Functions · Book or PDF

Functions 11

Chris Kirkpatrick, Barbara Alldred, Andrew Dmytriw, Shawn Godin, Angelo Lillo, David Pilmer, Susanne Trew, Noel Walker · Grade 11 · Study material generated by AI from this edition. Not affiliated with the publisher.

Chapter 1 · pp. 10–89

Chapter 1: Introduction to Functions

This chapter introduces relations and functions, the tools for telling them apart, and function notation. Later parts cover domain and range, parent functions, inverses and transformations. This part covers the first ideas: what a function is, how to recognize one, and how to read and use f(x).

Relations, functions, domain and range

A relation is a set of ordered pairs written (input, output). The inputs are values of the independent variable. The outputs are values of the dependent variable. For example, pairing students with their soccer practice day gives (Craig, Tuesday). (p. 15)

A function is a relation where each input is paired with exactly one output. If you know the input, you can predict a single output. Different inputs may share an output, but one input may never have two. (pp. 15, 16, 19)

The domain is the set of all inputs. The range is the set of all outputs. On a graph, these are usually the x-values and y-values. List repeated values only once. For {(15, Tue), (16, Tue), (15, Thu), (17, Sat)}, the domain is {15, 16, 17} and the range is {Tue, Thu, Sat}. (pp. 15, 16, 19)

Worked example (ages and practice days). In the age relation above, 15 pairs with both Tuesday and Thursday. Knowing a student's age does not tell you the practice day, so this is not a function. The student-name relation is a function, because each name appears once. (p. 16)

Ways to show a function

The same function can be given in words, a table of values, a set of ordered pairs, a mapping diagram, a graph or an equation. For example, "double x, then subtract 5" becomes y = 2x - 5. It can also be a table or a straight-line graph. (p. 9)

Mapping diagram. List the domain in a left oval and the range in a right oval. Draw an arrow from each input to its output. If any input has two or more arrows leaving it, the relation is not a function. In the age example, 15 has arrows to Tuesday and Thursday, so it fails. (pp. 16, 38)

Ordered pairs. Scan the first coordinates for repeats. {(0, 4), (3, 5), (5, -2), (0, 1)} is not a function, because 0 pairs with both 4 and 1. {(-5, 1), (-3, 2), (-1, 3), (1, 2)} is a function. The second coordinate 2 repeats, which is allowed. (pp. 16, 38)

The vertical-line test

A graph is a function if and only if no vertical line crosses it more than once. Hold a ruler upright and slide it across the graph. If it ever touches two points, one x-value has two y-values, so the graph is not a function. A circle fails, because most vertical lines cross it twice. A parabola opening up or down passes. (pp. 17, 19)

Recognizing functions from equations

To test an equation, substitute easy x-values, such as x = 0, and count the y-values. One x that gives two or more y-values means the relation is not a function. Finding single answers for a few x-values supports the idea that it is a function but does not prove it. (pp. 9, 19)

Circle. In $x^2 + y^2 = 9$ , putting x = 0 gives $y^2 = 9$ , so y = 3 or y = -3. One input gives two outputs. The graph is a circle of radius 3 centred at the origin. It is a relation but not a function. (pp. 18, 19)
Linear relations. y = mx + b is a function, and so is a horizontal line such as y = 3. A vertical line x = a is not a function, because one x-value is paired with every y-value. For y = 2x - 5, doubling x and subtracting 5 always gives one answer. (pp. 9, 19, 21)
Quadratic relations. y = ax^2 + bx + c and y = a(x - h)^2 + k are functions. Their graphs are parabolas opening up or down, so each vertical line meets them at most once. For example, y = 2x^2 - 3x + 1 is a function. A sideways parabola such as x = y^2 - 5y is not a function, because y is not written in terms of x. (pp. 18, 19, 20)

Real data and lines of best fit

A class recorded shoe size and height. Several students with shoe size 10 had different heights (158, 173, 170, 171, 168 cm). One input has many outputs, so the scatter plot is not a function. A line of best fit gives exactly one height for each shoe size. It is a function, so it is easier to use for prediction, but it is only an approximation. (pp. 14, 15)

Function notation

Function notation such as f(x) names the function f. It stands for the output when the input is x. It is read "f of x" or "f at x." Writing y = 2x + 1 as f(x) = 2x + 1 shows that x is the input. Other letters work too, such as v(t) for velocity at time t or C(n) for the cost of n items. (pp. 24, 32, 49)

