MATH 140 Calculus 1 • McGill University, Montreal

Revision sheet: functions, graphs and transformations (MATH 140)

This sheet is not a summary of sections 1.1 to 1.3 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks on functions and their graphs in MATH 140 at McGill University, and which precise gesture avoids each loss.

The chapter looks like high school review, and that is the danger: the same errors come back on every later chapter, a domain forgotten before differentiating, a graph shifted the wrong way in a curve sketch. Every value below is exact and done by hand, as on the exam, and every trap comes with the line that earns the mark.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next and tells you which chapter to tackle next. Create your space, an email is enough.

The thread of the chapter

Read a formula from the INSIDE out: inside the parentheses acts on $x$ and runs backwards, $f(x + 3)$ moves LEFT and $f(2x)$ COMPRESSES; outside acts on $y$ and runs as written, multiplication before addition; and the same reading decides the domain of a composition (inner function first), the parity test ( $-x$ inside, compare outside) and the value of a piecewise function (which piece first).

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

Domain and range
Piecewise defined function
Absolute value function
Even and odd functions
Transformations of graphs
Horizontal and vertical shifts
Stretching and compressing a graph
Reflections of a graph
Composition of functions
Completing the square
Power functions
Linear models

The essentials

Inside acts on x and runs backwards, outside acts on y as written

• $f(x - h)$ : right $h$ ; $f(x + h)$ : LEFT $h$ . $f(x) + k$ : up $k$ . The inside moves the graph against the sign you read.
• $f(bx)$ with $b > 1$ : horizontal COMPRESSION by $b$ ; $af(x)$ with $a > 1$ : vertical STRETCH by $a$ . The same number has opposite effects inside and outside.
• $f(-x)$ : reflection in the $y$ -axis. $-f(x)$ : reflection in the $x$ -axis. $-f(-x)$ : half-turn about the origin.
• Inside with a coefficient, FACTOR first: $f(2x + 4) = f(2(x + 2))$ , compress by $2$ then left $2$ .
• Outside, follow the order of operations of the formula: in $3 - 2f(x)$ , multiply by $-2$ first, then add $3$ .
• The point method never fails: for $(u, w)$ on $f$ , solve inside $= u$ for $x$ , then apply the outside to $w$ .

The

+3

inside moves

\sqrt x

three units LEFT, to start at

(-3, 0)

; the same

+3

outside moves it three units UP, to start at

(0, 3)

A marker checks one point. Write the image of one named point of the graph next to your sketch: it earns the mark even if the drawing is rough.

Domain, pieces and symmetry: three computations, always in the same order

• Domain: denominator $\ne 0$ , even root of a quantity $\ge 0$ (and $> 0$ if the root is in a denominator). Read it on the formula AS GIVEN, before any simplification.
• Composition $f(g(x))$ : $x$ in the domain of $g$ first, then $g(x)$ in the domain of $f$ .
• Piecewise: find the piece of the input first. To solve, solve each piece and REJECT the candidates outside their piece.
• $\sqrt{a^2} = |a|$ , never $a$ ; the sign to study in $|E|$ is the sign of $E$ .
• Even: $f(-x) = f(x)$ for EVERY $x$ ; odd: $f(-x) = -f(x)$ for every $x$ ; the domain must be symmetric. One point can disprove a symmetry, never prove it.

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 each formula does to the graph of f

Read a line as: the formula of the first column changes the variable of the second, with the effect of the third, shown on the point $(2, 5)$ of the graph of $f$ . The red lines are readings that students write and that are false.

