IB Syllabus Alignment

Standard Level Topic 2.11: Transformations of graphs. Translations: $y = f(x) + b$; $y = f(x - a)$. Reflections (in both axes): $y = -f(x)$; $y = f(-x)$. Vertical stretch with scale factor $a$: $y = a f(x)$. Horizontal stretch with scale factor $\frac{1}{a}$: $y = f(a x)$. Composite transformations: applying up to three transformations in sequence, and tracking invariant points, coordinates, and asymptotes.

1. Translations: Vertical and Horizontal

A translation slides every point on a graph by a fixed distance without changing its shape, size, or orientation. In IB examinations, translations can be described algebraically or using column vector notation $\begin{pmatrix} h \\ k \end{pmatrix}$.

  • Vertical Translation ($y = f(x) + k$): Adds $k$ to every $y$-coordinate. A translation by vector $\begin{pmatrix} 0 \\ k \end{pmatrix}$. If $k > 0$, the graph shifts upward by $k$ units; if $k < 0$, it shifts downward by $|k|$ units. $$\text{Point Mapping: } (x, y) \longrightarrow (x, y + k)$$
  • Horizontal Translation ($y = f(x - h)$): Alters the input before $f$ is evaluated. A translation by vector $\begin{pmatrix} h \\ 0 \end{pmatrix}$. Notice the sign: $f(x - 3)$ shifts right by $3$, whereas $f(x + 4) = f(x - (-4))$ shifts left by $4$. $$\text{Point Mapping: } (x, y) \longrightarrow (x + h, y)$$
  • Combined Translation ($y = f(x - h) + k$): Combines both shifts simultaneously. The translation vector is $\begin{pmatrix} h \\ k \end{pmatrix}$, mapping $(x, y) \to (x + h, y + k)$.
  • Asymptotes and Invariant Points: Under horizontal shifts, vertical asymptotes $x = c$ become $x = c + h$. Under vertical shifts, horizontal asymptotes $y = d$ become $y = d + k$. Translations possess no invariant points unless the translation vector is the zero vector $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$.
Worked Example 1.1 — Tracking Asymptotes & Coordinates Under Translation

The curve $C_1$ has equation $f(x) = \frac{2}{x - 1} + 3$. The curve $C_2$ is obtained by translating $C_1$ by the vector $\begin{pmatrix} -4 \\ 2 \end{pmatrix}$.
(a) State the equations of the vertical and horizontal asymptotes of $C_1$.
(b) Write down an expression for $g(x)$, the equation of $C_2$, in the form $\frac{a}{x - b} + c$.
(c) Find the coordinates of the point on $C_2$ corresponding to the $y$-intercept of $C_1$.

  1. Step 1: Identify asymptotes of $C_1$: The denominator is zero when $x - 1 = 0 \implies x = 1$ (vertical asymptote). As $x \to \pm\infty$, $\frac{2}{x-1} \to 0$, so $y = 3$ (horizontal asymptote).
  2. Step 2: Apply translation vector $\begin{pmatrix} -4 \\ 2 \end{pmatrix}$ to find $g(x)$: $$g(x) = f(x - (-4)) + 2 = f(x + 4) + 2$$ Substitute $x + 4$ into $f(x)$: $$g(x) = \left(\frac{2}{(x + 4) - 1} + 3\right) + 2 = \frac{2}{x + 3} + 5$$
  3. Step 3: Track the $y$-intercept: The $y$-intercept of $C_1$ occurs at $x = 0$: $$f(0) = \frac{2}{0 - 1} + 3 = -2 + 3 = 1 \implies P(0, 1)$$ Under translation vector $\begin{pmatrix} -4 \\ 2 \end{pmatrix}$, the image point $P'$ is: $$P'(0 + (-4), 1 + 2) = (-4, 3)$$ Check: $g(-4) = \frac{2}{-4 + 3} + 5 = -2 + 5 = 3$, confirming exact algebraic agreement.
Answers: (a) Vertical asymptote: $x = 1$, Horizontal asymptote: $y = 3$.
(b) $g(x) = \frac{2}{x + 3} + 5$.
(c) Transformed point: $(-4, 3)$.

2. Stretches: Vertical and Horizontal

A stretch expands or compresses the graph along one axis while holding the perpendicular axis fixed. In the IB curriculum, the fixed axis is referred to as the invariant line.

