Sketching graphs by transformation - TCE Mathematics Specialised (Tasmania)

Syllabus dot point

“Sketch graphs using transformations and the addition of ordinates.”

TCESpecialist MathematicsUnit 38 min readUpdated 2 Jun 2026

Quick answer

Build a graph from a known parent by applying transformations: $y = af(b(x - h)) + k$ dilates vertically by $a$ , dilates horizontally by $\tfrac1b$ , translates right by $h$ and up by $k$ . Changes outside $f$ act on $y$ directly; changes inside $f$ act on $x$ in the opposite way. Addition of ordinates builds $f(x) + g(x)$ by adding heights.

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.

Outside acts on y, inside acts on x in reverse

Transformations applied outside the function $f$ affect the output $y$ in the natural way: multiply by $a$ to dilate vertically, add $k$ to translate vertically. Transformations applied inside $f$ affect the input $x$ in the opposite way: dividing $x$ by $b$ dilates horizontally by factor $\tfrac1b$ , and subtracting $h$ translates right by $h$ . A negative $a$ reflects in the $x$ axis; a negative $b$ reflects in the $y$ axis.

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.

Reversing the direction of a horizontal change

The most common error is reading $f(x - 3)$ as a shift left by $3$ . Inside changes act in the opposite sense, so $f(x - 3)$ shifts the graph right by $3$ , and $f(2x)$ compresses horizontally by factor $\tfrac12$ rather than stretching. Always factor the inside to the form $f(b(x - h))$ and remember that the input direction is reversed before stating the transformation.

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 questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks

The 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.

Practice5 marks

Describe 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.

Practise this

Sources & how we know this

Mathematics Specialised Level 4 (MTS415118) course document (2018 onwards) — TASC