Transformations — vertical - horizontal shifts, reflections, stretches - compressions: Exercises

Before we start, one shared vocabulary reminder, so no symbol appears unearned.

Definition The point-mapping table (our master tool)

Every transformation is really a rule that moves each point $(x, y)$ of the old graph to a new place. " $(x, y)$ " just means "a dot at horizontal position $x$ and height $y$ ." Here is the complete map — memorise the directions, and every problem below is just bookkeeping.

Transformation	New point
$f (x) + k$ (outside, add)	$(x, y + k)$ — up if $k > 0$
$a f (x)$ (outside, multiply)	$(x, a y)$ — stretch if $∣ a ∣ > 1$
$- f (x)$	$(x, - y)$ — flip over the $x$ -axis
$f (x - h)$ (inside, subtract)	$(x + h, y)$ — right if $h > 0$
$f (b x)$ (inside, multiply)	$(\frac{x}{b}, y)$ — squeeze if $∣ b ∣ > 1$
$f (- x)$	$(- x, y)$ — flip over the $y$ -axis

Outside the function → vertical, behaves as expected. Inside the function → horizontal, behaves backwards.

Level 1 — Recognition

L1.1

The graph of $f (x) = x^{2}$ is changed to $g (x) = x^{2} - 5$ . Describe the transformation in words.

Recall Solution

WHAT: The change is outside the function — a constant is added to the output ( $- 5$ ). WHY it's vertical: whatever the machine spits out, we push it down by 5; output = height, so height drops. Using the map, $(x, y) \to (x, y - 5)$ . Answer: a vertical shift down by 5 units. The vertex moves from $(0, 0)$ to $(0, - 5)$ ; the shape is unchanged.

L1.2

Identify each transformation of $f$ : (a) $f (x + 7)$ (b) $4 f (x)$ (c) $f (- x)$ .

Recall Solution

(a) Inside the function, and it's $f (x - (- 7))$ , so $h = - 7$ . Inside behaves backwards → shift LEFT 7 units. Map: $(x, y) \to (x - 7, y)$ . (b) Outside, multiply by $4 > 1$ → vertical stretch by factor 4. Map: $(x, y) \to (x, 4 y)$ . (c) Inside, negate the input → reflection across the $y$ -axis. Map: $(x, y) \to (- x, y)$ .

Level 2 — Application

L2.1

Let $f (x) = ∣ x ∣$ . Find $g (x) = ∣ x - 3∣ + 2$ and give the coordinates of its corner (vertex).

Recall Solution

The corner of $∣ x ∣$ sits at $(0, 0)$ (where the inside is $0$ ). See the accent-red corner in the figure.

Figure — Transformations — vertical - horizontal shifts, reflections, stretches - compressions

Inside:

x - 3

→ shift right 3 (solve

x - 3 = 0 \Rightarrow x = 3

). Outside:

+ 2

→ shift up 2. Corner moves

(0, 0) \to (0 + 3, 0 + 2) = (3, 2)

. Answer: vertex at

(3, 2)

g (x) = ∣ x - 3∣ + 2

L2.2

Given $f (x) = sin x$ with period $2 π$ and amplitude $1$ , find the amplitude and period of $g (x) = \frac{1}{2} sin (4 x)$ .

Recall Solution

Amplitude comes from the outside multiplier $a = \frac{1}{2}$ : amplitude $= ∣ a ∣ = \frac{1}{2}$ . (Output halved.) Period comes from the inside multiplier $b = 4$ : period $= \frac{2 π}{∣ b ∣} = \frac{2 π}{4} = \frac{π}{2}$ . Answer: amplitude $\frac{1}{2}$ , period $\frac{π}{2}$ . The wave is squeezed sideways (runs 4× faster) and squashed to half height.

L2.3

$f (x) = x$ starts at $(0, 0)$ . Find where the start point of $g (x) = x + 5 - 1$ lands, and state its domain.

Recall Solution

Inside $x + 5 = 0 \Rightarrow x = - 5$ → shift left 5. Outside $- 1$ → shift down 1. Start point $(0, 0) \to (- 5, - 1)$ . Domain: we need $x + 5 \geq 0 \Rightarrow x \geq - 5$ , so $[- 5, \infty)$ . (See Domain and Range.) Answer: start at $(- 5, - 1)$ , domain $[- 5, \infty)$ .