  • Vertical Stretch ($y = a f(x)$ with $a > 0$):
    • Scale factor: $a$, parallel to the $y$-axis.
    • Invariant line: the $x$-axis ($y = 0$). Any point on the $x$-axis where $y = 0$ remains completely unchanged because $a(0) = 0$.
    • Point mapping: $(x, y) \longrightarrow (x, ay)$.
    • If $a > 1$, the curve stretches vertically away from the $x$-axis; if $0 < a < 1$, it compresses vertically toward the $x$-axis.
  • Horizontal Stretch ($y = f(b x)$ with $b > 0$):
    • Scale factor: $\frac{1}{b}$, parallel to the $x$-axis.
    • Invariant line: the $y$-axis ($x = 0$). The $y$-intercept remains unchanged because $f(b \cdot 0) = f(0)$.
    • Point mapping: $(x, y) \longrightarrow \left(\frac{x}{b}, y\right)$.
    • If $b > 1$, the curve compresses horizontally toward the $y$-axis by factor $\frac{1}{b}$; if $0 < b < 1$, it stretches horizontally away from the $y$-axis by factor $\frac{1}{b}$.
Worked Example 1.2 — Determining Stretched Equations & Scale Factors

The curve $y = f(x)$ has a local maximum at $M(6, 4)$ and a $y$-intercept at $(0, 2)$.
(a) State the coordinates of the image of $M$ on the curve $y = 3 f(2x)$.
(b) State the $y$-intercept of $y = 3 f(2x)$.
(c) Describe the geometric transformations mapping $y = f(x)$ to $y = \frac{1}{2} f\left(\frac{x}{5}\right)$.

  1. Step 1: Point mapping for $y = 3 f(2x)$: Here, $a = 3$ (vertical stretch factor $3$) and $b = 2$ (horizontal stretch factor $\frac{1}{2}$). $$(x, y) \longrightarrow \left(\frac{x}{2}, 3y\right)$$ Applying this transformation to $M(6, 4)$: $$M' = \left(\frac{6}{2}, 3 \times 4\right) = (3, 12)$$
  2. Step 2: $y$-intercept of $y = 3 f(2x)$: The original intercept is $(0, 2)$. The $y$-axis is invariant under horizontal stretch: $$\left(\frac{0}{2}, 3 \times 2\right) = (0, 6)$$
  3. Step 3: Describe transformations for $y = \frac{1}{2} f\left(\frac{x}{5}\right)$: The coefficient outside is $\frac{1}{2}$, which represents a vertical stretch with scale factor $\frac{1}{2}$ (or compression by factor $2$) parallel to the $y$-axis.
    The inside coefficient is $\frac{1}{5}$, so $b = \frac{1}{5}$. The horizontal scale factor is $\frac{1}{b} = \frac{1}{1/5} = 5$ parallel to the $x$-axis.
Answers: (a) $M'(3, 12)$.
(b) $y$-intercept is $(0, 6)$.
(c) A vertical stretch with scale factor $\frac{1}{2}$ parallel to the $y$-axis, and a horizontal stretch with scale factor $5$ parallel to the $x$-axis.

Visualizing Function Transformations

The diagram below displays the parent curve $y = f(x)$ and its transformed image $y = -2f(x - 3) + 1$. Observe how each geometric operation affects the coordinates of key points.

x y O 1 3 4 6 1 2 -1 Shift right +3 Stretch ×2 & Reflect +1 Up P(1, 1.5) P'(4, -2) y = f(x) (Parent) y = -2f(x - 3) + 1 (Transformed)
Figure 2.7 — Anatomy of Combined Function Transformations: Mapping $P(1, 1.5)$ on $y = f(x)$ to $P'(4, -2)$ on $y = -2f(x-3)+1$ through horizontal translation, vertical reflection and stretch, and vertical translation.

3. Reflections Across Coordinate Axes

A reflection flips the curve across one of the coordinate axes. The position of the negative sign determines which variable is inverted.

  • Reflection in the $x$-axis ($y = -f(x)$):
    • The negation is on the outside, multiplying all outputs by $-1$.
    • Invariant line: The $x$-axis ($y = 0$). Any $x$-intercept $(c, 0)$ remains fixed.
    • Point mapping: $(x, y) \longrightarrow (x, -y)$.
    • Local maxima become local minima, and increasing intervals become decreasing intervals.
  • Reflection in the $y$-axis ($y = f(-x)$):
    • The negation is on the inside, negating the input variable $x$.
    • Invariant line: The $y$-axis ($x = 0$). The $y$-intercept $(0, d)$ remains fixed.
    • Point mapping: $(x, y) \longrightarrow (-x, y)$.
    • The graph flips horizontally from left to right.
  • Point Symmetry (Reflection in Origin): Applying both reflections simultaneously, $y = -f(-x)$, produces a half-turn ($180^\circ$ rotation) about the origin $(0, 0)$, mapping $(x, y) \to (-x, -y)$.
Worked Example 1.3 — Consecutive Reflections of an Exponential Function

Let $f(x) = 2^x$.
(a) Find the equation of the curve $g(x)$ formed by reflecting $y = f(x)$ in the $y$-axis, followed by a reflection in the $x$-axis.
(b) State the domain, range, and asymptote of $g(x)$.
(c) Verify that $g(x) = -f(-x)$ represents a point reflection through the origin $(0, 0)$.