Formula	Acts on	Effect
$f(x) + 3$	$y$	up $3$
Example: $(2, 5) \mapsto (2, 8)$ .
$f(x + 3)$	$x$	left $3$
Example: $(2, 5) \mapsto (-1, 5)$ : solve $x + 3 = 2$ .
$3f(x)$	$y$	stretch $\times 3$
Example: $(2, 5) \mapsto (2, 15)$ ; points on the $x$ -axis do not move.
$f(3x)$	$x$	compress $\div 3$
Example: $(2, 5) \mapsto \left(\frac{2}{3}, 5\right)$ : solve $3x = 2$ .
$-f(x)$	$y$	flip in $x$ -axis
Example: $(2, 5) \mapsto (2, -5)$ .
$f(-x)$	$x$	flip in $y$ -axis
Example: $(2, 5) \mapsto (-2, 5)$ .
$3 - 2f(x)$	$y$	$w \mapsto 3 - 2w$
Example: $(2, 5) \mapsto (2, -7)$ : times $-2$ first, then plus $3$ .
$f(2x + 4)$	$x$	compress, left $4$ rule that does not exist
Example: $(2, 5) \mapsto (-1, 5)$ , since $2x + 4 = 2$ gives $x = -1$ ; the false reading gives $(1 - 4, 5) = (-3, 5)$ . What to do: Factor the inside, $f(2(x + 2))$ : compress by $2$ , THEN left $2$ . Or solve inside $= u$ .
$f(x + 3)$	$x$	$f(x) + f(3)$ rule that does not exist
Example: $f(x) = x^2$ at $x = 1$ : $f(4) = 16$ , while $f(1) + f(3) = 10$ . What to do: $f$ is not a number: replace EVERY $x$ in the formula of $f$ by the whole block $x + 3$ .

Every line of this table is the same statement: the inside is solved for $x$ , the outside is computed on $y$ .

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. Shifting the wrong way

1 mark, and the whole sketch that follows

What not to write

“ $y = (x + 3)^2$ is the parabola $y = x^2$ moved $3$ units to the right.”

What to write

“The inside $x + 3$ vanishes at $x = -3$ , so the vertex is $(-3, 0)$ : the parabola is moved $3$ units to the LEFT.”

Why: The new graph must take at $x$ the value that $f$ takes at $x + 3$ , further right, so it lags behind. Asking where the inside vanishes settles the direction in two seconds.

2. Reading f(2x + 4) as a shift of 4

2 marks: every point of the graph is in the wrong place

What not to write

“ $y = f(2x + 4)$ : compress by $2$ , then move $4$ units left.”

What to write

“ $f(2x + 4) = f(2(x + 2))$ : compress by $2$ , then move $2$ units left. Check: $(u, w) \mapsto \left(\frac{u - 4}{2}, w\right)$ .”

Why: After compressing, the shift acts on the compressed graph, so it must be read on $x + 2$ , not on $2x + 4$ . Solving $2x + 4 = u$ avoids the question of order entirely.

3. Adding before multiplying outside

1 to 2 marks: the vertex and the range are both wrong

What not to write

“ $y = -2(x + 2)^2 + 3$ : move up $3$ , then stretch by $2$ and reflect, so the vertex is at $(-2, -6)$ .”

What to write

“Outside, the formula multiplies by $-2$ first, then adds $3$ : the vertex $(0, 0)$ goes to $(-2, 0)$ , stays at height $0$ after the stretch, then rises to $(-2, 3)$ .”

Why: Outside the parentheses, the transformations follow the order of operations of the formula, exactly as when you compute a number: $-2w + 3$ is not $-2(w + 3)$ .

4. Reading the domain on the simplified formula

1 mark, and every later question that uses the domain

What not to write

“ $F(x) = \frac{x^2 - 9}{x - 3} = x + 3$ , so the domain of $F$ is $\mathbb{R}$ .”

What to write

“ $F$ is not defined at $x = 3$ . For $x \ne 3$ , $F(x) = x + 3$ : the graph is the line $y = x + 3$ with a hole at $(3, 6)$ , and the domain is $\mathbb{R} \setminus \{3\}$ .”

The graph of

\frac{x^2 - 9}{x - 3}

is the line

y = x + 3

with ONE point missing: the open dot at

(3, 6)

is the domain restriction made visible.

Why: Cancelling $x - 3$ is legal only because $x \ne 3$ was assumed; the simplified formula forgets that assumption, the function does not. The same happens with $v(v(x))$ for $v(x) = \frac{1}{x - 2}$ , whose simplified form is defined at $x = 2$ .

