§-Syllabus dot point

TASSpecialist MathematicsSyllabus dot point

How do translations, dilations and reflections build complex graphs from simple ones?

Sketch graphs using transformations and the addition of ordinates.

Translations, dilations and reflections of graphs, the order of transformations, and addition of ordinates, with worked examples for TCE Mathematics Specialised Unit 3.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

What this dot point is asking

This dot point asks you to sketch unfamiliar graphs by recognising a simple parent function and tracking the sequence of transformations applied to it, rather than plotting points. It also covers the addition of ordinates, a graphical method for adding two functions.

The general transformed form

For a parent $y = f(x)$ , the transformed graph

y = a\,f\big(b(x - h)\big) + k

is read as follows.

Order of transformations

When more than one transformation acts, order can matter. A safe approach is to deal with horizontal changes (inside $f$ ) by factorising so the coefficient of $x$ is clearly separated, and to apply dilations before translations when reading the standard form. The reliable check is to track one or two key points (such as an intercept or a turning point) through each step.

Transforming a square root graph

Worked example

Sketch $y = -2\sqrt{x - 1} + 3$ , starting from $y = \sqrt{x}$ .

Begin with the parent $y = \sqrt{x}$ , which starts at the origin and rises to the right. Apply the transformations in turn.

Inside: $x - 1$ translates the graph right by $1$ , so the endpoint moves to $(1, 0)$ .
Outside multiplier $-2$ : dilate vertically by factor $2$ and reflect in the $x$ axis, so the curve now falls to the right.
Outside $+3$ : translate up by $3$ , moving the endpoint to $(1, 3)$ .

The final graph starts at $(1, 3)$ and curves downward to the right, for example passing through $(2, 3 - 2) = (2, 1)$ . The domain is $x \ge 1$ and the range is $y \le 3$ .

Addition of ordinates

To sketch $y = f(x) + g(x)$ graphically, sketch $f$ and $g$ on the same axes, then at chosen $x$ values add the two heights (ordinates). Key anchor values to use: where either function is zero (the sum equals the other function there), and where the two graphs cross.

Addition of ordinates

Worked example

Sketch $y = x + \sin x$ for $0 \le x \le 2\pi$ using addition of ordinates.

Sketch the line $y = x$ and the curve $y = \sin x$ on the same axes. At each $x$ , add the height of $\sin x$ to the height of the line.

At $x = 0$ , $\sin x = 0$ , so the sum sits on the line at $(0, 0)$ .
At $x = \tfrac{\pi}{2}$ , $\sin x = 1$ , so the sum is one unit above the line.
At $x = \pi$ , $\sin x = 0$ , so the sum is back on the line at $(\pi, \pi)$ .
At $x = \tfrac{3\pi}{2}$ , $\sin x = -1$ , so the sum is one unit below the line.

The result is a curve that oscillates gently above and below the line $y = x$ , rising overall because the line dominates. This is the classic shape of a steadily increasing quantity with a periodic ripple.

Working back to the rule

Examiners sometimes give a transformed graph and ask for its equation. Identify the parent shape, locate the new position of a recognisable feature to read off $h$ and $k$ , then use a second point to solve for the dilation factor $a$ . Confirm any reflection by checking whether the graph opens or falls the opposite way to the parent.

Why this matters

Transformation thinking lets you sketch any graph that is a reshaped version of a known parent, which covers most functions in the exam. Combined with the reciprocal, absolute value and rational techniques, it means you rarely need to plot points by hand.

Exam-style practice questions

Practice questions written in the style of TASC exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

TCE 20234 marksThe graph of

y = f(x)

has a maximum turning point at

(2, 5)

. State the coordinates of the corresponding turning point on the graph of

y = -3f(x - 1) + 2

, and state whether it is a maximum or a minimum.

Show worked answer →

Track the point $(2, 5)$ through each transformation in order.

The inside change $x - 1$ translates right by $1$ , moving the $x$ -coordinate to $2 + 1 = 3$ . The factor $-3$ multiplies the $y$ -coordinate by $-3$ and reflects in the $x$ -axis, sending $5$ to $-3 \times 5 = -15$ . The $+2$ translates up by $2$ , giving $y = -15 + 2 = -13$ .

So the image point is $(3, -13)$ . Because the multiplier $-3$ is negative, the reflection turns the original maximum into a minimum. The corresponding turning point is a minimum at $(3, -13)$ . Markers reward applying the inside translation to $x$ , the outside scaling and reflection to $y$ , and identifying the maximum-to-minimum swap.

TCE 20245 marksDescribe the sequence of transformations that maps

y = x^2

y = 2(x + 3)^2 - 4

, and hence state the vertex and the equation of the axis of symmetry of the transformed graph.

Show worked answer →

Write the target in the standard form $y = a(x - h)^2 + k$ with $a = 2$ , $h = -3$ , $k = -4$ .

The transformations, applied in order, are: a horizontal translation left by $3$ (from the $x + 3$ , since inside changes act in reverse), a vertical dilation by factor $2$ (from $a = 2$ ), and a vertical translation down by $4$ (from $k = -4$ ). There is no reflection because $a > 0$ .

The vertex is at $(h, k) = (-3, -4)$ , and the axis of symmetry is the vertical line $x = -3$ . Markers reward reading $h$ and $k$ from the completed-square form, the reversed horizontal direction, and the vertex and axis of symmetry. A common slip is reading $x + 3$ as a shift right.