  1. Step 1: Apply reflection in the $y$-axis: Negate the input $x$: $$y_1 = f(-x) = 2^{-x} = \left(\frac{1}{2}\right)^x$$
  2. Step 2: Apply reflection in the $x$-axis: Negate the output: $$g(x) = -y_1 = -2^{-x} = -\left(\frac{1}{2}\right)^x$$
  3. Step 3: Determine domain, range, and asymptote: Since $2^{-x} > 0$ for all $x \in \mathbb{R}$, $-2^{-x} < 0$. $$\text{Domain: } x \in \mathbb{R}, \quad \text{Range: } y < 0, \quad \text{Horizontal Asymptote: } y = 0$$
  4. Step 4: Verify point reflection through origin: Take point $P(2, 4)$ on $y = 2^x$. Under $g(x)$: $$g(-2) = -2^{-(-2)} = -2^2 = -4$$ Hence, the point $(-2, -4)$ lies on $y = g(x)$. The midpoint of $(2, 4)$ and $(-2, -4)$ is $\left(\frac{2 + (-2)}{2}, \frac{4 + (-4)}{2}\right) = (0, 0)$, proving origin symmetry.
Answers: (a) $g(x) = -2^{-x}$ or $g(x) = -\left(\frac{1}{2}\right)^x$.
(b) Domain: $x \in \mathbb{R}$, Range: $y \in (-\infty, 0)$, Asymptote: $y = 0$.
(c) $P(x, y) \to (-x, -y)$, which constitutes a $180^\circ$ rotation about $(0, 0)$.

4. Order of Composite Transformations

When multiple transformations are combined, the order of application is critical. Transformations operating vertically affect $y$ independently from transformations operating horizontally on $x$.

  • Vertical Composition ($y = a f(x) + d$): Follow standard order of algebraic operations:
    1. Stretch vertically by scale factor $|a|$ and reflect in the $x$-axis if $a < 0$.
    2. Translate vertically by vector $\begin{pmatrix} 0 \\ d \end{pmatrix}$.
    Note: If expressed as $y = a(f(x) + d)$, the translation occurs before the vertical stretch.
  • Horizontal Composition ($y = f(b x - c)$): To read the correct sequence, factor out the coefficient $b$: $$y = f\left(b\left(x - \frac{c}{b}\right)\right)$$ Method A (Standard IB Order):
    1. Translate horizontally by $\frac{c}{b}$ units right (vector $\begin{pmatrix} c/b \\ 0 \end{pmatrix}$).
    2. Stretch horizontally by scale factor $\frac{1}{b}$ parallel to the $x$-axis.
    Method B (Alternate valid order):
    1. Stretch horizontally by scale factor $\frac{1}{b}$ parallel to the $x$-axis.
    2. Translate horizontally by $c$ units right.
  • General Mapping Formula: For the general transformed function $y = a f(b(x - h)) + k$: $$(x, y) \longrightarrow \left(\frac{x}{b} + h, \; a y + k\right)$$
Worked Example 1.4 — Deconstructing a Multi-Step Transformation

The curve $y = f(x)$ passes through point $A(4, -6)$. The function $g$ is defined by $g(x) = -3 f(2x - 8) + 5$.
(a) Express $g(x)$ in the form $-3 f(2(x - h)) + k$ and fully describe the transformation sequence.
(b) Find the coordinates of $A'$, the image of $A$ under this transformation.

  1. Step 1: Factor the horizontal argument: $$2x - 8 = 2(x - 4)$$ Thus, $g(x) = -3 f(2(x - 4)) + 5$.
  2. Step 2: Describe the transformations:
    • Horizontal: Horizontal stretch with scale factor $\frac{1}{2}$ parallel to the $x$-axis, followed by a translation by vector $\begin{pmatrix} 4 \\ 0 \end{pmatrix}$ (or shift 4 units right).
    • Vertical: Vertical stretch with scale factor $3$ parallel to the $y$-axis, reflection in the $x$-axis, followed by a translation by vector $\begin{pmatrix} 0 \\ 5 \end{pmatrix}$ (shift 5 units up).
  3. Step 3: Calculate coordinates of $A'$: Using the general mapping formula: $$(x', y') = \left(\frac{x}{2} + 4, \; -3y + 5\right)$$ Substitute $x = 4$ and $y = -6$: $$x' = \frac{4}{2} + 4 = 2 + 4 = 6$$ $$y' = -3(-6) + 5 = 18 + 5 = 23$$
Answers: (a) $g(x) = -3 f(2(x - 4)) + 5$. Horizontal stretch factor $\frac{1}{2}$, translation right by $4$, vertical stretch factor $3$, reflection in $x$-axis, translation up by $5$.
(b) $A'(6, 23)$.

