NSWMaths AdvancedSyllabus dot point

What inputs is a function allowed to take, what outputs does it actually reach, and how do you read both straight off a graph, including a restricted piece and a graph read backwards with horizontal lines?

Find the natural domain of a function (avoiding division by zero and square roots of negatives), determine the range from a sketch, work with restricted domains, and read the range from a graph using horizontal lines

A Year 11 Maths Advanced answer on domain and range: finding the natural domain by avoiding division by zero and square roots of negatives, reading the range from a sketch, restricted domains, and reading a graph backwards with horizontal lines, with worked parabola and semicircle examples and practice questions.

Generated by Claude Opus 4.820 min answerUpdated 2026-06-21

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Quick answer

The domain of a function is its set of allowed inputs ( $x$ ); the range is its set of outputs ( $y$ ). Find the domain by algebra and the range from a sketch. With no restriction stated, the natural domain is everything you can substitute, ruled by two facts: a denominator must be non-zero and a radicand (under a square root) must be non-negative; a polynomial has domain all real $x$ . When the same expression is under a root and in a denominator, as in $\dfrac{1}{\sqrt{x - 2}}$ , the boundary is excluded and the domain is strict, $x > 2$ . Read the range from the graph by finding the highest and lowest points the curve reaches, or read it backwards: a height $y = b$ is in the range exactly when the horizontal line $y = b$ meets the curve. For a parabola, complete the square to get the vertex $(h, k)$ and write $y \ge k$ (opens up) or $y \le k$ (opens down). A semicircle $y = \sqrt{a^2 - x^2}$ has domain $-a \le x \le a$ and range $0 \le y \le a$ . On a restricted domain the range comes from the relevant piece of the curve, so check whether the vertex is inside the interval before quoting its value; if not, use the endpoints. Always state the range as a $y$ -inequality.

What this dot point is asking

Every function comes with two sets of numbers attached to it. The domain is the set of all inputs $x$ you are allowed to feed in; the range is the set of all outputs $y$ that actually come out. This dot point is about finding both. There are two distinct skills, and students who blur them lose easy marks.

Finding the domain is an algebra question: given a rule, which $x$ -values can you legally substitute? Two operations are forbidden in this course, dividing by zero and taking the square root of a negative, so the natural domain is whatever is left once you rule those out. Finding the range is usually a graph question: once you have a sketch, the range is the spread of heights the curve covers, which is easiest to read after drawing. On top of that, NESA expects you to handle a restricted domain (a function deliberately cut down to part of the line, where the range can change) and to read a graph backwards using horizontal lines to decide which outputs are reached. The whole topic is short, but it underpins inverse functions, every graph you will sketch, and the language of "for what values" that recurs across the paper.

The answer

Domain and range: inputs and outputs

The domain of a function is the set of all possible $x$ -coordinates (the inputs); the range is the set of all possible $y$ -coordinates (the outputs). Picture a graph and project it onto the axes: the shadow on the $x$ -axis is the domain, and the shadow on the $y$ -axis is the range. Anywhere the curve sits above or below a point on the $x$ -axis, that $x$ is in the domain; anywhere the curve sits left or right of a point on the $y$ -axis, that $y$ is in the range.

The graph below shows the curve $y = x^2 - 4x + 7$ drawn only on $0 \le x \le 5$ . Its domain is shaded as a band along the $x$ -axis and its range as a band up the $y$ -axis. Reading the picture: the inputs run across $0 \le x \le 5$ , and the outputs run from the lowest point at the vertex up to the highest endpoint, $3 \le y \le 12$ .

The natural domain: the two forbidden operations

When a function is given by a rule with no domain stated, the convention is that the domain is every $x$ you can legally substitute. This is the natural domain. In Year 11 Advanced there are exactly two operations that fail for some inputs, and almost every domain question is one or both of them:

You cannot divide by zero. So any denominator must be non-zero: set it $\ne 0$ and exclude the values that make it zero.
You cannot take the square root of a negative number. So any expression under a square root (the radicand) must be non-negative: set it $\ge 0$ and solve.

For a polynomial such as $y = x^3 - 2x + 1$ there is no denominator and no root, so nothing is forbidden and the natural domain is all real $x$ . The domain only shrinks when a denominator or a root appears.