5. Taking the range of a quadratic from its two ends

1 mark

What not to write

“On $[-3, 1]$ , $q(x) = x^2 + 4x + 1$ goes from $q(-3) = -2$ to $q(1) = 6$ , so its range is $[-2, 6]$ .”

What to write

“ $q(x) = (x + 2)^2 - 3$ , and $x = -2$ is in $[-3, 1]$ , so the minimum is $-3$ and the maximum is $q(1) = 6$ : the range is $[-3, 6]$ .”

The arc on

[-3, 1]

goes down to

-3

x = -2

, lower than both ends: the range

[-3, 6]

lies between the two dashed levels, not between

q(-3)

and

q(1)

Why: The two ends give the range only when the function moves in one direction on the whole interval. A vertex inside the interval is lower (or higher) than both ends: complete the square and look where the vertex is.

6. Keeping a candidate that is not in its own piece

1 mark, for an extra solution that does not solve the equation

What not to write

“ $|x - 1| = 2x + 4$ : $x - 1 = 2x + 4$ gives $x = -5$ , and $-(x - 1) = 2x + 4$ gives $x = -1$ . Solutions: $-5$ and $-1$ .”

What to write

“Case $x \ge 1$ : $x = -5$ , not $\ge 1$ , rejected. Case $x < 1$ : $x = -1$ , kept. Check: $|-2| = 2 = 2(-1) + 4$ . The only solution is $x = -1$ .”

Why: Each formula of a piecewise function is valid on its own interval only. A candidate that solves the formula of a piece it does not belong to is not a solution: $|-6| = 6$ while $2(-5) + 4 = -6$ .

7. Proving a symmetry with one point

the whole question

What not to write

“ $p(x) = x^3 - x$ : $p(-1) = 0 = p(1)$ , so $p$ is even.”

What to write

“ $p(-x) = -x^3 + x = -p(x)$ for every $x$ , so $p$ is odd. It is not even: $p(-2) = -6 \ne 6 = p(2)$ .”

Why: Even and odd are statements about EVERY $x$ : they are proved by computing $f(-x)$ in general. One point can only disprove, as $x = 2$ does here.

8. Treating a composition as a product, or as commutative

2 marks, on the formula and on its domain

What not to write

“ $f(x) = x^2$ and $g(x) = x + 1$ , so $(f \circ g)(x) = x^2(x + 1)$ , and also $(g \circ f)(x)$ .”

What to write

“ $(f \circ g)(x) = f(x + 1) = (x + 1)^2$ , while $(g \circ f)(x) = g(x^2) = x^2 + 1$ ; at $x = 1$ they give $4$ and $2$ .”

Why: $f \circ g$ feeds the OUTPUT of $g$ into $f$ : every $x$ of $f$ is replaced by the whole block $g(x)$ . Multiplication is another operation, and the order of a composition is read from the inside.

Which method to choose

Which transformation, by the FORM of the formula

Look at where each number sits, inside the parentheses of f or outside

If a number added inside, $f(x + h)$ → horizontal shift, AGAINST the sign: $+h$ moves left

Example: $\sqrt{x + 3}$ starts at $x = -3$
If a number added outside, $f(x) + k$ → vertical shift, WITH the sign

Example: $\sqrt x + 3$ starts at $(0, 3)$
If a factor inside, $f(bx)$ → horizontal compression by $b$ (stretch if $0 < b < 1$ ); a negative $b$ also reflects in the $y$ -axis

Example: $f(2x)$ : $(4, 1) \mapsto (2, 1)$
If a factor outside, $af(x)$ → vertical stretch by $|a|$ , and a reflection in the $x$ -axis if $a < 0$

Example: $-2f(x)$ : $(4, 1) \mapsto (4, -2)$
If an inside of the form $bx + c$ with $b \ne 1$ → factor $b$ out first, $f(b(x + \frac{c}{b}))$ ; or solve $bx + c = u$

Example: $\sqrt{8 - 2x} = \sqrt{-2(x - 4)}$ : it starts at $x = 4$ , not $8$
If several operations outside → apply them in the order of operations of the formula

Example: $-2(x + 2)^2 + 3$ : times $-2$ , then plus $3$ , vertex $(-2, 3)$

The quadratic, the root and the quotient of two linear expressions are brought to these forms by completing the square, by factoring under the root, and by dividing: $\frac{x + 1}{x - 1} = 1 + \frac{2}{x - 1}$ .

