Functions and Notations
A function is a rule that gives exactly one output for each input, written with notation such as \(f(x)=x^2+4x\), so \(f(-3)\) is its value when \(x=-3\). The domain is the set of allowed inputs and the range is the set of outputs. In NSW Year 10 Advanced maths you also use the vertical line test and graph linear inequalities (NESA outcome MA5‑FNC‑P‑01).
You will decide which relations and graphs are functions, evaluate and simplify expressions such as \(f(a+1)\), solve \(f(x)=g(x)\), and find a linear function from two values. You will state domains and ranges for exponential, hyperbolic, square root and semicircle graphs, test which points lie in a region such as \(x+2y>4\) and shade it, and describe how \(y=f(x)\) moves under a transformation. It is part of Year 10 Advanced Non Linear Relationships Functions and Graphs.
Theory
A function pairs each input \(x\) with exactly one output, written \(f(x)\). Its domain is the set of allowed \(x\)-values and its range is the set of \(y\)-values it produces. This Year 10 Advanced maths page covers the vertical line test, evaluating and substituting with function notation, finding domain and range, transforming \(y = f(x)\), and shading the region for a linear inequality.
A relation is any set of ordered pairs \((x,\ y)\). A function is a relation in which every \(x\)-value has exactly one \(y\)-value. Two pairs may share a \(y\)-value, but never an \(x\)-value with different \(y\)-values.
On a graph, use the vertical line test: if any vertical line cuts the graph more than once, it is not a function. A circle and a sideways parabola fail; a parabola \(y = ax^2 + bx + c\) passes.
Function notation \(f(x)\) names the output for input \(x\). \(f(3)\) means the value of the function when \(x = 3\).
The domain is the set of all allowable \(x\)-values. The range is the set of \(y\)-values the function takes as \(x\) varies over the domain.
A linear inequality such as \(2x + y \lt 4\) describes a region of the plane, bounded by the line \(2x + y = 4\).
It comes up if you are heading for Advanced or Extension maths in Years 11 and 12.
Common restrictions on the domain:
| Function contains | Restriction | Example |
|---|---|---|
| a fraction | denominator \(\ne 0\) | \(\dfrac{1}{x + 5}\): \(x \ne -5\) |
| a square root | inside \(\ge 0\) | \(\sqrt{x - 7}\): \(x \ge 7\) |
| a polynomial or \(a^x\) | none | all real \(x\) |
Transformations of \(y = f(x)\):
| Equation | Effect | Point \((x,\ y)\) moves to |
|---|---|---|
| \(y = f(x) + c\) | \(c\) units up | \((x,\ y + c)\) |
| \(y = f(x - b)\) | \(b\) units right | \((x + b,\ y)\) |
| \(y = kf(x)\) | vertical dilation, factor \(k\) | \((x,\ ky)\) |
| \(y = f(ax)\) | horizontal dilation, factor \(\dfrac{1}{a}\) | \(\left(\dfrac{x}{a},\ y\right)\) |
How to work with a function
- Evaluate: to find \(f(c)\), replace every \(x\) with \((c)\) in brackets and simplify.
- Solve: to find \(x\) when \(f(x) = k\), set the rule equal to \(k\) and solve the equation.
- Domain: start with all real numbers and remove values that make a denominator zero or a square root negative.
- Range: sketch the graph, or reason from the rule (squares, roots and powers of a positive number), and read off the \(y\)-values reached.
Problem types
- Is it a function? For ordered pairs, look for a repeated \(x\)-value with different \(y\)-values. For a graph, use the vertical line test.
- Finding a rule: if \(f(x) = ax + b\) and two values are given, substitute both to get two equations, then solve simultaneously. The same works for \(f(x) = ax^2 + bx\).
- Expressions as inputs: \(f(a + 1)\), \(f(x + h) - f(x)\) or \(G(x + 1)\): substitute the whole expression in brackets, expand, and collect like terms. A fraction may simplify after substitution.
- Two functions: solve \(f(x) = g(x)\) to find where their graphs meet, then substitute back into either rule for the \(y\)-values.
- Semicircles: \(y = \sqrt{r^2 - x^2}\) is the top half and \(y = -\sqrt{r^2 - x^2}\) the bottom half of the circle of radius \(r\), centre \(O\).
- Points on transformed graphs: apply the change to each point (see the table), then check by substituting.
- Inequalities: find the boundary line's intercepts, choose solid or dashed, test a point (the origin, if it is not on the line), and shade.
(i) Find \(f(-3)\).
(ii) Find and simplify \(f(k - 1)\).
(iii) Solve \(f(x) = 5\).