A subtle but heavily examined case is when the root and the denominator are the same expression. In $f(x) = \dfrac{1}{\sqrt{x - 2}}$ the quantity $x - 2$ is under a root and in the denominator, so it must be both $\ge 0$ (root) and $\ne 0$ (denominator). Together these force $x - 2 > 0$ , a strict inequality, so the domain is $x > 2$ , not $x \ge 2$ . The boundary $x = 2$ is excluded because it would make the denominator zero. The graph below decays away from a vertical asymptote at $x = 2$ , and the open circle on the $x$ -axis marks that $2$ itself is not allowed.

This open-circle, included-versus-excluded distinction is the same one you met with intervals on the number line: a closed (filled) circle means the endpoint is in the set, an open (hollow) circle means it is not. We write intervals here in inequality form such as $x > 2$ or $-3 \le x \le 3$ , not as coordinate-style brackets.

Reading the range off a sketch, and reading it backwards

Once a curve is drawn, the range is the set of heights it covers. The reliable way to find it is to look for the highest and lowest points the curve reaches, since the range is everything between (and including, if reached) those extremes. For a parabola this is just the vertex on one side, with the curve running off to infinity on the other; for a bounded curve like a semicircle it is the top and the ends.

There is a second, sharper way to think about the range that pays off later: read the graph backwards with horizontal lines. To test whether a particular output $y = b$ is in the range, draw the horizontal line $y = b$ and see whether it meets the curve. If it does, that height is reached, so $b$ is in the range; if the line misses the curve entirely, $b$ is not an output. Sweeping the horizontal line up and down, the range is exactly the set of heights at which it touches the curve.

The figure below applies this to $y = x^2 - 4x + 7$ . The line $y = 7$ meets the curve (twice, in fact), so $7$ is in the range. The line $y = 2$ passes underneath the curve and never touches it, so $2$ is not in the range. The lowest the curve goes is the vertex at $y = 3$ , so the range is everything from $3$ upward, $y \ge 3$ .

The range of a parabola from its vertex

A parabola is the most common range question, and you do not need to plot it point by point. Put the quadratic into completed-square (vertex) form $y = (x - h)^2 + k$ , read off the vertex $(h, k)$ , and use the direction it opens:

If it opens upward (positive coefficient of $x^2$ ), the vertex is the minimum, and the range is $y \ge k$ .
If it opens downward (negative coefficient), the vertex is the maximum, and the range is $y \le k$ .

For $y = x^2 - 4x + 7$ , completing the square gives $(x - 2)^2 + 3$ , so the vertex is $(2, 3)$ and, opening upward, the range over the whole real line is $y \ge 3$ . The domain, with no denominator or root, is all real $x$ . The vertex does the work: its $y$ -coordinate is the boundary of the range.

The range of a semicircle

A semicircle is the standard bounded example, and it is the reason square-root functions matter here. The full circle $x^2 + y^2 = a^2$ is not a function (a vertical line cuts it twice), but solving for $y$ splits it into two halves that are functions: $y = \sqrt{a^2 - x^2}$ is the upper semicircle (outputs $\ge 0$ ), and $y = -\sqrt{a^2 - x^2}$ is the lower one (outputs $\le 0$ ).

Take $y = \sqrt{25 - x^2}$ , the upper semicircle of radius $5$ . For the domain, the radicand must be non-negative: $25 - x^2 \ge 0$ gives $x^2 \le 25$ , so $-5 \le x \le 5$ . For the range, read the shape: the curve climbs from $y = 0$ at the ends to a highest point $y = 5$ at the top $(0, 5)$ , and never goes negative, so $0 \le y \le 5$ . The figure shows both bands.

The lower semicircle $y = -\sqrt{25 - x^2}$ has the same domain $-5 \le x \le 5$ (the radicand is unchanged) but the range flips to $-5 \le y \le 0$ , since every output is the negative of the upper one. Changing the sign of the whole function reflects it in the $x$ -axis and flips the range, but leaves the domain alone.

Restricted domains

A function can be deliberately given on only part of the line, for instance " $y = x^2 - 6x + 5$ for $0 \le x \le 2$ ". This is a restricted domain: the rule is the same, but you only use the inputs in the stated interval, and the range is read from that piece of the curve, not the whole parabola. Restricted domains appear naturally in real problems (a height only makes sense while an object is in the air; a length must be positive) and are the key idea behind inverting a parabola later.