Common mistake: f(x) does not mean f multiplied by x. The letter f is the function's name, and f(x) is another symbol for the output, the same as y. For f(x) = 2 - 3x, f(2) means put 2 into the rule: 2 - 3(2) = -4. It is not f times 2. (pp. 24, 32)

Evaluating f(a). Substitute a for every x in the equation and simplify. Use brackets around the substituted value. For f(x) = x^2 - 3x:

$f(2) = 2^2 - 3(2) = -2$

The input can also be a variable or an expression. For g(x) = 1 - 2x, g(3b) = 1 - 2(3b) = 1 - 6b. (pp. 25, 31, 32)

Function notation and graphs

f(a) is the y-coordinate of the point on the graph of f whose x-coordinate is a. If f(2) = 3, the point (2, 3) is on the graph. To read a value, go up from x = a to the curve, then across to the y-axis. On the worked graph, g(3) = 2 and g(-1) = 0. (pp. 30, 32)

To solve f(x) = k from a graph, reverse this. Find the height y = k on the y-axis, move across to the curve, and read the x-coordinate(s) below. In the worked graph, g(x) = 1 at x = 0, because the curve crosses the y-axis at y = 1. Solving f(x) = 0 gives the x-intercepts. (p. 30)

A steady rate of change

A steady rate of change gives a linear function. In a deep mine, the temperature at the top of the shaft is 11 °C and it rises 0.015 °C for each metre of depth d. This gives:

$T(d) = 11 + 0.015d$

Each depth gives one temperature, so T is a function of d. At the bottom of East Rand mine, d = 3585:

$T(3585) = 11 + 53.775 = 64.775 \approx 65\text{ °C}$

At Western Deep, d = 4100, so T = 11 + 61.5 = 72.5, about 73 °C. (pp. 24, 25)

Interpolation and extrapolation

A graph can be used to estimate outputs. Take the mine temperature function $T(d) = 11 + 0.015d$ , where $d$ is depth in metres. Its graph was drawn through points for depths 0 to 4000 m.

Interpolation means reading a value between known points, inside the plotted range. A mine 3585 m deep lies inside the range, so reading $T(3585)$ from the graph is interpolation. It gives about 65 °C. The equation gives $T(3585) = 64.775$ .
Extrapolation means extending the graph past the known points to estimate a value outside them. A mine 4100 m deep lies beyond 4000 m. Extending the line gives about 73 °C, and the equation gives $T(4100) = 72.5$ .

Interpolation is usually safer. Extrapolation assumes the trend keeps going the same way. (p. 26)

Building a function from a number game

To turn a "think of a number" game into a function, follow the number through each step. A flow chart helps. Call the starting number $x$ , the input.

Double it: $2x$
Subtract that from 12: $12 - 2x$
Multiply by the original number: $x(12 - 2x)$

So $f(x) = x(12 - 2x) = 12x - 2x^2$ . The result is quadratic, and each $x$ gives only one answer, so $f$ is a function of $x$ . (p. 28)

To see who wins, evaluate $f$ for each player's number. For example, $f(5) = 60 - 50 = 10$ , $f(7) = 84 - 98 = -14$ and $f(10) = 120 - 200 = -80$ . The number 5 beats these, so that player wins. (p. 28)

What is the best number to choose? The coefficient of $x^2$ is negative, so the parabola opens down and the vertex is the maximum. Factor to $f(x) = -2x(x - 6)$ . The zeros are 0 and 6. The vertex lies halfway between the zeros, at $x = \frac{0+6}{2} = 3$ . Then $f(3) = 18$ , which beats every other output, so 3 is the best choice. (p. 29)

Comparing two functions at one input

Comparing outputs at the same $x$ compares heights on the graphs. Let $f(x) = x^2 - 3x$ and $g(x) = 1 - 2x$ . Then $f(2) = 4 - 6 = -2$ and $g(2) = 1 - 4 = -3$ . Since $-2 > -3$ , the point of $f$ at $x = 2$ is above the point of $g$ at $x = 2$ . (p. 31)

Families and parent functions

A family of functions is a group of functions that share common features. The parent function is the simplest function in the family. (pp. 35, 37)

Linear family: $f(x) = mx + b$ , parent $f(x) = x$ . Example: $f(x) = 3x + 2$ .
Quadratic family: $f(x) = a(x-h)^2 + k$ , parent $f(x) = x^2$ . Example: $f(x) = 5(x-3)^2 - 1$ .