(i) Replace every \(x\) with \((-3)\), brackets included:
| \(f(-3)\) | \(=\) | \(2(-3)^2 + 3(-3)\) |
| \(\) | \(=\) | \(18 - 9 = \mathbf{9}\) |
(ii) Replace every \(x\) with \((k - 1)\):
| \(f(k - 1)\) | \(=\) | \(2(k - 1)^2 + 3(k - 1)\) |
| \(\) | \(=\) | \(2k^2 - 4k + 2 + 3k - 3\) |
| \(\) | \(=\) | \(\boldsymbol{2k^2 - k - 1}\) |
(iii) This time the output is given. Set the rule equal to \(5\) and solve:
| \(2x^2 + 3x - 5\) | \(=\) | \(0\) |
| \((2x + 5)(x - 1)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\mathbf{1}\ \text{or}\ \mathbf{-\dfrac{5}{2}}\) |
Check: \(f(1) = 2 + 3 = 5\).
(i) \(f(x) = \sqrt{10 - 2x}\)
(ii) \(g(x) = \dfrac{4}{x + 1}\)
(iii) \(h(x) = -\sqrt{16 - x^2}\).
(i) The number under a square root cannot be negative:
| \(10 - 2x\) | \(\ge\) | \(0\) |
| \(x\) | \(\le\) | \(5\) |
Domain \(\boldsymbol{x \le 5}\). A square root is never negative, and it takes every value from \(0\) up: range \(\boldsymbol{y \ge 0}\).
(ii) The denominator cannot be zero, so \(x \ne -1\). A fraction with numerator \(4\) is never \(0\). Domain: all real \(x\), \(x \ne -1\). Range: all real \(y\), \(y \ne 0\).
(iii) Need \(16 - x^2 \ge 0\), so \(-4 \le x \le 4\). Squaring \(y = -\sqrt{16 - x^2}\) gives \(x^2 + y^2 = 16\) with \(y \le 0\): the lower half of a circle of radius \(4\).
Domain \(\boldsymbol{-4 \le x \le 4}\), range \(\boldsymbol{-4 \le y \le 0}\).
(i) \(y = f(x + 3)\)
(ii) \(y = f(x) - 4\)
(iii) \(y = f(3x)\), and describe each transformation.
(i) \(y = f(x + 3)\): the input is \(3\) more, so the graph moves \(3\) units left. Subtract \(3\) from each \(x\): \(\mathbf{(-4,\ 3)}\) and \(\mathbf{(-1,\ -2)}\).
(ii) \(y = f(x) - 4\): every output drops by \(4\), a translation \(4\) units down: \(\mathbf{(-1,\ -1)}\) and \(\mathbf{(2,\ -6)}\).
(iii) \(y = f(3x)\): the same outputs happen at one third of the \(x\)-value, a horizontal dilation by factor \(\dfrac{1}{3}\) (towards the \(y\)-axis). Divide each \(x\) by \(3\): \(\mathbf{\left(-\dfrac{1}{3},\ 3\right)}\) and \(\mathbf{\left(\dfrac{2}{3},\ -2\right)}\).
| \(\text{Check: } f\left(3 \times \dfrac{2}{3}\right)\) | \(=\) | \(f(2) = -2\) |
Changes inside the bracket act on \(x\) and work the opposite way to what they look like; changes outside act on \(y\) and work as they look.
Boundary \(2x + y = 4\). Intercepts: \(x = 0\) gives \(y = 4\); \(y = 0\) gives \(x = 2\). The sign is \(\lt \), so points on the line are not included: draw it dashed.
Test a point not on the line, such as \(O(0,\ 0)\):
| \(2(0) + 0\) | \(=\) | \(0 \lt 4\ \ \text{true}\) |
Shade the side containing \(O\).
| \((3,\ 1):\ \ 2(3) + 1\) | \(=\) | \(7 \lt 4\ \ \text{false}\) |
So \((3,\ 1)\) is not in the region.
Common pitfalls
Frequently asked questions
What is the difference between a relation and a function?
A relation is any set of ordered pairs. A function is a special relation where each x-value is paired with exactly one y-value. Every function is a relation, but not every relation is a function.
How do you use the vertical line test?
Imagine vertical lines drawn across the whole graph. If any vertical line meets the graph more than once, it is not a function. If every vertical line meets it at most once, it is a function.
What does f(x) mean?
f(x) is the output of the function f when the input is x. It is read as f of x, and f(2) means the value of the function when x equals 2. It does not mean f times x.
How do you find the domain and range of a function?
For the domain, start with all real numbers and remove any x-values that make a denominator zero or put a negative number under a square root. For the range, find which y-values the graph actually reaches, often by sketching it.
How do you know which side to shade for a linear inequality?
Pick a test point not on the boundary line, often the origin, and substitute it into the inequality. If the statement is true, shade the side containing the test point; if false, shade the other side.