The trap is to quote the vertex value out of habit. Always check whether the vertex lies inside the restricted domain. If it does, it is the maximum or minimum as usual. If it does not, the parabola is steadily increasing or decreasing across the whole piece, so the range runs between the two endpoint values instead. For $y = x^2 - 6x + 5 = (x - 3)^2 - 4$ on $0 \le x \le 2$ , the vertex is at $x = 3$ , outside the interval, so the curve is decreasing across $[0, 2]$ : the range is from $f(2) = -3$ up to $f(0) = 5$ , that is $-3 \le y \le 5$ , and the vertex value $-4$ never appears.

How exam questions ask about domain and range

The wording is a reliable cue to which skill is wanted:

"Find the natural domain" or "state the largest possible domain" means apply the two rules: set every denominator $\ne 0$ and every radicand $\ge 0$ , solve each, and combine. No graph needed.
"For what values of $x$ is the function defined?" is the same domain question in words.
"Write down the range" of a sketched graph means read the heights off the picture, from lowest to highest. State it as an inequality such as $y \ge 3$ or $-5 \le y \le 5$ .
"State the domain and range" of a circle or semicircle wants both: the domain from the root condition (or the $x$ -extent of the circle), the range from the shape.
A restricted domain in the question ("for $0 \le x \le 2$ ") is a flag to check whether the vertex is inside the interval before quoting it, and otherwise to use the endpoints.
A real context (height of a projectile, area of a rectangle, cost over a number of items) usually restricts the domain to sensible values (time $\ge 0$ , length $> 0$ ), so state that restriction as part of the answer.
"Is $y = b$ a possible value?" is reading the range backwards: test whether the horizontal line $y = b$ meets the curve.

Worked example

Five examples: the natural domain of a function with a single root in the denominator, the range of a parabola from its vertex, the domain and range of an upper semicircle, the natural domain of a function with both a root and a denominator, and the range of a parabola on a restricted domain.

Find the natural domain of $f(x) = \dfrac{1}{\sqrt{x - 2}}$

The expression $x - 2$ is under a square root and in a denominator, so it must satisfy both conditions.

Apply the root condition: The radicand must be non-negative: $x - 2 \ge 0$ , i.e. $x \ge 2$ .
Apply the denominator condition: The denominator cannot be zero, and $\sqrt{x - 2} = 0$ when $x - 2 = 0$ , so $x \ne 2$ .
Combine the two: Both must hold: $x \ge 2$ and $x \ne 2$ together force the strict inequality

x > 2.

State the answer. The natural domain is $x > 2$ . The boundary $x = 2$ is excluded because it makes the denominator zero, even though the root alone would allow it. This is why the curve has a vertical asymptote at $x = 2$ rather than touching down there.

Find the range of $y = x^2 - 4x + 7$

Use the vertex; do not plot points.

Complete the square. Halve the coefficient of $x$ (which is $-4$ ) to get $-2$ , then:

y = x^2 - 4x + 7 = (x - 2)^2 - 4 + 7 = (x - 2)^2 + 3.

Read the vertex and direction. The vertex is $(2, 3)$ , and the coefficient of the squared bracket is $+1 > 0$ , so the parabola opens upward and the vertex is the minimum.

Find the range. The lowest output is $y = 3$ (at $x = 2$ ) and the curve rises without bound, so

\text{range: } y \ge 3.

State the domain too. There is no denominator or root, so the natural domain is all real $x$ . Answer: domain all real $x$ , range $y \ge 3$ . Check the vertex value: $f(2) = 4 - 8 + 7 = 3$ , matching $k = 3$ .

Find the domain and range of the semicircle $y = \sqrt{25 - x^2}$

This is the upper half of the circle $x^2 + y^2 = 25$ .

Find the domain from the root. The radicand must be non-negative:

25 - x^2 \ge 0 \;\Rightarrow\; x^2 \le 25 \;\Rightarrow\; -5 \le x \le 5.

Find the range from the shape. Because of the $+$ square root, every output is $\ge 0$ . The curve is highest at the top of the semicircle, $(0, 5)$ , and lowest at the two ends on the $x$ -axis where $y = 0$ . So the outputs run from $0$ to $5$ :

\text{range: } 0 \le y \le 5.