Level 3 — Analysis

L3.1

$f (x) = x^{2}$ is transformed to $g (x) = 2 (x - 1)^{2} - 3$ . A point on $f$ is $(2, 4)$ . Where does it go?

Recall Solution

Apply the map in order (inside first, then outside).

Inside shift right 1: $x$ -coordinate $2 \to 2 + 1 = 3$ .
Outside multiply by 2: $y$ -coordinate $4 \to 2 \cdot 4 = 8$ .
Outside shift down 3: $y$ -coordinate $8 \to 8 - 3 = 5$ . Answer: $(2, 4) \to (3, 5)$ . Check directly: $g (3) = 2 (3 - 1)^{2} - 3 = 2 \cdot 4 - 3 = 5$ . ✓

L3.2

$g (x) = - 3 f (x) + 1$ where $f (x)$ has range $[0, 4]$ . Find the range of $g$ . (Uses Domain and Range.)

Recall Solution

Take the extreme outputs of $f$ , which are $0$ and $4$ , and push them through the outside operations $y \mapsto - 3 y + 1$ .

$y = 0 \to - 3 (0) + 1 = 1$
$y = 4 \to - 3 (4) + 1 = - 11$ The $- 3$ flips the order (bottom becomes top), so the new range runs from the smaller to the larger: $[- 11, 1]$ . Answer: range $= [- 11, 1]$ .

L3.3

The order matters. Compare $p (x) = (2 x) - 6$ style thinking: for $f (x) = x$ , is $g_{1} (x) = 2 x - 6$ the same as "shift left/right then squeeze"? Rewrite $g_{1}$ in standard form $f (b (x - h))$ and find the true horizontal shift.

Recall Solution

WHAT: we factor the inside so the $x$ is bare: $2 x - 6 = 2 (x - 3)$ . So $g_{1} (x) = 2 (x - 3)$ , i.e. $b = 2$ , $h = 3$ . Interpretation: shift right 3, then compress horizontally by factor $\frac{1}{2}$ . The start point: inside $= 0 \Rightarrow 2 (x - 3) = 0 \Rightarrow x = 3$ , landing at $(3, 0)$ . Answer: $g_{1} (x) = 2 (x - 3)$ ; horizontal shift is right 3 (not 6!).

Level 4 — Synthesis

L4.1

Write the equation of a parabola that is $f (x) = x^{2}$ reflected over the $x$ -axis, stretched vertically by 3, shifted left 4 and up 1.

Recall Solution

Build outward using $g (x) = a f (b (x - h)) + k$ .

Reflect over $x$ -axis + stretch 3 → $a = - 3$ .
Shift left 4 → $h = - 4$ , so inside is $(x - (- 4)) = (x + 4)$ .
Up 1 → $k = + 1$ .
No horizontal scaling → $b = 1$ . Answer: $g (x) = - 3 (x + 4)^{2} + 1$ . Vertex at $(- 4, 1)$ , opening downward.

L4.2

A cosine curve has amplitude 5, period $π$ , is shifted right $\frac{π}{4}$ and down 2, starting from $f (x) = cos x$ . Write $g (x)$ .

Recall Solution

Amplitude 5 (outside): $a = 5$ .
Period $π$ → $\frac{2 π}{∣ b ∣} = π \Rightarrow ∣ b ∣ = 2$ , take $b = 2$ .
Shift right $\frac{π}{4}$ : $h = \frac{π}{4}$ , inside $= 2 (x - \frac{π}{4})$ .
Down 2: $k = - 2$ . Answer: $g (x) = 5 cos (2 (x - \frac{π}{4})) - 2$ . Quick check at the shifted peak $x = \frac{π}{4}$ : inside $= 0$ , $cos 0 = 1$ , $g = 5 (1) - 2 = 3$ — the maximum height, as expected.

L4.3

Express $g (x) = 4 - - x$ as a chain of transformations of $f (x) = x$ , in the correct order.

Recall Solution

Rewrite: $g (x) = - - x + 4$ , so $a = - 1$ , inside factor $b = - 1$ , $k = + 4$ . Order (inside first, outside last):

$f (- x) = - x$ : reflect across the $y$ -axis (domain becomes $x \leq 0$ ).
$- - x$ : reflect across the $x$ -axis ( $a = - 1$ ).
$+ 4$ : shift up 4. Start point $(0, 0)$ : $y$ -reflection keeps it $(0, 0)$ ; $x$ -reflection keeps it $(0, 0)$ ; up 4 → $(0, 4)$ . Answer: reflect over $y$ -axis, then over $x$ -axis, then up 4; start point $(0, 4)$ , domain $x \leq 0$ .

