MATH 140 Calculus 1 • McGill University, Montreal

Revision sheet: curve sketching (MATH 140)

Q: What are the steps to sketch a curve in calculus?

Find the domain first, then the intercepts and any symmetry. Compute the asymptotes with one-sided limits and with the limits at both ends. Factor the first derivative and study its sign for increase, decrease and extrema, then factor the second derivative for concavity and inflection points. Put everything in one sign table and draw the graph from the table only.

Q: Can a graph cross its horizontal asymptote?

Yes. A horizontal asymptote only describes what the function does as x goes to plus or minus infinity. At finite values of x the graph can cross the line once, several times or infinitely often, as sin x over x does. What is true is that far enough out, the graph stays as close to the line as you like.

Q: Is there an inflection point at a vertical asymptote?

No. The concavity can change across a vertical asymptote, as it does for one over x at zero, but an inflection point is a point of the graph, and the function has no value there. Check that the point is in the domain before calling a change of concavity an inflection point.

Q: How do I tell a hole from a vertical asymptote in a rational function?

Factor the numerator and the denominator and cancel the common factors. If the factor that vanishes cancels, the limit at that point is finite and the graph has a hole, drawn as an open circle. If a zero of the reduced denominator remains while the numerator is not zero there, the limit is infinite and the line is a vertical asymptote.

Q: What is the difference between a cusp and a vertical tangent?

Both happen at a point of the domain where the derivative does not exist because it becomes infinite. If the derivative tends to plus infinity on one side and minus infinity on the other, the graph turns back on itself in a sharp point: a cusp, and an extremum. If it tends to the same infinity on both sides, the graph passes through with a vertical tangent and keeps going.

This sheet is not a summary of section 4.5 of Stewart: you already have the course notes and the checklist. It answers one question only, what makes students lose marks when they sketch a graph in MATH 140 at McGill University, and which precise gesture avoids each loss.

Almost every lost mark on this chapter is lost at a point that is not in the domain, or at a point where $f'$ does not exist. The computations are those of the previous chapters; what is new is putting them together without contradiction, by hand, exactly as on the final.

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The thread of the chapter

A sketch is a SYNTHESIS: every feature comes from a computed sign or limit, gathered in ONE sign table built on the DOMAIN, and the domain decides what each sign change means. No extremum, no inflection point and no asymptote is declared before checking whether the point is in the domain.

This chapter is part of MATH 140, Calculus 1 (McGill)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Curve sketching
Domain of a function
Vertical asymptote
Horizontal asymptote
Slant asymptote
Removable discontinuity
Sign table
Critical number
Local extrema
Concavity
Inflection point
Cusp and vertical tangent

The essentials

The domain first: what an excluded point can and cannot be

• The domain is computed BEFORE any derivative: every excluded point cuts every row of the sign table, and is marked there by a double bar.
• At a zero $a$ of the denominator of a rational function, cancel the common factors first. A factor that cancels gives a HOLE, drawn as an open circle; a zero of the reduced denominator gives a vertical asymptote.
• At an excluded point there is no extremum and no inflection point, even when $f'$ or $f''$ changes sign across it: there is no point $(a, f(a))$ to name.
• At a point of the domain where $f'$ does not exist, there IS a critical number: a corner, a cusp or a vertical tangent.
• A vertical asymptote needs an infinite limit of $f$ itself. Each side is computed separately, and its sign decides where the branch goes.

Two zeros of the denominator of

\frac{x^2 - x - 2}{x^2 - 4}

: at

x = 2

the factor cancels and leaves a hole at

\left(2, \frac{3}{4}\right)

, at

x = -2

it does not, and the branches follow an asymptote.

Write the domain on the first line of the copy and put a double bar in the table under each excluded point: most of the marks lost on this chapter are lost at those points.

