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Transformations of Functions Review

Try each problem first. You don’t need to finish everything.

Core path ~55 min Full sheet ~1 h 50 min

Recommended: K1–K5, then one A and one T problem

K = Knowledge & Understanding · A = Application · T = Thinking · Boss = Challenge

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Core Skills (K)

Translations · Reflections · Stretches · Mapping Notation · Equations from Graphs

K1 Translations

The table lists four points on the graph of $y=f(x)$ . For each function, give the images of the four points.

$x$	$-2$	$0$	$1$	$4$
$f(x)$	$3$	$-1$	$0$	$2$

a)
$y=f(x)+4$
b)
$y=f(x+3)$
c)
$y=f(x-2)-5$

Stuck? Peek at a hint

For $f(x+3)$ , find the $x$ that makes the bracket equal each table input: solve $x+3=-2$ , and so on.

K2 Reflections

Let $f(x)=\sqrt{x+4}-1$ .

a)
Write the equation of $y=-f(x)$ and state its range in interval notation.
b)
Write the equation of $y=f(-x)$ and state its domain in interval notation.
c)
State the invariant points of $y=-f(x)$ and of $y=f(-x)$ .

Stuck? Peek at a hint

Put the whole of $f(x)$ in brackets before negating it. For $f(-x)$ , every $x$ becomes $(-x)$ .

K3 Stretches & Compressions

The graph of $y=f(x)$ is shown. Each grid square is 1 unit.

a)
State the images of $P$ , $Q$ , $R$ and $S$ on $y=2f(x)$ .
b)
State the images of $P$ , $Q$ , $R$ and $S$ on $y=f(2x)$ .
c)
State the invariant points of $y=2f(x)$ and of $y=f(2x)$ .

Stuck? Peek at a hint

For $y=f(2x)$ , find which new input gives the old output: solve $2x=-4$ , $2x=-2$ , and so on.

K4 Mapping Notation

a)
Write the mapping rule for $y=-2f(3(x-4))+5$ .
b)
The point $(-3,2)$ is on $y=f(x)$ . Find its image on $y=-2f(3(x-4))+5$ .
c)
A function $f$ has domain $[-4,8]$ and range $[-3,5]$ . State the domain and range of $y=-2f(2x+6)+1$ in interval notation.

Stuck? Peek at a hint

Read $a$ , $k$ , $d$ and $c$ before mapping anything. In (c), factor $2x+6$ first.

K5 Reading Equations & Graphs

a)
State the transformations, in order, that map $y=f(x)$ onto $y=\frac{1}{2}f(-3x+9)-2$ .
b)
The graph is a transformation of $y=x^2$ . Determine its equation. Each grid square is 1 unit.
c)
For $y=-\dfrac{2}{x+4}+1$ , state the equations of the asymptotes and the image of the point $(1,1)$ on $y=\dfrac{1}{x}$ .

Stuck? Peek at a hint

In (a), factor $-3$ out of $-3x+9$ . In (b), substitute a point other than the vertex to find $a$ .

Real World (A)

Translations in Time · Stretches in Context · Square-Root Models

A1 Second Kick

Priya kicks a ball from the ground. Its height, in metres, after $t$ seconds is $h(t)=-5t^2+20t$ . Her friend kicks an identical ball the same way from the same spot, 1.5 s later. Let $g(t)$ be the height of the friend's ball, with $t$ still measured from Priya's kick.

a)
Write $g(t)$ in terms of $h$ , then in vertex form.
b)
When does each ball reach its maximum height, and when does each land?
c)
State the domain of $g$ in this context. The formula gives $g(1)=-11.25$ . Where is the friend's ball at $t=1$ ?

Stuck? Peek at a hint

Make a mini-table: the friend's ball at $t=1.5$ , $2.5$ and $3.5$ is where Priya's ball was at which times?

A2 Fair Ferris Wheel

At a summer fair in Ottawa, a Ferris wheel seat starts at its lowest point and makes one turn in 40 s. The table gives its height $h(t)$ , in metres, after $t$ seconds.

$t\text{ (s)}$	$0$	$10$	$20$	$30$	$40$
$h(t)\text{ (m)}$	$2$	$10$	$18$	$10$	$2$

a)
The operator speeds the wheel up so one turn takes 30 s. Write the new height function in terms of $h$ , and find when a rider first reaches the top.
b)
A bigger wheel turns in 40 s, and every height is 1.5 times the height in the table. Write its height function in terms of $h$ and state its range.
c)
Riders board from a platform 2 m high, at the seat's lowest point. Which of the two new rides can still use this platform? Explain.

Stuck? Peek at a hint

In (a), the new ride reaches in 30 s what the old one reached in 40 s. Which input of $h$ matches $t=30$ ?

A3 Horizon from the Cliff

On a clear day, the distance to the horizon, in kilometres, from eyes $h$ metres above the water is $d(h)=3.57\sqrt{h}$ . A lighthouse on Lake Superior stands on a cliff 45 m above the water, and its tower is 36 m tall. Let $D(h)$ be the horizon distance for a visitor whose eyes are $h$ metres above the cliff top.

a)
Write $D(h)$ in terms of $d$ , then as an equation. Which transformations of $y=\sqrt{h}$ give $D$ ?
b)
Visitors stand anywhere from the cliff top to the top of the tower. State the domain and range of $D$ in this context, to one decimal place.
c)
A guide wants to know the height at which the horizon is exactly 20 km away. Solve $D(h)=20$ and interpret the result.