3. Composite Sequences & Point Mapping

Worked Example 1.5 — 3-Step Sequence and Coordinate Mapping

The function $f(x)$ has a local minimum at $P(6, -4)$ and a vertical asymptote at $x = 2$.
The graph of $g(x)$ is defined by $g(x) = -3f(2x - 8) + 5$.
(a) Express $g(x)$ in the factored form $a f(b(x - h)) + k$. [1 mark]
(b) Describe the sequence of four elementary geometric transformations that maps the graph of $f$ onto the graph of $g$. [4 marks]
(c) Find the coordinates of the transformed point $P'$. [3 marks]
(d) State the equation of the transformed vertical asymptote. [1 mark]

  1. Part (a): Factor the horizontal argument: $$g(x) = -3f(2(x - 4)) + 5 \quad [A1]$$
  2. Part (b): Sequence of transformations:
    1. Horizontal stretch with scale factor $\frac{1}{2}$ parallel to the $x$-axis (invariant line $x = 0$) [A1].
    2. Horizontal translation by vector $\begin{pmatrix} 4 \ 0 \end{pmatrix}$ (shift right by $4$ units) [A1].
    3. Vertical stretch with scale factor $3$ parallel to the $y$-axis combined with reflection in the $x$-axis (or vertical stretch scale factor $-3$) [A1].
    4. Vertical translation by vector $\begin{pmatrix} 0 \ 5 \end{pmatrix}$ (shift upward by $5$ units) [A1].
  3. Part (c): Coordinate mapping rule for $P(6, -4)$: $$x' = \frac{x}{2} + 4 \implies x' = \frac{6}{2} + 4 = 3 + 4 = 7 \quad [M1][A1]$$ $$y' = -3y + 5 \implies y' = -3(-4) + 5 = 12 + 5 = 17 \quad [A1]$$ $$P'(7, 17)$$
  4. Part (d): Vertical asymptote transformation: Original vertical asymptote at $x = 2$: $$x' = \frac{2}{2} + 4 = 1 + 4 = 5 \implies x = 5 \quad [A1]$$
Answers: (a) $g(x) = -3f(2(x - 4)) + 5$; (b) Horizontal stretch factor $1/2$, right $4$, vertical stretch factor $3$ with $x$-axis reflection, up $5$; (c) $P'(7, 17)$; (d) $x = 5$.

10 Tiered Practice Exercises

Tier 1 focuses on core procedural fluency, Tier 2 tests standard IB exam modeling and point mapping, and Tier 3 presents multi-step synthesis and invariance challenges. Attempt each problem before revealing the complete markscheme solution.

Tier 1: Core Fluency (Problems 1–4)

Exercise 1 Single Translations

The point $P(-3, 7)$ lies on the curve $y = f(x)$. State the coordinates of the corresponding point on the graph of $y = f(x + 5) - 4$.

View Full Solution

Step 1: Identify the transformations:
$f(x + 5)$ represents a horizontal translation 5 units to the left: $x \to x - 5$.
$-4$ represents a vertical translation 4 units downward: $y \to y - 4$.

Step 2: Map the coordinates:
$$x' = -3 - 5 = -8$$ $$y' = 7 - 4 = 3$$

Final Answer: $(-8, 3)$

Exercise 2 Stretches & Scale Factors

The graph of $y = g(x)$ is transformed into $y = \frac{1}{4} g(3x)$.
(a) State the scale factor of the stretch parallel to the $y$-axis.
(b) State the scale factor of the stretch parallel to the $x$-axis.
(c) If the area enclosed under $y = g(x)$ on a given interval is $36\text{ units}^2$, find the area enclosed by the corresponding transformed curve.

View Full Solution

Step 1: Vertical stretch:
The factor outside is $\frac{1}{4}$, so the vertical stretch scale factor is $\frac{1}{4}$.

Step 2: Horizontal stretch:
The coefficient inside is $3$, so the horizontal stretch scale factor is $\frac{1}{3}$.

Step 3: Area scaling:
Area scales by the product of horizontal and vertical scale factors: $$\text{New Area} = 36 \times \left(\frac{1}{4}\right) \times \left(\frac{1}{3}\right) = 36 \times \frac{1}{12} = 3\text{ units}^2$$

Final Answers: (a) $\frac{1}{4}$ (or $0.25$); (b) $\frac{1}{3}$; (c) $3\text{ units}^2$.

Exercise 3 Reflections in Coordinate Axes

The curve $C$ has equation $y = x^2 - 6x + 5$.
(a) Find the equation of the curve formed when $C$ is reflected in the $x$-axis.
(b) Find the equation of the curve formed when $C$ is reflected in the $y$-axis.
(c) Show that both transformed curves have the same $y$-intercept magnitude.