Check key points. At $x = 0$ , $y = \sqrt{25} = 5$ (the top); at $x = 3$ , $y = \sqrt{25 - 9} = \sqrt{16} = 4$ ; at $x = 5$ , $y = \sqrt{0} = 0$ (an end). All lie on the curve, confirming domain $-5 \le x \le 5$ and range $0 \le y \le 5$ .

Find the natural domain of $f(x) = \dfrac{\sqrt{x + 4}}{x - 1}$

Here the root and the denominator are different expressions, so set up one condition for each.

Condition from the root. The radicand $x + 4$ must be non-negative:

x + 4 \ge 0 \;\Rightarrow\; x \ge -4.

Condition from the denominator. The denominator $x - 1$ cannot be zero:

x - 1 \ne 0 \;\Rightarrow\; x \ne 1.

Combine them. Both must hold at once. Start from $x \ge -4$ , then remove the single point $x = 1$ (which lies inside that set):

x \ge -4, \; x \ne 1.

State the answer. The natural domain is $x \ge -4$ with $x = 1$ excluded. Because the two conditions involve different expressions, they give two separate restrictions; the only overlap to watch is that $x = 1$ does sit inside $x \ge -4$ , so it genuinely must be punched out.

Find the range of $y = x^2 - 6x + 5$ on the restricted domain $0 \le x \le 2$

The rule is a parabola, but only the piece over $0 \le x \le 2$ is used.

Complete the square to find the vertex.

y = x^2 - 6x + 5 = (x - 3)^2 - 9 + 5 = (x - 3)^2 - 4,

so the vertex is $(3, -4)$ .

Check whether the vertex is in the domain. The domain is $0 \le x \le 2$ , but the vertex is at $x = 3$ , which is outside. So the turning point is never reached on this piece, and the endpoints control the range.

Determine the behaviour and evaluate the endpoints. Left of the vertex an upward parabola is decreasing, so on $[0, 2]$ the curve falls from $x = 0$ to $x = 2$ :

f(0) = 0 - 0 + 5 = 5, \qquad f(2) = 4 - 12 + 5 = -3.

State the range. The outputs run from $-3$ (at $x = 2$ ) up to $5$ (at $x = 0$ ):

\text{range: } -3 \le y \le 5.

The vertex value $-4$ does not appear, because $x = 3$ is not in the restricted domain.

Final answers: the natural domain of $\dfrac{1}{\sqrt{x - 2}}$ is $x > 2$ ; $y = x^2 - 4x + 7$ has domain all real $x$ and range $y \ge 3$ ; the semicircle $y = \sqrt{25 - x^2}$ has domain $-5 \le x \le 5$ and range $0 \le y \le 5$ ; the natural domain of $\dfrac{\sqrt{x + 4}}{x - 1}$ is $x \ge -4$ , $x \ne 1$ ; and $y = x^2 - 6x + 5$ on $0 \le x \le 2$ has range $-3 \le y \le 5$ .

Common traps

Confusing domain with range: Domain is the set of inputs ( $x$ , the $x$ -axis shadow); range is the set of outputs ( $y$ , the $y$ -axis shadow). Find the domain by algebra, the range from the graph. Swapping them is the most common slip.
Allowing the boundary when a denominator is involved: For $\dfrac{1}{\sqrt{x - 2}}$ the domain is $x > 2$ , not $x \ge 2$ : the value $x = 2$ makes the denominator zero. A root alone gives $\ge$ ; a root inside a denominator gives a strict $>$ .
Forgetting to exclude a point that lies inside the domain: In $\dfrac{\sqrt{x + 4}}{x - 1}$ the domain is $x \ge -4$ with $x \ne 1$ . Because $1$ sits inside $x \ge -4$ , it really must be removed; do not write just $x \ge -4$ .
Quoting the vertex value on a restricted domain without checking: If the vertex is outside the given interval, its $y$ -value is not in the range. For $y = x^2 - 6x + 5$ on $0 \le x \le 2$ , the range is $-3 \le y \le 5$ , not anything involving the vertex value $-4$ .
Treating the full circle as a function: $x^2 + y^2 = 25$ fails the vertical line test, so it is a relation, not a function. Only the half-circles $y = \pm\sqrt{25 - x^2}$ are functions; quote the range of the half you are given.
Giving the range as an $x$ -inequality: The range must be written in terms of $y$ (the outputs), for example $y \ge 3$ or $0 \le y \le 5$ , not in terms of $x$ .