The five parent functions in this chapter are $f(x) = x$ , $f(x) = x^2$ , $f(x) = \sqrt{x}$ , $f(x) = \frac{1}{x}$ and $f(x) = |x|$ . They are the linear, quadratic, square root, reciprocal and absolute value functions. Each graph has its own shape, and later they are the building blocks for sketching harder functions. (pp. 35–37)

Features of each parent function

$f(x) = x$ . It is a straight line through the origin with slope 1. It lies in quadrants 1 and 3 and cuts the plane diagonally in half. Points like $(2, 2)$ and $(-4, -4)$ are on it. The line never ends, so the domain and range are all real numbers. (p. 36)

$f(x) = x^2$ . It is a parabola opening up, with its vertex at the origin and the $y$ -axis as its axis of symmetry. Squares are never negative, so the graph lies in quadrants 1 and 2 and the minimum is at $(0, 0)$ . The points $(2, 4)$ and $(-2, 4)$ mirror each other. The domain is all real numbers and the range is $y \ge 0$ . (p. 36)

$f(x) = \sqrt{x}$ . The symbol means the positive root only. A negative $x$ gives no real root, so the domain is $x \ge 0$ . Outputs are never negative, so the range is $y \ge 0$ . The graph starts at the origin and rises through $(1, 1)$ and $(4, 2)$ . It looks like half of a parabola lying on its side, but it still passes the vertical-line test, so it is a function. (pp. 36, 37, 44)

$|x|$ and $f(x) = |x|$ . The absolute value $|x|$ is the distance of $x$ from 0. It equals $x$ when $x \ge 0$ and $-x$ when $x < 0$ . So $|3| = 3$ and $|-3| = 3$ . The output is never negative. The graph is a V with its corner at the origin, and the range is $y \ge 0$ . (pp. 36, 37)

$f(x) = \frac{1}{x}$ . You cannot divide by 0, so $x = 0$ is not allowed, and a table of values shows an error there. Near $x = 0$ , the outputs become very large, positive or negative. Far from 0, the outputs get close to 0 but never reach it. This splits the graph into two separate parts. The domain and range are all real numbers except 0. (p. 37)

An asymptote is a line that a graph gets closer and closer to without meeting it, over some part of its domain. The reciprocal function has two: $x = 0$ (vertical) and $y = 0$ (horizontal). The other four parent functions have none. (p. 37)

Finding domain and range

The domain and range can be found from a table, a graph or an equation. A table shows only a few points. A graph shows where the curve starts, stops and spreads, so it is usually the easiest. The equation reveals restrictions, such as a square root of a negative number or a maximum at a vertex. Using two sources to check each other is wise. (p. 44)

Real-world limits. A pebble dropped from the CN Tower's 346 m lookout has speed $v(d) = \sqrt{19.6d}$ , where $d$ is the distance fallen. The pebble can only fall from 0 to 346 m. So the domain is $\{d \in \mathbb{R} \mid 0 \le d \le 346\}$ . The speed starts at $v(0) = 0$ and grows to $v(346) \approx 82.4$ m/s. The range is $0 \le v \le 82.4$ . (pp. 39, 40, 44)

Set notation. The set $\{x \in \mathbb{R} \mid x \ge 3\}$ reads "all real numbers $x$ such that $x$ is at least 3." The symbol $\in$ means "belongs to," $\mathbb{R}$ is the set of real numbers (rational and irrational, such as $\sqrt{2}$ and $\pi$ ), and $\mid$ means "such that." (pp. 40, 42, 44)

Open and closed circles. On a graph, an open circle at an endpoint means the point is excluded. A closed circle means it is included. A segment with an open circle at $x = -5$ and a closed circle at $x = 11$ has domain $\{x \in \mathbb{R} \mid -5 < x \le 11\}$ . Use $<$ for an open end and $\le$ for a closed end. (p. 42)