The checklist, and what each line puts in the table

• Domain; intercepts; symmetry ( $f(-x) = f(x)$ even, $f(-x) = -f(x)$ odd, a period for a trigonometric function).
• Asymptotes: one-sided limits at each excluded point, limits at $-\infty$ AND $+\infty$ separately, slant asymptote by long division when the degree of the numerator exceeds that of the denominator by one.
• $f'$ factored and its sign on each interval of the domain: increase, decrease, extrema where $f'$ changes sign at a point of the domain.
• $f''$ factored and its sign: concavity, inflection points where the concavity changes at a point of the domain.
• Then ONE table with the rows $f'$ , $f''$ and $f$ , and a sketch drawn from the table only.

The table is the proof and the sketch is its picture. A feature of the drawing that is not in the table, or a line of the table that the drawing ignores, costs marks either way.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

What the sign table says at a point a

Read a line as: what happens at $x = a$ , under the condition of the middle column, gives the conclusion of the last one. The red lines are the features that a careless sketch shows and the true graph does not have.

At $x = a$	Condition	Conclusion
$f'$ changes sign	$a$ in the domain	local extremum
Example: $x^{2/3}(x - 5)$ : $f'$ goes from $-$ to $+$ at $2$ , local min $f(2) = -3\sqrt[3]{4}$ .
$f'(a) = 0$ , same sign both sides	$a$ in the domain	no extremum not an extremum
Example: $\frac{x^3}{x^2 - 3}$ has $f'(0) = 0$ and $f' < 0$ on both sides of $0$ . Same form, other result: The same function has $f'(3) = 0$ too, and there $f'$ changes sign: local min $f(3) = \frac{9}{2}$ . What to do: Read the sign of $f'$ on BOTH sides before writing max or min.
$f''$ changes sign	$a$ in the domain	inflection point
Example: $\frac{x}{x^2 - 1}$ : $f''$ changes sign at $0$ , inflection point $(0, 0)$ .
$f''$ changes sign	$a$ not in the domain	no point no inflection point
Example: $\frac{x}{x^2 - 1}$ : $f''$ changes sign at $x = 1$ , a vertical asymptote. Same form, other result: At $x = 0$ the same function does have an inflection point, because $0$ is in the domain. What to do: Check the domain: the concavity changes across an asymptote, but there is no point $(a, f(a))$ to mark.
$f'(a)$ undefined	$a$ in the domain	critical number
Example: $x^{2/3}(x - 5)$ at $0$ : $f' \to +\infty$ on the left, $-\infty$ on the right; a cusp and a local max $f(0) = 0$ .
denominator $= 0$	the factor cancels	hole
Example: $\frac{x^2 - x - 2}{x^2 - 4}$ at $x = 2$ : a hole at $\left(2, \frac{3}{4}\right)$ , no asymptote.

The middle column is the one students skip. Each red line is a feature of a careless sketch that the true graph does not have.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. An interval of increase or decrease written across an asymptote

1 mark on the table, and every comparison drawn from it

What not to write

“ $f(x) = \frac{x}{x^2 - 1}$ has $f'(x) = -\frac{x^2 + 1}{(x^2 - 1)^2} < 0$ , so $f$ is decreasing on its domain and $f(-2) > f(2)$ .”

What to write

“ $f$ is decreasing on each of $(-\infty, -1)$ , $(-1, 1)$ and $(1, \infty)$ .” Indeed $f(-2) = -\frac{2}{3} < f(2) = \frac{2}{3}$ .

Why: Increase and decrease are properties of an INTERVAL. The asymptotes cut the domain into three intervals, and the curve restarts from $\pm\infty$ on each of them.

2. An inflection point placed on a vertical asymptote

1 to 2 marks, and points drawn on the asymptotes

What not to write

“ $f''(x) = \frac{2x(x^2 + 3)}{(x^2 - 1)^3}$ changes sign at $-1$ , $0$ and $1$ : three inflection points.”

What to write

“ $f''$ changes sign at $-1$ , $0$ and $1$ , but $\pm 1$ are not in the domain: the only inflection point is $(0, 0)$ .”

The concavity of

\frac{x}{x^2 - 1}

changes at

-1

0

and

1

, but only the origin is a point of the graph: there is nothing to mark on the dashed lines.

Why: An inflection point is a POINT $(a, f(a))$ of the graph where the concavity changes. Where $f$ is undefined there is nothing to name, whatever $f''$ does across it.