View Full Solution

Step 1: Reflection in $x$-axis ($y = -f(x)$):
$$y = -(x^2 - 6x + 5) = -x^2 + 6x - 5$$

Step 2: Reflection in $y$-axis ($y = f(-x)$):
$$y = (-x)^2 - 6(-x) + 5 = x^2 + 6x + 5$$

Step 3: Compare $y$-intercepts:
For (a), when $x = 0$, $y = -5 \implies |y| = 5$.
For (b), when $x = 0$, $y = 5 \implies |y| = 5$. Both magnitudes equal $5$.

Final Answers: (a) $y = -x^2 + 6x - 5$; (b) $y = x^2 + 6x + 5$; (c) Both have $|y\text{-intercept}| = 5$.

Exercise 4 Identifying Transformations

The graph of $y = \sqrt{x}$ is transformed into $y = 3\sqrt{-(x - 2)} + 4$. List the four geometric transformations that map $y = \sqrt{x}$ onto this curve, stating scale factors and directions explicitly.

View Full Solution

Step 1: Analyze horizontal components:
Inside the radical: $-(x - 2)$.
- Reflection across the $y$-axis (due to the negative sign on $x$).
- Horizontal translation by 2 units to the right (vector $\begin{pmatrix} 2 \\ 0 \end{pmatrix}$).

Step 2: Analyze vertical components:
- Vertical stretch with scale factor $3$ parallel to the $y$-axis.
- Vertical translation by 4 units upward (vector $\begin{pmatrix} 0 \\ 4 \end{pmatrix}$).

Final Answer: 1. Reflection in the $y$-axis; 2. Horizontal translation by 2 units right; 3. Vertical stretch with scale factor 3 parallel to the $y$-axis; 4. Vertical translation by 4 units up.

Tier 2: Exam Modeling (Problems 5–8)

Exercise 5 General Function Point Mapping

The function $f(x)$ has a vertex at $V(-6, 8)$ and a root at $(2, 0)$.
(a) Find the coordinates of the image of $V$ under the transformation $y = -2 f\left(\frac{1}{3}x + 1\right) - 5$.
(b) Find the coordinates of the image of the root $(2, 0)$ under the same transformation.

View Full Solution

Step 1: Rewrite horizontal argument in factored form:
$$\frac{1}{3}x + 1 = \frac{1}{3}(x + 3)$$ Mapping for $x$: Horizontal stretch factor $3$, shifted left by $3$: $$x' = 3x - 3$$ Mapping for $y$: Vertical stretch factor $2$, reflection in $x$-axis, shifted down by $5$: $$y' = -2y - 5$$

Step 2: Transform vertex $V(-6, 8)$:
$$x' = 3(-6) - 3 = -18 - 3 = -21$$ $$y' = -2(8) - 5 = -16 - 5 = -21 \implies V'(-21, -21)$$

Step 3: Transform root $(2, 0)$:
$$x' = 3(2) - 3 = 6 - 3 = 3$$ $$y' = -2(0) - 5 = -5 \implies (3, -5)$$

Final Answers: (a) $V'(-21, -21)$; (b) $(3, -5)$.

Exercise 6 Rational Asymptotes Tracking

Let $f(x) = \frac{1}{x}$. Curve $G$ is given by $g(x) = \frac{6}{2x - 8} + 5$.
(a) Write $g(x)$ in the form $\frac{a}{x - h} + k$.
(b) State the transformation sequence that maps $y = \frac{1}{x}$ onto $y = g(x)$.
(c) Write down the equations of the asymptotes of $y = g(x)$.

View Full Solution

Step 1: Simplify $g(x)$:
$$g(x) = \frac{6}{2(x - 4)} + 5 = \frac{3}{x - 4} + 5$$

Step 2: Describe transformation sequence:
From $y = \frac{1}{x}$ to $y = \frac{3}{x - 4} + 5$:
1. Vertical stretch with scale factor $3$ parallel to the $y$-axis (or horizontal compression by $\frac{1}{3}$).
2. Translation by vector $\begin{pmatrix} 4 \\ 5 \end{pmatrix}$ (4 units right, 5 units up).

Step 3: Asymptotes:
Original asymptotes of $f(x) = \frac{1}{x}$ are $x = 0$ and $y = 0$.
Under translation $\begin{pmatrix} 4 \\ 5 \end{pmatrix}$: $$\text{Vertical Asymptote: } x = 4, \quad \text{Horizontal Asymptote: } y = 5$$

Final Answers: (a) $g(x) = \frac{3}{x - 4} + 5$; (b) Vertical stretch by scale factor 3, translation by vector $\begin{pmatrix} 4 \\ 5 \end{pmatrix}$; (c) $x = 4$ and $y = 5$.

Exercise 7 Reversing Transformations

The function $h(x) = 5\ln(x - 2) + 7$ is obtained from $f(x) = \ln x$ by a sequence of three transformations.
(a) State the three transformations mapping $f(x)$ onto $h(x)$.
(b) State the exact sequence of transformations required to map $h(x)$ back onto $f(x)$.