Stuck? Peek at a hint

A visitor $h$ metres above the cliff top has eyes $h+45$ metres above the water.

Think It Through (T)

Error Analysis · Counterexamples

T1 Order Matters

A student wants the graph of $y=\sqrt{x}$ compressed horizontally by a factor of $\frac{1}{3}$ , then translated 2 units right. They write $y=\sqrt{3x-2}$ .

a)
Find the endpoint of the student's graph and the endpoint the description requires.
b)
Write the correct equation, both factored and expanded under the root.
c)
The student's equation does match a translation and a compression of $y=\sqrt{x}$ , in some order. Name them in the order applied.

Stuck? Peek at a hint

Set the expression under each root equal to 0 to find where each graph starts.

T2 Do the Negatives Cancel?

A student makes two claims.
Claim 1: " $y=-f(-x)$ is always the same graph as $y=f(x)$ , because the two negatives cancel."
Claim 2: " $g(x)=x^3-2x+1$ is odd, because every power of $x$ is odd."

a)
Use $f(x)=(x-1)^2$ to show Claim 1 is false: write $-f(-x)$ and give a point on $y=f(x)$ that is not on $y=-f(-x)$ .
b)
Describe the functions for which Claim 1 is true, and give one example other than $y=x$ .
c)
Evaluate $g(-1)$ and $-g(1)$ . Is $g$ even, odd, or neither?

Stuck? Peek at a hint

For (a), replace every $x$ in $f$ with $(-x)$ , then negate the whole result.

Boss Level

Optional challenge

Mirror and Slide

Let $R$ be a reflection in the $y$ -axis and $S$ a translation 4 units right. Transformations are applied to the graph of $y=f(x)$ one after another, from left to right.

a)
Apply $R$ , then $S$ , then $R$ . Write the final equation in terms of $f$ , and state the single transformation it equals.
b)
The point $(1,6)$ is on $y=f(x)$ . Apply the pair " $R$ then $S$ " 50 times in a row. Where does the point end up?
c)
Find every point that one pair " $R$ then $S$ " leaves where it is.

Stuck? Peek at a hint

Track one point: $R$ sends $(x,y)$ to $(-x,y)$ , and $S$ sends it to $(x+4,y)$ .

Answer key

Final answers only. Try first, then check.

Core Skills

a) $(-2,7)$ , $(0,3)$ , $(1,4)$ , $(4,6)$
b) $(-5,3)$ , $(-3,-1)$ , $(-2,0)$ , $(1,2)$
c) $(0,-2)$ , $(2,-6)$ , $(3,-5)$ , $(6,-3)$

a) $y=-\sqrt{x+4}+1$ ; $R=(-\infty,1]$
b) $y=\sqrt{4-x}-1$ ; $D=(-\infty,4]$
c) $y=-f(x)$ : $(-3,0)$ ; $y=f(-x)$ : $(0,1)$

a) $(-4,0)$ , $(-2,4)$ , $(0,2)$ , $(4,-6)$
b) $(-2,0)$ , $(-1,2)$ , $(0,1)$ , $(2,-3)$
c) $y=2f(x)$ : $(-4,0)$ and $(1,0)$ ; $y=f(2x)$ : $(0,1)$

a) $(x,y)\to\left(\dfrac{x}{3}+4,\ -2y+5\right)$
b) $(3,1)$
c) $D=[-5,1]$ , $R=[-9,7]$

a) Vertical compression by a factor of $\frac{1}{2}$ ; horizontal compression by a factor of $\frac{1}{3}$ and reflection in the $y$ -axis; translation 3 units right and 2 units down
b) $y=\frac{1}{2}(x+1)^2-3$
c) $x=-4$ and $y=1$ ; image $(-3,-1)$

Real World

a) $g(t)=h(t-1.5)=-5(t-3.5)^2+20$
b) Priya's: top at 2 s, lands at 4 s. Friend's: top at 3.5 s, lands at 5.5 s (both reach 20 m).
c) $D=[1.5,5.5]$ . At $t=1$ the ball has not been kicked yet: it is resting on the ground. $t=1$ is outside the domain, so $-11.25$ m means nothing.

a) $H(t)=h\left(\frac{4}{3}t\right)$ ; at $t=15$ s
b) $W(t)=1.5h(t)$ ; $R=[3,27]$ (metres)
c) Only the faster ride: a horizontal compression keeps every height, so its lowest point stays at 2 m. On the bigger wheel the lowest seat is at 3 m, 1 m above the platform.

a) $D(h)=d(h+45)=3.57\sqrt{h+45}$ ; a vertical stretch by a factor of 3.57 and a translation 45 units left
b) $D=[0,36]$ (metres), $R\approx[23.9,32.1]$ (kilometres)
c) $h\approx-13.6$ : 13.6 m below the cliff top, outside the domain. Every visitor already sees at least 23.9 km, so no place on the cliff or tower gives 20 km.