3. A vertical asymptote at every zero of the denominator

1 mark, and a branch drawn where none exists

What not to write

“ $x^2 - 4 = 0$ at $x = \pm 2$ , so $\frac{x^2 - x - 2}{x^2 - 4}$ has two vertical asymptotes.”

What to write

“ $\frac{x^2 - x - 2}{x^2 - 4} = \frac{(x - 2)(x + 1)}{(x - 2)(x + 2)} = \frac{x + 1}{x + 2}$ for $x \ne 2$ : a hole at $\left(2, \frac{3}{4}\right)$ and one asymptote, $x = -2$ .”

Why: A vertical asymptote needs an INFINITE limit. At $x = 2$ the limit is $\frac{3}{4}$ , finite, so the graph simply has a missing point there. The figure of the first block shows both cases on one graph.

4. Writing the square root of x squared as x at minus infinity

2 marks, and the whole left half of the sketch

What not to write

“ $\frac{x + 2}{\sqrt{x^2 + 1}} = \frac{1 + 2/x}{\sqrt{1 + 1/x^2}} \to 1$ at both ends: one horizontal asymptote, $y = 1$ .”

What to write

“For $x < 0$ , $\sqrt{x^2 + 1} = -x\sqrt{1 + 1/x^2}$ , so the limit at $-\infty$ is $-1$ : two horizontal asymptotes, $y = 1$ on the right and $y = -1$ on the left.”

\frac{x + 2}{\sqrt{x^2 + 1}}

settles on

y = -1

on the left and on

y = 1

on the right, and crosses

y = 1

x = -\frac{3}{4}

before its maximum

\sqrt 5

Why: $\sqrt{x^2} = |x|$ , which is $-x$ for $x < 0$ . The two ends of a graph are two different limits, and must be written as two lines.

5. Refusing to let the curve cross its asymptote

1 mark, and a sketch that contradicts the maximum

What not to write

“A curve never crosses its asymptote, so left of its maximum the graph of $\frac{x + 2}{\sqrt{x^2 + 1}}$ stays below $y = 1$ .”

What to write

“ $f(x) = 1 \iff x + 2 = \sqrt{x^2 + 1} \iff x = -\frac{3}{4}$ (and $-\frac{3}{4} + 2 > 0$ ): the curve crosses $y = 1$ once, then approaches it from above.”

Why: An asymptote describes the ENDS of the graph only. Nothing prevents a crossing at finite $x$ , and here the maximum $\sqrt 5 > 1$ forces one.

6. Forgetting the critical number where the derivative does not exist

1 to 2 marks, and a smooth curve where the graph turns vertical

What not to write

“ $f(x) = x^{1/3}(x - 4)$ has $f'(x) = \frac{4(x - 1)}{3x^{2/3}}$ , which vanishes only at $x = 1$ : one critical number.”

What to write

“ $f'(1) = 0$ , and $f'(0)$ does not exist while $f(0) = 0$ : two critical numbers. At $0$ , $f' \to -\infty$ on both sides: a vertical tangent, no extremum. At $1$ , a local min $f(1) = -3$ .”

Why: A critical number is a point of the DOMAIN where $f' = 0$ OR $f'$ does not exist. Forgetting the second kind loses every cusp and every vertical tangent of the chapter.

7. L'Hospital's rule on a form that is not indeterminate

the limit, and the whole left side of the sketch

What not to write

“ $\lim_{x\to 0^+} \frac{\ln x}{x} = \lim_{x\to 0^+} \frac{1/x}{1} = +\infty$ .”

What to write

“As $x \to 0^+$ the form is $\frac{-\infty}{0^+}$ , which is not indeterminate: $\frac{\ln x}{x} \to -\infty$ .”

Why: The rule needs $\frac{0}{0}$ or $\frac{\infty}{\infty}$ ; outside those forms its conclusion is simply false. Name the form before every use, as the rule itself requires.

8. Taking a local maximum below a local minimum for an error

time, and often a correct answer crossed out

What not to write

“For $\frac{x^3}{x^2 - 3}$ I find a local max $-\frac{9}{2}$ and a local min $\frac{9}{2}$ : a maximum below a minimum is impossible, so I must have a sign error.”