Which restriction, by the FORM of the expression

Before simplifying anything, list every place where the formula could fail

If a denominator → it must not be $0$ : factor it and remove each root

Example: $\frac{1}{x^2 - 4x}$ : $x \ne 0$ and $x \ne 4$
If a square root (or any even root) in a numerator → what is under it is $\ge 0$

Example: $\sqrt{x + 3}$ : $x \ge -3$
If an even root in a denominator → what is under it is STRICTLY positive

Example: $\frac{1}{\sqrt{(x - 1)(x - 4)}}$ : $x < 1$ or $x > 4$
If an odd root, $\sqrt[3]{\ }$ → no restriction at all

Example: $\sqrt[3]{x - 8}$ at $x = 0$ is $-2$
If a composition $f(g(x))$ → domain of $g$ first, then $g(x)$ in the domain of $f$

Example: $\frac{1}{\sqrt x - 2}$ : $x \ge 0$ and $x \ne 4$

In a model, add the restrictions of reality: a volume, a length, a price are positive, even where the formula would accept a negative value.

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.

Proving that a function is even, odd, or neither

When to use it: Any question that asks whether $f$ is even, odd or neither, or to justify a symmetry of its graph

1 State the domain and check that it is symmetric about $0$ . If it is not, conclude NEITHER at once.
2 Compute $f(-x)$ in general, replacing EVERY $x$ by $(-x)$ with parentheses, and simplify.
3 Compare with $f(x)$ and with $-f(x)$ . If one of them matches for every $x$ , conclude.
4 If neither matches, give ONE numerical counterexample for each symmetry: a value $a$ with $f(-a) \ne f(a)$ and a value $b$ with $f(-b) \ne -f(b)$ .

Concluding sentence

“The domain $\mathbb{R}$ is symmetric about $0$ , and for every real $x$ , $f(-x) = (-x)^3 - 4(-x) = -(x^3 - 4x) = -f(x)$ . Hence $f$ is odd, and its graph is symmetric about the origin.”

The trap: Checking $f(-1) = f(1)$ and concluding even: one point proves nothing, as $x^3 - x$ shows.

Marking: Typically 1 mark for the domain, 2 for the general computation of f(-x), 1 for the conclusion; for NEITHER, the two counterexamples carry the marks.

Finding the domain of a composition

When to use it: Any question that gives $f$ and $g$ and asks for $f \circ g$ with its domain

1 Write the domain of the INNER function $g$ .
2 Write the condition for $g(x)$ to be a legal input of $f$ , and solve it.
3 Intersect the two conditions: that is the domain.
4 Only then write and simplify the formula of $f(g(x))$ , and say that the simplified formula is valid on that domain only.

Concluding sentence

“ $x$ must be in the domain of $v$ , so $x \ne 2$ , and $v(x)$ must be in the domain of $v$ , so $\frac{1}{x - 2} \ne 2$ , that is $x \ne \frac{5}{2}$ . The domain of $v \circ v$ is $\mathbb{R} \setminus \{2, \frac{5}{2}\}$ , and $(v \circ v)(x) = \frac{x - 2}{5 - 2x}$ on it.”

The trap: Reading the domain on the simplified formula, which is defined at $x = 2$ although $v(2)$ does not exist.

Marking: Typically 2 marks for the formula and 2 for the domain, of which 1 for the condition on the inner function.

Check before you hand in

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

One point through the formula

After a transformation, take one named point of the original graph, compute where it should go, and plug that x into the new formula.