Parabolas. Every quadratic function has all real numbers as its domain. The vertex limits the range. A parabola opening down with vertex $(1, 3)$ has no output above 3, so its range is $y \le 3$ . A parabola opening up has range $y \ge k$ , where $k$ is the vertex's $y$ -value. (pp. 42, 44)

A circle. The circle $x^2 + y^2 = 25$ has centre $(0, 0)$ and radius 5. It reaches 5 units left, right, up and down, so the domain is $-5 \le x \le 5$ and the range is $-5 \le y \le 5$ . Many vertical lines cross it twice, so it is not a function, though it still has a domain and range. (p. 42)

Linear and constant functions. If $m \ne 0$ , then $f(x) = mx + b$ has domain and range of all real numbers. For $f(x) = 2x - 3$ , any $x$ works and any $y$ can be reached. A constant function $f(x) = b$ gives the same output every time, so its range is just $\{b\}$ . (pp. 42, 44)

Square root functions. The expression under the root must be at least 0. For $h(x) = \sqrt{2 - x}$ , solve $2 - x \ge 0$ to get $x \le 2$ . The domain is $\{x \in \mathbb{R} \mid x \le 2\}$ . The root is never negative, so the range is $\{y \in \mathbb{R} \mid y \ge 0\}$ . Similarly, $g(x) = \sqrt{x - 3}$ has domain $x \ge 3$ . (pp. 43, 44)

Fixed fencing. Suppose 24 m of fencing goes on three sides of a rectangular garden, with the house forming the fourth side. Let the width be $x$ . The two widths use $2x$ of fence, so the length is $24 - 2x$ . The area is

$A(x) = x(24 - 2x) = -2x(x - 12).$

The zeros are 0 and 12. A width of 0 or 12 gives zero area, and widths outside that interval give negative area, so the domain is $0 < x < 12$ . The vertex is halfway between the zeros, at $x = 6$ , where $A(6) = -2(6)(-6) = 72$ . This is the maximum, so the range is $0 < A \le 72$ . (pp. 43, 44)

What an inverse function is

The inverse of a function reverses it. If the original takes an input and gives an output, the inverse takes that output and gives back the original input. "Add 3" is undone by "subtract 3" (p. 51).

For a linear function the idea is the same, and the roles of the variables swap. Take $f(x)=5x+8$ . It sends 2 to 18, so its inverse sends 18 back to 2. The independent variable of the inverse is the dependent variable of the original, and the reverse also holds (pp. 52, 56).

Example: the paving problem. A company charges $10 per square foot of paving plus a$ 50 delivery fee, so cost is $C=10A+50$ . Forty square feet costs $10(40)+50=\$ 450 $. Tom has a budget and wants to know the area, so he needs the inverse. Swapping the columns of the table gives \$ 450 → 40 sq ft. His budget of $850 buys 80 sq ft. Both relations are functions (pp. 52–53).

Finding the inverse algebraically

Method 1: reverse the operations. List the operations applied to $x$ in order. Then apply the opposite operations in the opposite order.

For $f(x)=5x+8$ , the steps are multiply by 5, then add 8. Reversing gives subtract 8, then divide by 5, so $f^{-1}(x)=\frac{x-8}{5}$ .
For $f(x)=2-5x$ , the steps are multiply by $-5$ , then add 2. Reversing gives subtract 2, then divide by $-5$ , so $f^{-1}(x)=\frac{x-2}{-5}$ (pp. 52–53, 56, 84).

Method 2: interchange the variables. Write $y=f(x)$ , swap $x$ and $y$ , then solve for $y$ . This works because every point $(x,y)$ on the function becomes $(y,x)$ on the inverse. For $y=3x-4$ , swap to get $x=3y-4$ . Add 4 and divide by 3 to get $y=\frac{x+4}{3}$ (pp. 53, 56, 84).

Both methods give the same answer. For the paving function, reversing the operations on $C=10A+50$ gives $A=\frac{C-50}{10}$ , which is Tom's relation (p. 53).

Two common mistakes.

Notation. In $f^{-1}$ , the $-1$ only names the inverse. It is not an exponent and does not mean a reciprocal. For $f(x)=5x+8$ , $f^{-1}(x)=\frac{x-8}{5}$ , not $\frac{1}{5x+8}$ . A reciprocal is written $\frac{1}{f(x)}$ or $[f(x)]^{-1}$ (p. 53).
Order. The steps must be reversed in order, not just replaced by opposites. Let $f$ be "multiply by 3, then add 1," so $f(2)=7$ . The correct inverse is subtract 1, then divide by 3, which sends 7 back to 2. If you divide by 3 first and then subtract 1, you get $\frac73-1=\frac43$ , which is wrong (pp. 46–47).