View Full Solution

Step 1: Forward transformations $f(x) \to h(x)$:
1. Horizontal translation 2 units right (vector $\begin{pmatrix} 2 \\ 0 \end{pmatrix}$).
2. Vertical stretch with scale factor $5$ parallel to the $y$-axis.
3. Vertical translation 7 units up (vector $\begin{pmatrix} 0 \\ 7 \end{pmatrix}$).

Step 2: Reverse transformations $h(x) \to f(x)$:
Reverse operations in the exact reverse order:
1. Vertical translation 7 units down (vector $\begin{pmatrix} 0 \\ -7 \end{pmatrix}$).
2. Vertical stretch with scale factor $\frac{1}{5}$ parallel to the $y$-axis.
3. Horizontal translation 2 units left (vector $\begin{pmatrix} -2 \\ 0 \end{pmatrix}$).

Final Answers: (a) Shift 2 right, vertical stretch factor 5, shift 7 up; (b) Shift 7 down, vertical stretch factor $\frac{1}{5}$, shift 2 left.

Exercise 8 Determining Parameters from Features

The curve $y = f(x) = a (x - h)^2 + k$ has its vertex at $(3, -8)$ and passes through $(1, 0)$.
(a) Determine the values of $a$, $h$, and $k$.
(b) The curve is transformed to $y = g(x)$ by a reflection in the $x$-axis followed by a translation by $\begin{pmatrix} -1 \\ 4 \end{pmatrix}$. Find the coordinates of the new vertex.

View Full Solution

Step 1: Find $a, h, k$:
Vertex is $(h, k) = (3, -8) \implies h = 3, k = -8$.
$$f(x) = a(x - 3)^2 - 8$$ Passes through $(1, 0)$: $$0 = a(1 - 3)^2 - 8 \implies 0 = 4a - 8 \implies a = 2$$

Step 2: Transform vertex $(3, -8)$:
1. Reflection in $x$-axis: $(3, -8) \to (3, -(-8)) = (3, 8)$.
2. Translation by $\begin{pmatrix} -1 \\ 4 \end{pmatrix}$: $$(3 + (-1), 8 + 4) = (2, 12)$$

Final Answers: (a) $a = 2$, $h = 3$, $k = -8$; (b) New vertex is $(2, 12)$.

Tier 3: Synthesis & Problem-Solving (Problems 9–10)

Exercise 9 Domain & Range Under Multi-Step Transforms

A continuous function $f$ has domain $[-4, 8]$ and range $[-3, 11]$. The function $p$ is defined by $p(x) = -4 f\left(\frac{1}{2}x - 2\right) + 5$.
(a) Determine the domain of $p$.
(b) Determine the range of $p$.
(c) If $f(2) = -3$, find the coordinates of the corresponding point on the graph of $p$.

View Full Solution

Step 1: Domain of $p$:
The input to $f$ must lie in $[-4, 8]$: $$-4 \le \frac{1}{2}x - 2 \le 8$$ Add 2 to all sides: $$-2 \le \frac{1}{2}x \le 10$$ Multiply by 2: $$-4 \le x \le 20 \implies \text{Domain: } [-4, 20]$$

Step 2: Range of $p$:
Since $-3 \le f(u) \le 11$:
Multiply by $-4$ (reverses inequality signs): $$-4(11) \le -4 f(u) \le -4(-3) \implies -44 \le -4 f(u) \le 12$$ Add 5: $$-44 + 5 \le p(x) \le 12 + 5 \implies -39 \le p(x) \le 17$$ $$\text{Range: } [-39, 17]$$

Step 3: Point mapping for $(2, -3)$:
Set $\frac{1}{2}x - 2 = 2 \implies \frac{1}{2}x = 4 \implies x = 8$.
$y$-value: $p(8) = -4(-3) + 5 = 12 + 5 = 17$.
The point is $(8, 17)$.

Final Answers: (a) Domain: $[-4, 20]$; (b) Range: $[-39, 17]$; (c) $(8, 17)$.

Exercise 10 Trigonometric Transformation Parameter Synthesis

The graph of $g(x) = a \cos(b(x - c)) + d$ for $x \in [0, 2\pi]$ has a maximum at $\left(\frac{\pi}{6}, 9\right)$ and the very next minimum at $\left(\frac{2\pi}{3}, 1\right)$, with $a > 0$ and $b > 0$.
(a) Find the values of $a$ and $d$.
(b) Find the period of $g$ and hence determine the value of $b$.
(c) Find the smallest positive value of $c$.
(d) Fully describe the geometric transformations mapping $y = \cos x$ onto $y = g(x)$.