For $y = f(2x + 4)$ and $B(0, 4)$ on $f$ : the image is $(-2, 4)$ , and indeed $f(2(-2) + 4) = f(0) = 4$ . An image at $(-4, 4)$ would give $f(-4)$ , not defined: the error is caught.
Each candidate in its own piece

Before writing the solutions of a piecewise or absolute value equation, test each one in the ORIGINAL equation.

$x = -5$ in $|x - 1| = 2x + 4$ gives $6 = -6$ : rejected in two seconds.
The domain against a test value

Pick a number just inside and just outside each boundary you found, and see whether the formula computes.

$\frac{\sqrt{x + 3}}{x^2 - 4x}$ at $x = -3$ gives $\frac{0}{21} = 0$ : the bracket at $-3$ is closed. At $x = 4$ the denominator is $0$ : excluded.
Signs that cannot be

A square root is never negative, an absolute value is never negative, a price or a volume is positive. A negative answer for one of them means a lost sign.

$\sqrt{x^2 - 6x + 9} = x - 3$ gives $-3$ at $x = 0$ : impossible, the correct simplification is $|x - 3| = 3 - x$ there.

The typical problem, taken apart

A transformed graph with an inside to factor and two outside operations

The graph of $f$ is made of three segments through $A(-2, 1)$ , $B(0, 3)$ , $C(2, -1)$ and $D(4, 1)$ . Sketch $y = 1 - f\left(\frac{x}{2} + 1\right)$ and give its domain and range.

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

The graph of

f

: domain

[-2, 4]

, range

[-1, 3]

, with its four named points. The question asks for the image of each named point, not for a new drawing from scratch.

Step 1

Inside: $\frac{x}{2} + 1 = u$ gives $x = 2(u - 1) = 2u - 2$ . Outside: $y = 1 - w$ . So $(u, w) \mapsto (2u - 2,\ 1 - w)$ .

Why

Solving inside $= u$ gives the $x$ -coordinate directly, whatever the order of the transformations: this is the line that earns the method mark.

Step 2

In words: $\frac{x}{2} + 1 = \frac{1}{2}(x + 2)$ , a horizontal STRETCH by $2$ , then a shift LEFT $2$ ; outside, a reflection in the $x$ -axis ( $-w$ ), then up $1$ .

Why

The factor $\frac{1}{2}$ is inside, so it stretches; factoring it out shows that the shift is $2$ , not $1$ . Outside, the minus sign acts before the $+1$ .

Step 3

$A(-2, 1) \mapsto (-6, 0)$ , $B(0, 3) \mapsto (-2, -2)$ , $C(2, -1) \mapsto (2, 2)$ , $D(4, 1) \mapsto (6, 0)$ . Join them by segments in the same order.

Why

Segments stay segments under these transformations, so four points suffice for the sketch. Naming them makes the answer checkable.

Step 4

Domain: $-2 \le \frac{x}{2} + 1 \le 4$ gives $-6 \le x \le 6$ . Range: $f$ takes every value of $[-1, 3]$ , so $1 - f$ takes every value of $[1 - 3, 1 + 1] = [-2, 2]$ .

Why

The domain comes from the inside, the range from the outside. Multiplying by $-1$ swaps the ends of the interval, which is where the order of the bracket is lost.

Step 5

Check with $B$ : at $x = -2$ , $1 - f\left(\frac{-2}{2} + 1\right) = 1 - f(0) = 1 - 3 = -2$ , as found.

Why

One point through the formula confirms both the inside and the outside at once; it costs ten seconds.

The conclusion, written out

“The graph of $y = 1 - f\left(\frac{x}{2} + 1\right)$ is the polygonal line through $(-6, 0)$ , $(-2, -2)$ , $(2, 2)$ and $(6, 0)$ ; its domain is $[-6, 6]$ and its range is $[-2, 2]$ .”

The classic mistake on this problem: Reading $\frac{x}{2} + 1$ as a compression by $2$ and a shift of $1$ to the left, which sends $A$ to $(-2, 0)$ ; or adding $1$ before reflecting, which gives the range $[-4, 0]$ .