Exam technique

Markers reward the correct condition set up for the domain, a range read cleanly from the sketch, and the inequality stated in the right variable. To bank the method marks:

Show each domain condition explicitly. Write " $x - 2 \ge 0$ " or " $x - 1 \ne 0$ " before solving. The condition line is where the method mark sits, especially when there are two of them.
Watch the strict-versus-inclusive boundary. A radicand gives $\ge 0$ ; a denominator gives $\ne 0$ ; the same expression under a root and in a denominator gives a strict inequality. State whether each endpoint is in or out.
Sketch first, then read the range. A quick sketch (vertex, ends, direction) makes the highest and lowest points obvious. State the range as a $y$ -inequality.
For a parabola, complete the square and use the vertex. Read $(h, k)$ , decide up or down, and write $y \ge k$ or $y \le k$ . On a restricted domain, check whether $x = h$ is inside the interval before using $k$ .
Use inequality form for intervals. Write $-5 \le x \le 5$ and $x > 2$ , the convention for this course, and mark endpoints as included or excluded.
In a worded problem, state the sensible domain. Time $\ge 0$ , length $> 0$ , "for the flight $0 \le t \le 6$ "; the restriction is part of the answer and often part of the marks.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksFind the natural domain of each function: (a)

g(x) = \dfrac{1}{x + 3}

, (b)

h(x) = \sqrt{x - 5}

Show worked solution →

State the rule for each. A denominator cannot be zero; a square root cannot take a negative number.

(a) $g(x) = \dfrac{1}{x + 3}$ . The only danger is dividing by zero, so set the denominator not equal to $0$ :

x + 3 \ne 0 \;\Rightarrow\; x \ne -3.

The natural domain is all real $x$ except $x = -3$ .

(b) $h(x) = \sqrt{x - 5}$ . The radicand must be non-negative:

x - 5 \ge 0 \;\Rightarrow\; x \ge 5.

The natural domain is $x \ge 5$ .

State the answers. (a) all real $x$ , $x \ne -3$ ; (b) $x \ge 5$ . Note the contrast: a root alone allows the boundary value ( $x = 5$ gives $\sqrt{0} = 0$ ), but a denominator alone excludes it.

foundation2 marksThe parabola

y = (x - 1)^2 - 4

has its turning point at

(1, -4)

and opens upward. State its range, and explain in one line why the domain is all real

x

Show worked solution →

Read the vertex. The form $y = (x - 1)^2 - 4$ is already completed-square, so the vertex is $(1, -4)$ and the parabola opens upward (the coefficient of the squared bracket is $+1$ ).

Find the range from the vertex. An upward parabola has its lowest point at the vertex, so the smallest output is $y = -4$ , reached at $x = 1$ , and the curve rises forever above it:

\text{range: } y \ge -4.

Check the lowest value. At $x = 1$ , $y = (1 - 1)^2 - 4 = -4$ , the minimum, confirming the range starts at $-4$ .

Domain. You can substitute any real number into $(x - 1)^2 - 4$ (no denominator, no root), so the natural domain is all real $x$ . The range is the restricted side; the domain is unrestricted.

core3 marksSketch the relationship

y = -\sqrt{9 - x^2}

in your head (a lower semicircle of radius

3

), and state its domain and range with reasons.

Show worked solution →

Identify the curve. Squaring gives $y^2 = 9 - x^2$ , i.e. $x^2 + y^2 = 9$ , a circle of radius $3$ centred at the origin. The minus sign in $y = -\sqrt{9 - x^2}$ keeps only the lower half (outputs are $\le 0$ ).

Find the domain (where the root is defined). The radicand must be non-negative:

9 - x^2 \ge 0 \;\Rightarrow\; x^2 \le 9 \;\Rightarrow\; -3 \le x \le 3.

So the domain is $-3 \le x \le 3$ .