View Full Solution

Step 1: Calculate amplitude $a$ and midline $d$:
$$\text{Max} = 9, \quad \text{Min} = 1$$ $$a = \frac{\text{Max} - \text{Min}}{2} = \frac{9 - 1}{2} = 4$$ $$d = \frac{\text{Max} + \text{Min}}{2} = \frac{9 + 1}{2} = 5$$

Step 2: Calculate period and $b$:
The distance between a consecutive maximum and minimum is half a period: $$\frac{T}{2} = \frac{2\pi}{3} - \frac{\pi}{6} = \frac{4\pi - \pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2}$$ $$T = \pi$$ Since $T = \frac{2\pi}{b} \implies b = \frac{2\pi}{\pi} = 2$.

Step 3: Find horizontal shift $c$:
A standard unshifted cosine curve has its first maximum at $x = 0$. Here, the maximum is at $x = \frac{\pi}{6}$, so $c = \frac{\pi}{6}$.

Step 4: Geometric description:
1. Horizontal stretch with scale factor $\frac{1}{2}$ parallel to the $x$-axis.
2. Horizontal translation by $\frac{\pi}{6}$ units right (vector $\begin{pmatrix} \pi/6 \\ 0 \end{pmatrix}$).
3. Vertical stretch with scale factor $4$ parallel to the $y$-axis.
4. Vertical translation by $5$ units up (vector $\begin{pmatrix} 0 \\ 5 \end{pmatrix}$).

Final Answers: (a) $a = 4$, $d = 5$; (b) Period is $\pi$, $b = 2$; (c) $c = \frac{\pi}{6}$; (d) Horizontal stretch factor $\frac{1}{2}$, translation right by $\frac{\pi}{6}$, vertical stretch factor $4$, translation up by $5$.

Transformation Reference & Revision Sheet

In the IB Formula Booklet (Topic 2.11)

The official IB Mathematics AA SL formula booklet provides the following transformation summaries:

Vertical Translation $$y = f(x) + b \quad \longleftrightarrow \quad \text{Translation by } \begin{pmatrix} 0 \\ b \end{pmatrix}$$
Horizontal Translation $$y = f(x - a) \quad \longleftrightarrow \quad \text{Translation by } \begin{pmatrix} a \\ 0 \end{pmatrix}$$
Vertical Stretch $$y = a f(x) \quad \longleftrightarrow \quad \text{Scale factor } a, \text{ parallel to } y\text{-axis}$$
Horizontal Stretch $$y = f(a x) \quad \longleftrightarrow \quad \text{Scale factor } \frac{1}{a}, \text{ parallel to } x\text{-axis}$$
Reflection in $x$-axis $$y = -f(x) \quad \longleftrightarrow \quad (x, y) \mapsto (x, -y)$$
Reflection in $y$-axis $$y = f(-x) \quad \longleftrightarrow \quad (x, y) \mapsto (-x, y)$$
Must Memorize & Core Structural Insights
  • The "Inside is Opposite" Principle: Operations inside the argument $f(\dots)$ act counter-intuitively on $x$:
    • $f(x - h)$ shifts right by $h$.
    • $f(x + h)$ shifts left by $h$.
    • $f(b x)$ compresses horizontally by scale factor $\frac{1}{b}$ (not $b$).
  • Horizontal Factoring Rule: In an expression like $f(ax + b)$, you must factor out $a$: $$f(ax + b) = f\left(a\left(x + \frac{b}{a}\right)\right)$$ The horizontal translation is $\frac{b}{a}$ units left, NOT $b$ units left!
  • Point Mapping Matrix: $$y = A \cdot f(B(x - H)) + K \implies (x, y) \longmapsto \left(\frac{x}{B} + H, \; A y + K\right)$$
  • Asymptote Transformation Rules:
    • Vertical asymptotes $x = c$ are affected only by horizontal operations: $x_{new} = \frac{c}{B} + H$.
    • Horizontal asymptotes $y = d$ are affected only by vertical operations: $y_{new} = A d + K$.
Critical IB Examination Pitfalls
  • Pitfall 1: Confusing Horizontal Stretch Scale Factor: Stating that $y = f(3x)$ is a stretch of scale factor $3$. It is a scale factor of $\mathbf{\frac{1}{3}}$. An answer stating "scale factor 3" loses both the method and accuracy marks.
  • Pitfall 2: Forgetting to Factor Out Coefficient of $x$: When describing $y = f(2x - 6)$, writing "shift right 6 then stretch by $\frac{1}{2}$". The shift is $3$ units right because $2x - 6 = 2(x - 3)$.
  • Pitfall 3: Reversing the Coordinate Axes for Reflections: Mixing up $y = -f(x)$ (reflect in $x$-axis, line $y=0$) and $y = f(-x)$ (reflect in $y$-axis, line $x=0$). Mnemonic: $-f(x)$ negates $y$, so points jump across the $x$-axis.
  • Pitfall 4: Violating Vertical Order of Operations: For $y = 2f(x) + 3$, applying the translation first yields $2(f(x) + 3) = 2f(x) + 6$, which is incorrect. Stretch precedes translation when unparenthesized.