Points, graphs, domain and range

Swapped points. If $(a,b)$ is on the graph of $f$ , then $(b,a)$ is on the graph of $f^{-1}$ . If $f(2)=7$ , then $f^{-1}(7)=2$ . In a table of values for $f$ , swapping the two columns gives a table for the inverse (pp. 54, 56).

Reflection in $y=x$ . Swapping coordinates mirrors each point across the line $y=x$ . If you fold the paper along that line, the graph of $f$ lands on the graph of $f^{-1}$ . Any point where the graph meets $y=x$ lies on both the function and its inverse (pp. 42, 44, 56, 84).

Graphing an inverse from a graph. List points on the function and swap each pair. Plot the new points and join them. Then draw $y=x$ and check that the two graphs mirror each other. For a line through $(5,0)$ and $(0,3)$ , the inverse passes through $(0,5)$ and $(3,0)$ (pp. 44, 56).

Domain and range swap. Inputs and outputs trade places, so the domain of the inverse is the range of the original, and the range of the inverse is the domain of the original (pp. 55–56).

Example. The temperature in a mine is $T(d)=11+0.015d$ , with domain $0\le d\le5000$ . The end values of the domain give the range. At $d=0$ , $T=11$ . At $d=5000$ , $T=11+0.015(5000)=86$ . The function is linear and increasing, so the range is $11\le T\le86$ . The inverse therefore has domain $11\le T\le86$ and range $0\le d\le5000$ (pp. 45, 55).

Is the inverse a function?

The inverse of a function is not always a function. To check, apply the vertical-line test to the inverse graph (pp. 44, 46).

Suppose $f=\{(1,4),(2,4),(3,5)\}$ . It is a function. Swapping gives $\{(4,1),(4,2),(5,3)\}$ , where $x=4$ appears twice. A vertical line at $x=4$ hits two points, so the inverse is not a function.
A horizontal line such as $y=3$ has points $(a,3)$ . Swapping gives $(3,a)$ , which forms the vertical line $x=3$ . That fails the vertical-line test.
Every other line has an inverse that is also a line and also a function. So the inverse of a linear function is linear unless the original is horizontal (pp. 43, 56, 84).

Slopes and intercepts

For a linear function and its inverse, the slopes are reciprocals. Also, the y-intercept of the inverse equals the x-intercept of the original (pp. 42, 52).

For $f(x)=10x+50$ , the slope is 10. The inverse is $y=\frac{x-50}{10}=\frac1{10}x-5$ , with slope $\frac1{10}$ . The original crosses the x-axis at $-5$ , and the inverse crosses the y-axis at $-5$ .

Using an inverse to solve a problem

To use an inverse, find the inverse function. Then substitute the known output of the original function as its input (p. 45).

For the mine, the inverse is $d(T)=\frac{T-11}{0.015}$ . To reach $22\,^\circ\text{C}$ , $d(22)=\frac{22-11}{0.015}\approx733$ m. Someone planning a geothermal heating system could use this to decide how deep to go.

Self-inverse functions

A self-inverse function is its own inverse. Among linear functions, $f(x)=x$ and $f(x)=-x+b$ qualify. For $f(x)=-x+4$ , swap to get $x=-y+4$ , so $y=4-x$ , which is the same function. Its graph is symmetric about $y=x$ . Any line with slope $-1$ , and the line $y=x$ itself, works (p. 49).

Transformations of parent functions

A transformation of a parent function $f$ has the form $g(x)=a\,f(x-d)+c$ . The constants $a$ , $d$ and $c$ change the location or shape of the parent graph. For $f(x)=x^2$ , the function $g(x)=3(x-2)^2+1$ is a parabola with vertex $(2,1)$ , stretched vertically by the factor 3 (pp. 50–51).

The parent functions here each have a recognizable shape (p. 50):

$y=x^2$ is a parabola.
$y=\sqrt{x}$ is half of a parabola on its side, starting at $(0,0)$ .
$y=\frac1x$ has two branches, with asymptotes at $x=0$ and $y=0$ .
$y=|x|$ is a V with its vertex at $(0,0)$ .