Find the range from the shape. The lower semicircle dips to its lowest point at the bottom, $(0, -3)$ , and rises to $y = 0$ at the two ends $(\pm 3, 0)$ :

\text{range: } -3 \le y \le 0.

Check the endpoints. At $x = 0$ , $y = -\sqrt{9} = -3$ (the bottom); at $x = 3$ , $y = -\sqrt{0} = 0$ (an end). Both match, so the domain is $-3 \le x \le 3$ and the range is $-3 \le y \le 0$ .

core3 marksFind the natural domain of

f(x) = \dfrac{\sqrt{x + 4}}{x - 1}

. Set out both conditions and combine them.

Show worked solution →

List the two restrictions. This function has both a square root (in the numerator) and a denominator, so it must satisfy two conditions at once.

Condition 1, the root. The radicand must be non-negative:

x + 4 \ge 0 \;\Rightarrow\; x \ge -4.

Condition 2, the denominator. It cannot be zero:

x - 1 \ne 0 \;\Rightarrow\; x \ne 1.

Combine them. Both must hold, so take the overlap: start from $x \ge -4$ and then punch out the single point $x = 1$ :

x \ge -4 \text{ and } x \ne 1.

State the answer. The natural domain is $x \ge -4$ , $x \ne 1$ . The value $x = 1$ lies inside $x \ge -4$ , so it really must be removed; that one excluded point is the trap.

exam3 marksA function is defined by

y = x^2 - 6x + 5

on the restricted domain

0 \le x \le 2

. Find its range. (Hint: check whether the vertex lies inside the domain.)

Show worked solution →

Locate the vertex: Completing the square, $x^2 - 6x + 5 = (x - 3)^2 - 4$ , so the (unrestricted) vertex is $(3, -4)$ .
Check whether the vertex is in the domain: The domain is $0 \le x \le 2$ , but the vertex is at $x = 3$ , which is outside it. So the turning point is never reached, and the range is governed by the endpoints, not the vertex.
Decide the behaviour on $0 \le x \le 2$: To the left of the vertex ( $x = 3$ ) an upward parabola is still decreasing, so on $[0, 2]$ the function falls steadily from $x = 0$ to $x = 2$ . The largest output is at $x = 0$ and the smallest at $x = 2$ .
Evaluate the endpoints

f(0) = 0 - 0 + 5 = 5, \qquad f(2) = 4 - 12 + 5 = -3.

State the range. The outputs run from $-3$ up to $5$ , so the range is $-3 \le y \le 5$ . The trap is to use the vertex value $-4$ ; it is not in the range because $x = 3$ is not in the domain.

exam4 marksA drone takes off from the ground and its height above the ground after

t

seconds is

h = 30t - 5t^2

metres, for the flight

0 \le t \le 6

. (a) State the natural restriction on

t

given the context. (b) Find the greatest height reached. (c) State the range of

h

over the flight, and say what it means.

Show worked solution →

(a) The domain in context. Time cannot be negative and the flight lasts until the drone lands, so the sensible domain is the one given, $0 \le t \le 6$ (a restricted domain set by the real situation, not by any division or root).

(b) Greatest height. The height $h = 30t - 5t^2$ is a parabola opening downward (coefficient of $t^2$ is $-5$ ), so its maximum is at the vertex. The axis of symmetry is at

t = \frac{-30}{2(-5)} = 3,

which lies inside $0 \le t \le 6$ . The greatest height is

h(3) = 30(3) - 5(3)^2 = 90 - 45 = 45 \text{ metres}.

(c) Range over the flight. At the ends, $h(0) = 0$ and $h(6) = 30(6) - 5(6)^2 = 180 - 180 = 0$ , so the drone starts and finishes on the ground. Between them it rises to $45$ m, so the height takes every value from $0$ up to $45$ :

\text{range: } 0 \le h \le 45.

Interpret it. The drone's height ranges from $0$ metres (ground, at take-off and landing) up to a maximum of $45$ metres at $t = 3$ seconds. The vertex is in the domain this time, so its value $45$ is the top of the range.

What this dot point is asking

The answer

Domain and range: inputs and outputs

The natural domain: the two forbidden operations

Reading the range off a sketch, and reading it backwards

The range of a parabola from its vertex

The range of a semicircle

Restricted domains

How exam questions ask about domain and range

Practice questions

Related dot points