What to write

“The local max belongs to the branch left of $x = -\sqrt 3$ and the local min to the branch right of $x = \sqrt 3$ : a local extremum is compared with its neighbours only.”

Why: LOCAL means on a small interval around the point. Asymptotes separate the branches, and nothing relates the height of one branch to the height of another.

Which method to choose

What to look for, by the FORM of f

Look at the formula before computing: its form tells you where the graph will do something

If a rational function → factor and cancel; zeros of the reduced denominator give vertical asymptotes, cancelled zeros give holes; compare the degrees for a horizontal or slant asymptote

Example: $\frac{x^3}{x^2 - 3}$ : asymptotes $x = \pm\sqrt 3$ and $y = x$ , by $x^3 = x(x^2 - 3) + 3x$
If the root of a quadratic, $\sqrt{x^2 + c}$ , next to a degree one term → compute the two ends separately with $\sqrt{x^2} = |x|$ ; expect two different horizontal asymptotes

Example: $\frac{x + 2}{\sqrt{x^2 + 1}}$ : $y = 1$ on the right, $y = -1$ on the left
If a fractional power $x^{p/q}$ with $p < q$ → $f'$ is undefined at $0$ : a critical number; compare the signs of $f'$ on both sides for a cusp or a vertical tangent

Example: $x^{2/3}(x - 5)$ : cusp at $0$ ; $x^{1/3}(x - 4)$ : vertical tangent at $0$
If $\ln x$ in the formula → one-sided domain; limit at $0^+$ often not indeterminate; at $\infty$ , the logarithm loses against every power

Example: $\frac{\ln x}{x}$ : $-\infty$ at $0^+$ , $0$ at $\infty$ , max $\frac{1}{e}$ at $x = e$
If $e^{x}$ or $e^{-x}$ times a power of $x$ → an asymptote $y = 0$ on ONE side only; L'Hospital's rule on the other side

Example: $xe^{-x}$ : $0$ at $+\infty$ , $-\infty$ at $-\infty$ , max $\frac{1}{e}$ at $x = 1$
If $\sin$ and $\cos$ only → one period, symmetry, exact angles from double angle identities; no asymptote

Example: $2\cos x + \sin 2x$ on $[0, 2\pi]$ : max $\frac{3\sqrt 3}{2}$ at $\frac{\pi}{6}$

A formula often belongs to two branches: $\frac{e^x}{x}$ is both a quotient with a zero denominator and an exponential. Apply both, and write in the copy which feature comes from which branch.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Writing up a complete curve sketch

When to use it: “Sketch the graph of $f$ , showing all asymptotes, local extrema and inflection points”, the long question of most finals

1 Domain on the first line, then intercepts and symmetry, each with its one line of computation.
2 Asymptotes: each one-sided limit written with its form and its sign, each end $\pm\infty$ written separately.
3 $f'$ in FACTORED form, its critical numbers (both kinds), and the sign of each factor on each interval of the domain.
4 $f''$ in factored form and its sign, again only on the domain.
5 One table with the rows $f'$ , $f''$ and $f$ , the values at every critical number and inflection point, and double bars at the excluded points.
6 The sketch, drawn from the table: asymptotes dashed, extrema and inflection points marked with their exact coordinates.

Concluding sentence

“ $f$ is decreasing on $(-\sqrt 3, \sqrt 3)$ with $f'(0) = 0$ ; since $f'$ does not change sign at $0$ , $f$ has no extremum there, and since $f''$ changes sign at $0$ , which is in the domain, $(0, 0)$ is an inflection point with a horizontal tangent.”

The trap: Computing $f'$ and $f''$ in expanded form: the sign table then needs a second, harder computation that most students skip.

Marking: Typically 2 marks for the domain, intercepts and asymptotes with their one-sided limits, 3 for f prime and its table, 3 for f double prime and its table, 2 for a sketch consistent with both.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The sketch passes through its intercepts

Put the intercepts on the axes before drawing anything. A branch that crosses the x-axis where the table found no zero means a sign is wrong somewhere.