Section A: Diagnostic Multiple Choice

Answer the questions to check your understanding and score.

Score: 0 / 4

Question 1: The curve $y = f(x)$ is transformed to $y = f(x + 4)$. Which single geometric transformation has occurred?

Question 2: What is the scale factor of the stretch that maps $y = f(x)$ onto $y = f(5x)$?

Question 3: The point $P(6, -2)$ lies on $y = f(x)$. What are the coordinates of the image of $P$ on $y = -3 f(2x) + 4$?

Question 4: To transform $y = f(x)$ to $y = f(3x - 12)$ by a stretch followed by a translation, the horizontal shift must be:

Section B: Exam-Style Free-Response Questions

Authentic IB style questions with detailed markschemes showing method ($M$), accuracy ($A$), and reasoning ($R$) marks.

Question 5 (Paper 1 Style) 6 Marks

The diagram shows the graph of a function $f$ with domain $[-3, 5]$. The curve has a local minimum at $A(-2, -4)$, a $y$-intercept at $(0, 2)$, and an endpoint at $B(5, 7)$.
The graph of $g$ is obtained from $f$ by the transformation $g(x) = -2 f(x - 1) + 3$.
(a) Find the coordinates of $A'$, the image of $A$ under this transformation. [2]
(b) State the domain and range of $g$. [2]
(c) Find the $y$-intercept of $y = g(x)$. [2]

View Full Markscheme

(a) Coordinates of $A'$: [2 marks]
Apply horizontal shift: $x' = -2 + 1 = -1$ (M1)
Apply vertical operations: $y' = -2(-4) + 3 = 8 + 3 = 11$ (A1)
$$A'(-1, 11)$$

(b) Domain and Range of $g$: [2 marks]
Domain shifted right by 1: $[-3 + 1, 5 + 1] = [-2, 6]$ (A1)
Original range is $[-4, 7]$ (min at $-4$, max endpoint at $7$).
Transformed range: Multiply by $-2$ and add 3: $$-2(7) + 3 \le y \le -2(-4) + 3 \implies -14 + 3 \le y \le 8 + 3 \implies [-11, 11] \quad (A1)$$

(c) $y$-intercept of $g$: [2 marks]
The $y$-intercept occurs where $x = 0$: $$g(0) = -2 f(0 - 1) + 3 = -2 f(-1) + 3 \quad (M1)$$ If given that $f(-1) = -1$ (or evaluating from coordinates), $g(0) = -2(-1) + 3 = 2 + 3 = 5$ (A1).

Question 6 (Paper 2 Style) 8 Marks

Let $f(x) = \frac{4x - 2}{x + 1}$ for $x \neq -1$.
(a) Show that $f(x)$ can be written in the form $4 - \frac{6}{x + 1}$. [2]
(b) Hence describe a sequence of three geometric transformations that maps the parent curve $y = \frac{1}{x}$ onto $y = f(x)$. [3]
(c) The line $L$ with equation $y = mx + c$ is the image of the vertical asymptote of $y = \frac{1}{x}$ under this sequence. Write down the equation of $L$. [1]
(d) Find the coordinates of the intersection of $y = f(x)$ with its inverse function $y = f^{-1}(x)$. [2]

View Full Markscheme

(a) Algebraic decomposition: [2 marks]
$$f(x) = \frac{4(x + 1) - 6}{x + 1} = \frac{4(x + 1)}{x + 1} - \frac{6}{x + 1} = 4 - \frac{6}{x + 1} \quad (M1)(A1)$$

(b) Sequence of three transformations: [3 marks]
Starting from $y = \frac{1}{x}$:
1. Vertical stretch with scale factor $6$ parallel to the $y$-axis (or horizontal compression) (A1).
2. Reflection in the $x$-axis (since the term is negative, $-\frac{6}{x+1}$) (A1).
3. Translation by vector $\begin{pmatrix} -1 \\ 4 \end{pmatrix}$ (1 unit left and 4 units up) (A1).

(c) Transformed vertical asymptote: [1 mark]
Vertical asymptote of $\frac{1}{x}$ is $x = 0$. Translated left by 1: $$x = -1 \quad (A1)$$

(d) Intersection with $y = f^{-1}(x)$: [2 marks]
A function intersects its inverse on the identity line $y = x$ (for continuous monotonic branches): $$x = \frac{4x - 2}{x + 1} \implies x(x + 1) = 4x - 2 \implies x^2 + x = 4x - 2 \quad (M1)$$ $$x^2 - 3x + 2 = 0 \implies (x - 1)(x - 2) = 0$$ $$x = 1 \implies y = 1 \implies (1, 1)$$ $$x = 2 \implies y = 2 \implies (2, 2) \quad (A1)$$