Level 5 — Mastery

L5.1

Fully analyse $g (x) = - 2 3 (x + 1) + 4$ (from Example 5 in the parent note). Find (a) the start point, (b) the image of the parent point $(3, 3)$ ... actually of $(3, 3)$ — track $(a, a)$ for $a = 3$ — and (c) the range.

Recall Solution

Here $a = - 2$ , $b = 3$ , $h = - 1$ , $k = 4$ , parent $f (x) = x$ .

(a) Start point. Parent start $(0, 0)$ . Inside $3 (x + 1) = 0 \Rightarrow x = - 1$ . Outside: $- 2 \cdot 0 + 4 = 4$ . Start $\to (- 1, 4)$ .

(b) Image of parent point $(3, 3)$ .

Inside scaling & shift move the $x$ : solve $3 (x + 1) = 3 \Rightarrow x + 1 = 1 \Rightarrow x = 0$ . So the new $x$ is $0$ .
Outside: new $y = - 23 + 4$ . Image $= (0, 4 - 23)$ . Numerically $4 - 23 \approx 0.536$ . Verify: $g (0) = - 2 3 (1) + 4 = - 23 + 4$ . ✓

(c) Range. Parent $x$ outputs $[0, \infty)$ . Multiply by $- 2$ : $(- \infty, 0]$ . Add 4: $(- \infty, 4]$ . Because of the reflection the graph goes down from its start height 4. Answer: start $(- 1, 4)$ , image $(0, 4 - 23)$ , range $(- \infty, 4]$ .

L5.2

Two students transform $f$ . Student A computes $g_{A} (x) = f (2 x - 6)$ ; Student B computes $g_{B} (x) = f (2 (x - 6))$ . If $f (x) = x$ , find the start-point $x$ -value for each and explain the difference.

Recall Solution

Start point is where inside $= 0$ .

A: $2 x - 6 = 0 \Rightarrow x = 3$ . (Factor: $2 (x - 3)$ , so shift right 3, then compress.)
B: $2 (x - 6) = 0 \Rightarrow x = 6$ . (Shift right 6, then compress.) They differ because A's " $- 6$ " is inside the scaling (gets divided by 2 in effect), while B's shift is applied to the bare $x$ . Answer: A starts at $x = 3$ , B starts at $x = 6$ ; the $b = 2$ scaling only "eats" A's shift, halving it.

L5.3 (capstone)

Start from $f (x) = x^{2}$ . Apply, in this literal order of writing: reflect over $y$ -axis, compress horizontally by factor $\frac{1}{2}$ (i.e. $b = 2$ ), shift right 3, stretch vertically by 4, shift down 5. Write the final $g (x)$ and locate the vertex.

Recall Solution

Careful: the written order names operations, but the algebra must respect $a f (b (x - h)) + k$ . Combine the horizontal pieces first.

Reflect over $y$ -axis: $b$ gains a sign → $b = - 2$ (magnitude 2 from the compress).
Shift right 3: $h = 3$ .
Inside becomes $b (x - h) = - 2 (x - 3)$ .
Vertical stretch 4: $a = 4$ .
Shift down 5: $k = - 5$ . So $g (x) = 4 (- 2 (x - 3))^{2} - 5$ . Since it's squared, $(- 2)^{2} = 4$ : $g (x) = 4 \cdot 4 (x - 3)^{2} - 5 = 16 (x - 3)^{2} - 5$ . Vertex: inside $= 0 \Rightarrow x = 3$ ; then $g (3) = - 5$ . Vertex $(3, - 5)$ . Answer: $g (x) = 16 (x - 3)^{2} - 5$ , vertex $(3, - 5)$ , opening up.

Recall Quick self-test

Inside a function changes which direction, and behaves how? ::: Horizontal, and backwards (opposite of the sign / divide by $b$ ). Amplitude of $\frac{1}{2} sin (4 x)$ ? ::: $\frac{1}{2}$ . Period of $\frac{1}{2} sin (4 x)$ ? ::: $\frac{π}{2}$ . True horizontal shift of $2 x - 6$ ? ::: right 3 (factor to $2 (x - 3)$ first).