$\frac{x^2 + 4}{x^2 - 4}$ has no $x$ -intercept: a middle branch drawn above the axis contradicts $f(0) = -1$ and the maximum $-1$ .
One monotone piece per interval of the table

Between two consecutive critical numbers or excluded points the curve must be monotone. Count the pieces of the drawing and compare with the columns of the table.

$\frac{x^3}{x^2 - 3}$ : the points $-3$ , $-\sqrt 3$ , $\sqrt 3$ , $3$ make five intervals, but $0$ adds no piece since $f'$ keeps its sign there.
The last concavity agrees with the asymptote

A curve that decreases toward a horizontal asymptote from above must end concave up; from below, concave down. If the table says otherwise, recheck the sign of f double prime.

$\frac{\ln x}{x}$ decreases to $0$ from above, and indeed $f'' > 0$ for $x > e^{3/2}$ .
One value plugged in

Compute f at one convenient point of each branch and check that the drawing is on the right side of the asymptote there.

$\frac{x + 2}{\sqrt{x^2 + 1}}$ at $x = 0$ gives $2 > 1$ : the curve is above $y = 1$ there, as the crossing at $-\frac{3}{4}$ predicts.

The typical problem, taken apart

A closed domain, vertical tangents at the ends, and zeros of f double prime to reject

Sketch the graph of $f(x) = x\sqrt{4 - x^2}$ , showing its extrema, its inflection points and its behaviour at the ends of the domain.

No calculator. Every step must be justified as on a MATH 140 final.

Step 1

Domain: $4 - x^2 \ge 0$ , so $[-2, 2]$ . $f(-x) = -f(x)$ : $f$ is odd. Zeros: $x = 0$ and $x = \pm 2$ . $f$ is continuous on a closed bounded interval: no asymptote at all.

Why

A closed domain rules out every asymptote at once, and makes the two endpoints lines of the table. Oddness will serve as a free check at the end.

Step 2

Product rule, and chain rule on the root (inner function $4 - x^2$ ): $f'(x) = \sqrt{4 - x^2} + x \cdot \frac{-x}{\sqrt{4 - x^2}} = \frac{4 - 2x^2}{\sqrt{4 - x^2}}$ on $(-2, 2)$ . $f' = 0$ at $x = \pm\sqrt 2$ , and $f' \to -\infty$ as $x \to 2^-$ and as $x \to -2^+$ .

Why

Putting everything over one denominator is what makes the sign readable. The infinite limits at the ends mean VERTICAL tangents at $(\pm 2, 0)$ , which the sketch must show.

Step 3

Sign of $4 - 2x^2$ : negative on $(-2, -\sqrt 2)$ , positive on $(-\sqrt 2, \sqrt 2)$ , negative on $(\sqrt 2, 2)$ . Local min $f(-\sqrt 2) = -\sqrt 2 \cdot \sqrt 2 = -2$ , local max $f(\sqrt 2) = 2$ ; they are absolute, the endpoints giving $0$ .

Why

On a closed interval, comparing with the endpoint values turns the local extrema into absolute ones for one extra line.

Step 4

$f''(x) = \frac{2x(x^2 - 6)}{(4 - x^2)^{3/2}}$ . On $(-2, 2)$ , $x^2 - 6 < 0$ , so $f''$ has the sign of $-x$ : concave up on $(-2, 0)$ , concave down on $(0, 2)$ . The zeros $\pm\sqrt 6$ of $x^2 - 6$ are outside the domain.

Why

This is where the domain decides: $\pm\sqrt 6 \approx \pm 2.45$ are not in $[-2, 2]$ , so they are NOT inflection points and must not appear in the table.

Step 5

Inflection point $(0, 0)$ , with slope $f'(0) = 2$ . The curve starts at $(-2, 0)$ with a vertical tangent, falls to $(-\sqrt 2, -2)$ , rises through $(0, 0)$ to $(\sqrt 2, 2)$ and falls to $(2, 0)$ , arriving vertically.

Why

Every element of the drawing is a line of the table, and the half turn about the origin checks the whole picture: the minimum is the image of the maximum.

