to its defining feature (answers below). $x^{3} - 3 x$ ::: two turning points, odd (cell A/F) $x^{3}$ ::: flat inflection, no extremum (cell B) $\frac{x ^{2}}{x ^{2} - 1}$ ::: horizontal + twin vertical asymptotes, even (C/D/F) $\frac{x ^{2} + 1}{x}$ ::: oblique asymptote $y = x$ (cell E) $\frac{x}{x ^{2} + 1}$ ::: crosses its horizontal asymptote at the origin (cell G) $\frac{x ^{2} - 1}{x - 1}$ ::: removable hole at $x = 1$ , not a pole (cell H)

Connections

Curve sketching — systematic approach — the parent checklist these examples exercise.
First derivative and monotonicity — turning points and climb/fall in Ex 1, 2, 4, 5, 8.
Second derivative test — confirming max vs min in Ex 1, 3, 4, 7.
Rational functions and asymptotes — horizontal, vertical and oblique walls in Ex 3, 4, 5.
Limits and continuity — hole vs blow-up in Ex 6.
Optimization — the average-cost minimum in Ex 7.
Mean Value Theorem — the theoretical reason the sign of $f^{'}$ dictates the trend.