Think It Through

a) Student's: $\left(\frac{2}{3},0\right)$ ; required: $(2,0)$
b) $y=\sqrt{3(x-2)}=\sqrt{3x-6}$
c) Translate 2 units right, then compress horizontally by a factor of $\frac{1}{3}$ (equivalently: compress by $\frac{1}{3}$ , then translate $\frac{2}{3}$ of a unit right)

a) $-f(-x)=-(x+1)^2$ ; e.g. $(1,0)$ is on $y=f(x)$ , but $-f(-1)=-4$
b) Odd functions, where $f(-x)=-f(x)$ for every $x$ ; e.g. $y=x^3$ or $y=\dfrac{1}{x}$
c) $g(-1)=2$ , $-g(1)=0$ : not odd. $g(1)=0\ne g(-1)$ : not even. Neither.

Boss Level

Boss

a) $y=f(x+4)$ : a translation 4 units left
b) $(1,6)$
c) Every point with $x=2$ : the pair is a reflection in the line $x=2$ .

Free cheat sheet

Transformations of Functions Cheat Sheet

The whole unit on one page: every rule, a worked example for each idea, and the traps that cost marks.

Unit Summary

One printable page with every rule.

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Most testedFind the image of $(4,5)$ under $y=-2f(3x+6)-1$ , and describe the transformations. (card 4)

1Translations

$y=f(x-d)+c\qquad(x,y)\to(x+d,\ y+c)$

Use $d>0$ : right $d$ . $c>0$ : up $c$ . Every point moves, not only the intercepts.

Example $y=(x+3)^2-4$ : left $3$ , down $4$ , so the vertex is $(-3,-4)$ .

Trap $f(x+3)$ moves left $3$ : the sign inside the brackets works the opposite way.

2Reflections

$-f(x)$ : in the $x$ -axis $\qquad f(-x)$ : in the $y$ -axis

UseInvariant points lie on the mirror axis. $-f(-x)$ does both: $(x,y)\to(-x,-y)$ . Even: $f(-x)=f(x)$ . Odd: $f(-x)=-f(x)$ .

TrapTest every term: $2x^3-5x+3$ is neither even nor odd, because the $+3$ never changes sign.

Trap $-f(x)$ negates the whole rule: $-(2\sqrt{x-1}-4)=-2\sqrt{x-1}+4$ .

3Stretches

$af(x)$ : $y\times a\qquad f(kx)$ : $x\div k$

Use $|a|>1$ stretches, $0<|a|<1$ compresses. $f(kx)$ scales horizontally by $1/k$ . Invariant points: $x$ -intercepts under $af(x)$ ; the $y$ -intercept under $f(kx)$ .

Trap $f(2x)$ is a horizontal compression by $1/2$ : divide the $x$ -coordinates by $2$ .

Trap $3f(x)$ multiplies every $y$ : for $f(x)=x^2-2$ , $3f(x)=3x^2-6$ , not $3x^2-2$ .

4Combined: the mapping rule

$y=af[k(x-d)]+c$

$(x,y)\to\left(\dfrac{x}{k}+d,\ ay+c\right)$

UseFactor $k$ out first. Stretches and reflections, then translations.

Example $y=3f(2x-6)+4=3f[2(x-3)]+4$ : $(4,1)\to(4/2+3,\ 3+4)=(5,7)$ .

Trap $f(2x-6)=f[2(x-3)]$ shifts right $3$ , not $6$ : factor before reading $d$ .

5Domain and range

UseDomain: map the $x$ -endpoints with $x/k+d$ . Range: map the $y$ -endpoints with $ay+c$ . A negative $a$ or $k$ swaps the ends.

ExampleDomain $[-2,6]$ , $y=f(2x-4)=f[2(x-2)]$ : new domain $[1,5]$ .

TrapRange $y\ge0$ under $y=-3f(x)+1$ becomes $y\le1$ : the reflection flips the inequality.

6Reading an equation

UseFactor, then read $a$ (vertical; $x$ -axis reflection if negative), $k$ (horizontal by $1/|k|$ ; $y$ -axis reflection if negative), $d$ , $c$ .

Example $3f(-2x+6)-4=3f[-2(x-3)]-4$ : stretch $3$ , compress $1/2$ , reflect in the $y$ -axis, right $3$ , down $4$ .

TrapCompress by $1/3$ , then right $2$ : $f[3(x-2)]$ , not $f(3x-2)$ , which moves right only $2/3$ .

7Equation from a graph

UseName the parent. Read $(d,c)$ from the vertex or endpoint. Substitute a second point to find $a$ .

ExampleVertex $(2,-3)$ , through $(4,5)$ : $5=a(4-2)^2-3$ , so $a=2$ : $y=2(x-2)^2-3$ .

TrapSubstituting the vertex gives $-3=-3$ and no $a$ : use another point.

Inside or outside?

a

and

c

, outside the brackets

y

, as written

k

and

d

, inside the brackets

x

, reversed

A negative sign Reflection

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