The conclusion, written out

“On $[-2, 2]$ , $f$ has an absolute minimum $-2$ at $-\sqrt 2$ , an absolute maximum $2$ at $\sqrt 2$ , a single inflection point $(0, 0)$ , and vertical tangents at $(-2, 0)$ and $(2, 0)$ .”

The classic mistake on this problem: Listing $x = \pm\sqrt 6$ as inflection points because $f''$ vanishes there, or drawing horizontal tangents at the endpoints because $f = 0$ there.

Learn by heart

• Domain FIRST: every excluded point cuts every row of the table.
• No extremum and no inflection point at a point outside the domain.
• Cancel common factors: a cancelled factor gives a hole, a remaining zero of the denominator an asymptote.
• Limits at $-\infty$ and $+\infty$ separately: $\sqrt{x^2} = |x|$ .
• A curve may cross its horizontal or slant asymptote, even infinitely often.
• Critical number: $f' = 0$ OR $f'$ undefined, at a point of the domain. Extremum only if $f'$ changes sign.
• $f'$ infinite with opposite signs on the two sides: a cusp; with the same sign: a vertical tangent.

Frequently asked questions

What are the steps to sketch a curve in calculus?

Find the domain first, then the intercepts and any symmetry. Compute the asymptotes with one-sided limits and with the limits at both ends. Factor the first derivative and study its sign for increase, decrease and extrema, then factor the second derivative for concavity and inflection points. Put everything in one sign table and draw the graph from the table only.

Can a graph cross its horizontal asymptote?

Yes. A horizontal asymptote only describes what the function does as x goes to plus or minus infinity. At finite values of x the graph can cross the line once, several times or infinitely often, as sin x over x does. What is true is that far enough out, the graph stays as close to the line as you like.

Is there an inflection point at a vertical asymptote?

No. The concavity can change across a vertical asymptote, as it does for one over x at zero, but an inflection point is a point of the graph, and the function has no value there. Check that the point is in the domain before calling a change of concavity an inflection point.

How do I tell a hole from a vertical asymptote in a rational function?

Factor the numerator and the denominator and cancel the common factors. If the factor that vanishes cancels, the limit at that point is finite and the graph has a hole, drawn as an open circle. If a zero of the reduced denominator remains while the numerator is not zero there, the limit is infinite and the line is a vertical asymptote.

What is the difference between a cusp and a vertical tangent?

Both happen at a point of the domain where the derivative does not exist because it becomes infinite. If the derivative tends to plus infinity on one side and minus infinity on the other, the graph turns back on itself in a sharp point: a cusp, and an extremum. If it tends to the same infinity on both sides, the graph passes through with a vertical tangent and keeps going.

Practise it

Corrected exercises: Curve sketching, MATH 140 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet L'Hospital's rule Next sheet Optimization problems

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math140-curve-sketching. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: curve sketching (MATH 140)

The essentials

The domain first: what an excluded point can and cannot be

The checklist, and what each line puts in the table

The rules in table form

What the sign table says at a point a

The mistakes that cost marks

1. An interval of increase or decrease written across an asymptote

2. An inflection point placed on a vertical asymptote

3. A vertical asymptote at every zero of the denominator

4. Writing the square root of x squared as x at minus infinity

5. Refusing to let the curve cross its asymptote

6. Forgetting the critical number where the derivative does not exist

7. L'Hospital's rule on a form that is not indeterminate

8. Taking a local maximum below a local minimum for an error

Which method to choose

What to look for, by the FORM of f

How the answer is expected to be written

Writing up a complete curve sketch

Check before you hand in

The typical problem, taken apart

A closed domain, vertical tangents at the ends, and zeros of f double prime to reject

Learn by heart

Frequently asked questions

What are the steps to sketch a curve in calculus?

Can a graph cross its horizontal asymptote?

Is there an inflection point at a vertical asymptote?

How do I tell a hole from a vertical asymptote in a rational function?

What is the difference between a cusp and a vertical tangent?

Corrected exercises: Curve sketching, MATH 140 at McGill

See also

Looking for a MATH 140 tutor in Montreal?