Learn by heart

• Inside acts on $x$ and runs BACKWARDS: $f(x + 3)$ left, $f(2x)$ compress. Outside acts on $y$ as written: $f(x) + 3$ up, $2f(x)$ stretch.
• $f(bx + c)$ : factor $b$ first, or solve $bx + c = u$ . Outside: multiply before adding, as the formula says.
• Domain: denominator $\ne 0$ , even root $\ge 0$ ( $> 0$ in a denominator), read BEFORE simplifying.
• $\sqrt{a^2} = |a|$ . Piecewise: which piece first; every candidate checked against its piece.
• Even $f(-x) = f(x)$ , odd $f(-x) = -f(x)$ , for EVERY $x$ of a symmetric domain.
• $(f \circ g)(x) = f(g(x))$ : $g$ first. Not a product, not commutative. Domain: $g$ first, then $g(x)$ in the domain of $f$ .
• Power model $kx^p$ : doubling $x$ multiplies $y$ by $2^p$ . Linear model: equal steps add the slope.

Frequently asked questions

Why does f(x + 3) shift the graph to the left and not to the right?

Because the new function must take at each x the value that f takes at x plus 3, three units further right. The whole graph therefore arrives three units earlier, on the left. The quick test is to ask where the inside of the parentheses equals zero: for x plus 3, that happens at x equal to minus 3, which is where the point at the origin of f has moved.

In what order should I apply transformations to a graph?

Handle the inside and the outside separately. Outside the parentheses, follow the order of operations of the formula: multiply first, then add. Inside, factor the coefficient of x first, then read the stretch and the shift. When in doubt, take a point of the original graph, solve for the x that makes the inside equal to its first coordinate, then apply the outside operations to its second coordinate.

How do I find the domain of a composition of two functions in MATH 140?

Start with the inner function: x must be in its domain. Then its output must be an allowed input of the outer function, which gives a second condition to solve. The domain is the set of x satisfying both. Never read the domain from the simplified final formula, because simplifying can erase a value that the inner function already excluded.

How do I prove that a function is odd or even?

Check that the domain is symmetric about zero, then compute f of minus x in general, replacing every x by minus x in parentheses, and simplify. If you get f of x, the function is even; if you get minus f of x, it is odd. If you get neither, give one numerical counterexample for each symmetry. Testing one point can show a function is not even, but it can never prove that it is.

Practise it

Corrected exercises: Functions, graphs and transformations, 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

Next sheet Exponential, inverse and logarithmic functions

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

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Get in touch for a first session. The review of functions decides more of the course than it seems: every derivative, limit and curve sketch of MATH 140 is done on these functions.

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Revision sheet: functions, graphs and transformations (MATH 140)

The essentials

Inside acts on x and runs backwards, outside acts on y as written

Domain, pieces and symmetry: three computations, always in the same order

The rules in table form

What each formula does to the graph of f

The mistakes that cost marks

1. Shifting the wrong way

2. Reading f(2x + 4) as a shift of 4

3. Adding before multiplying outside

4. Reading the domain on the simplified formula

5. Taking the range of a quadratic from its two ends

6. Keeping a candidate that is not in its own piece

7. Proving a symmetry with one point

8. Treating a composition as a product, or as commutative

Which method to choose

Which transformation, by the FORM of the formula

Which restriction, by the FORM of the expression

How the answer is expected to be written

Proving that a function is even, odd, or neither

Finding the domain of a composition

Check before you hand in

The typical problem, taken apart

A transformed graph with an inside to factor and two outside operations

Learn by heart

Frequently asked questions

Why does f(x + 3) shift the graph to the left and not to the right?

In what order should I apply transformations to a graph?

How do I find the domain of a composition of two functions in MATH 140?

How do I prove that a function is odd or even?

Corrected exercises: Functions, graphs and transformations, MATH 140 at McGill

See also

Looking for a MATH 140 tutor in Montreal?