The constants have the same effect on every parent function (p. 51):

$d$ shifts the graph horizontally.
$c$ shifts it vertically.
$a$ stretches it vertically.

The graph of $y=|x-1|+2$ moves the V one unit right and two up, just as $y=(x-1)^2+2$ moves the parabola. If $a<0$ , as in $y=-\sqrt{x}$ , the graph flips over the x-axis. If $|a|>1$ the graph is narrower than the parent, and if $0<|a|<1$ it is wider.

Horizontal stretches and compressions: y = f(kx)

Earlier transformations changed the output of a function (vertical stretches by $a$ , shifts by $c$ and $d$ ). Now we change the input by multiplying $x$ by a number $k$ .

In $y=f(kx)$ , the graph of $y=f(x)$ is stretched or compressed horizontally by the factor $\frac{1}{|k|}$ . A point $(x,y)$ on $f$ moves to $\left(\frac{x}{k},y\right)$ . The y-coordinate stays the same, because only the input changed. (pp. 52, 53, 58)

If $|k|>1$ , the graph is compressed.
If $0<|k|<1$ , the graph is stretched.
If $k<0$ , the graph is also reflected in the y-axis. (p. 58)

Example: $y=\sqrt{2x}$ . The point $(4,2)$ on $y=\sqrt{x}$ becomes $(2,2)$ , a horizontal compression by the factor $\frac12$ . (pp. 52, 53)

An invariant point is a point that a transformation leaves unchanged. For horizontal stretches of the parent functions in this chapter, the origin $(0,0)$ is invariant, because $0\div k=0$ . (p. 52)

Common mistake: many students think a number bigger than 1 multiplying $x$ stretches the graph. It compresses it. In $y=(4x)^2$ , the output 1 appears at $x=\pm\frac14$ , not $\pm1$ . You get the same output from a smaller input, so the graph is squeezed toward the y-axis. (pp. 54, 58)

Scale factors. For a vertical change $af(x)$ , the scale factor is $a$ . For a horizontal change $f(kx)$ , it is $\frac1k$ . In both cases, a factor above 1 is a stretch and a factor between 0 and 1 is a compression. So $y=3f(x)$ stretches vertically by 3, while $y=f(3x)$ compresses horizontally by $\frac13$ . These graphs are generally different. (pp. 53, 58)

Graphing and reading y = f(kx)

To graph $y=f(kx)$ : pick key points on $y=f(x)$ , divide each x-coordinate by $k$ , keep the y-coordinates, and join the points. For $y=(4x)^2$ , the points $(1,1),(2,4),(3,9)$ become $(\frac14,1),(\frac12,4),(\frac34,9)$ . Join them to $(0,0)$ and use symmetry for the left half. (p. 54)

For $y=|0.25x|$ , the factor is $\frac{1}{0.25}=4$ . The points $(1,1),(2,2),(3,3)$ become $(4,1),(8,2),(12,3)$ . (p. 54)

Even symmetry. If $f(-x)=f(x)$ , as for $y=|x|$ and $y=x^2$ , reflecting in the y-axis changes nothing. So $y=|-3x|$ looks the same as $y=|3x|$ . The reflection is there but cannot be seen. (pp. 55, 66)

To find $k$ from a graph: match a point on the parent with its partner on the transformed graph, then divide the parent's x-coordinate by the new one. If $(1,1)$ on $y=|x|$ matches $(4,1)$ , then $k=1\div4=\frac14$ . If $(1,1)$ on $y=\sqrt{x}$ matches $(-0.25,1)$ , then $k=1\div(-0.25)=-4$ . That means a compression and a reflection in the y-axis. (pp. 55, 56)

Real-life case. The pendulum time $p(L)=2\pi\sqrt{\frac{1}{10}L}$ has $k=\frac1{10}$ and $a=2\pi\approx6.3$ . It is $y=\sqrt{x}$ stretched horizontally by 10 and vertically by $2\pi$ . The point $(1,1)$ goes to $(10,1)$ and then to $(10,6.3)$ . (pp. 52, 57)

Combining all four transformations

The general form is $y=af[k(x-d)]+c$ . A key point $(x,y)$ on $f$ moves to $\left(\frac{x}{k}+d,\ ay+c\right)$ . (pp. 61, 62, 63, 69)

Order matters. Apply $a$ and $k$ (stretches, compressions, reflections) first, then $c$ and $d$ (translations). This is like the order of operations: multiplying comes before adding. For $y=2\sqrt{x}+3$ , $(4,2)$ goes to $(4,4)$ and then to $(4,7)$ . (pp. 61, 69)

Reading the parameters. (p. 69)

Parameter	Effect
$\lvert a\rvert>1$	vertical stretch
$a<0$	reflection in the x-axis
$\lvert k\rvert>1$	horizontal compression
$k<0$	reflection in the y-axis
$d>0$	move right
$c>0$	move up

For $y=-2f[\frac12(x+3)]-4$ : $a=-2$ (stretch by 2 and flip over the x-axis), $k=\frac12$ (horizontal stretch by 2), $d=-3$ (3 left), $c=-4$ (4 down). (p. 69)

Worked sketch: $f(x)=-3\sqrt{2(x+4)}-1$ . Starting from $y=\sqrt{x}$ : compress horizontally by $\frac12$ , stretch vertically by 3, reflect in the x-axis, then shift 4 left and 1 down. The point $(4,2)$ goes to $(2,2)$ , then $(2,6)$ , then $(2,-6)$ , then $(-2,-7)$ . The start $(0,0)$ ends at $(-4,-1)$ . The domain and range come from the final graph: $x\ge-4$ and $y\le-1$ . The reflection makes the graph go downward from the start point. (pp. 61–63)

Shortcut: you can do $a$ and $k$ together in one pass, then $c$ and $d$ together in a second pass. (pp. 63, 69, 85)

Writing equations and spotting traps

Writing an equation from a description. Assign values to $a$ , $k$ , $d$ , $c$ , then substitute. For $f(x)=\frac1x$ with a horizontal stretch by 3 and a y-axis reflection ( $k=-\frac13$ ), a vertical stretch by 2 ( $a=2$ ), 5 right and 4 up: $g(x)=\frac{2}{-\frac13(x-5)}+4$ . (p. 64)

Reciprocal function. Translations move the asymptotes, but stretches and reflections do not. Shifting $\frac1x$ right 5 and up 4 gives asymptotes $x=5$ and $y=4$ . The domain is $x\ne5$ and the range is $y\ne4$ . (pp. 64, 65)

Factor out $k$ first. If the equation looks like $f(kx+b)$ , rewrite it as $f[k(x-d)]$ before reading the shift. For $g(x)=f(-5x+10)$ with $f(x)=x^2$ , you get $f[-5(x-2)]$ . That is a horizontal compression by $\frac15$ , a y-axis reflection (invisible for a parabola), and a shift of 2 right. (p. 66)

Common mistake: $y=f(2x-6)$ does not shift 6 right. Factoring gives $f[2(x-3)]$ , so the shift is 3 right, plus a compression by $\frac12$ . (pp. 66, 69)

Matching equations to graphs. Check these in order:

The parent shape.
Whether it opens up or down, or is flipped.
The vertex or start point (this gives $d$ and $c$ ; for a reciprocal, use the asymptotes).
A specific point, such as $(1,1)$ , to find the stretch. (pp. 67–69)

Domain restrictions in design. Limiting the domain of a transformed parent graph keeps only the piece you need. In the cat-face task, absolute value pieces form the ears, parabolas form the head and chin, and half-parabolas from $\sqrt{x}$ form the eyes. Reflections give left-right symmetry. (p. 89)

Check yourself

Describe how the graph of $y=\sqrt{-4x}$ differs from $y=\sqrt{x}$ . Where does the point $(4,2)$ go?
For $g(x)=3f(x-2)+1$ , what happens to the point $(5,4)$ on $y=f(x)$ ? What if the function is $y=f(3x)$ instead?
Write $y=\frac{1}{2x-8}$ in the form $a\,f[k(x-d)]+c$ with $f(x)=\frac1x$ . State the asymptotes, the domain and the range.
If $f(x)=5x+8$ , find $f^{-1}(x)$ by reversing operations. How do the domain and range of $f^{-1}$ compare with those of $f$ ?
A graph is a function only if it passes the vertical-line test. Does the circle $x^2+y^2=9$ pass it? Explain.

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