Translate a known graph vertically and horizontally and reflect it in the $x$-axis and the $y$-axis, recognise and test even functions (symmetric about the $y$-axis, $f(-x) = f(x)$) and odd functions (symmetric about the origin, $f(-x) = -f(x)$), sketch absolute-value graphs as transformations of $y = |x|$, and form composite functions $f(g(x))$ and determine their domain

What this dot point is asking

Once you can sketch a basic curve, you almost never have to start from scratch again. Most graphs in the HSC are a known shape that has been moved or flipped: a parabola slid sideways, a reciprocal curve turned upside down, a V-shape dropped below the axis. This dot point gives you the four moves that turn one curve into a whole family, the symmetry that lets you halve the work, and the way two functions are chained together into a composite. The four skills are: translating a graph up/down and left/right, reflecting it in the $x$-axis or the $y$-axis, recognising even and odd functions (and proving it algebraically), and forming composite functions $f(g(x))$ together with their domain.

The single idea underneath all of it is that changing the equation in a fixed way moves the graph in a fixed way. Replace $x$ by $x - h$ and the whole picture slides right by $h$; add $k$ to the function and it rises by $k$; put a minus sign on the function and it flips top-to-bottom; put a minus sign on $x$ and it flips left-to-right. Even and odd functions are then just the special curves that a flip leaves unchanged, which is why a quick algebraic test ($f(-x)$ versus $f(x)$ and $-f(x)$) settles their symmetry without a single plotted point. The exam marks come from naming each transformation precisely, keeping the order and direction right (the horizontal ones run "backwards", which is the classic trap), and, for a composite, stating the domain rather than just the formula.

The answer

Translating a known graph

A translation slides a graph without rotating, reflecting or resizing it: every point moves the same distance in the same direction. There are two independent moves, vertical and horizontal, and the rules look pleasingly similar once you see where each one acts.

A vertical translation changes the output. To shift the graph of $y = f(x)$ up by $k$, add $k$ to the function: $y = f(x) + k$ (and down by $k$ is $y = f(x) - k$). This is the intuitive one: making every $y$-value bigger by $k$ lifts the whole curve by $k$. A horizontal translation changes the input, and it runs the opposite way to how it reads. To shift the graph right by $h$, replace $x$ by $x - h$: $y = f(x - h)$ (and left by $h$ is $y = f(x + h)$). The minus sign is right but feels backwards: $y = (x - 2)^2$ is $y = x^2$ moved right $2$, not left. The reason is that to get the same height the new curve reaches at $x$, the original only needed to reach at $x - 2$, so the picture has been dragged forward by $2$. Combining the two, $y = f(x - h) + k$ is $y = f(x)$ shifted $h$ right and $k$ up; for a parabola this is exactly the completed-square (vertex) form, with the vertex sitting at $(h, k)$.

For example, $y = (x - 2)^2 + 1$ is the parabola $y = x^2$ translated right $2$ and up $1$, so its vertex moves from $(0, 0)$ to $(2, 1)$. You can check a point rather than trust the rule: at $x = 0$ the image gives $y = (0 - 2)^2 + 1 = 4 + 1 = 5$, and indeed the original point $(-2, 4)$ on $y = x^2$ has been carried to $(0, 5)$, a move of $2$ right and $1$ up. The diagram overlays the two.

Reflecting in the x-axis and the y-axis

A reflection flips the graph across an axis, producing its mirror image. As with translations, one reflection acts on the output and the other on the input.

To reflect in the $x$-axis, negate the whole function: $y = -f(x)$. Each point $(x, y)$ goes to $(x, -y)$, so the graph flips top-to-bottom about the horizontal axis. For instance $y = -x^2$ is $y = x^2$ turned upside down, and $y = -\sqrt{x}$ is the square-root curve flipped below the axis. To reflect in the $y$-axis, replace $x$ by $-x$: $y = f(-x)$. Each point $(x, y)$ goes to $(-x, y)$, so the graph flips left-to-right about the vertical axis. For instance $y = \sqrt{-x}$ is $y = \sqrt{x}$ reflected into the second quadrant, defined now for $x \le 0$. Reflection is mutual: it maps each graph to the other, so reflecting twice returns the original.

A neat consequence ties this section to the next: reflecting in the $x$-axis and then in the $y$-axis (in either order) gives $y = -f(-x)$, which is the same as a $180^\circ$ rotation about the origin. Curves that this double flip leaves unchanged are exactly the odd functions. Curves unchanged by the single $y$-axis flip are the even functions.

Even and odd functions

Some functions are so symmetric that a reflection does nothing to them, and recognising this halves the work of sketching and lets you read off values you have not computed. There are two kinds.

A function is even if its graph has line symmetry in the $y$-axis, that is, the left half is the mirror image of the right half. Reflecting in the $y$-axis sends $y = f(x)$ to $y = f(-x)$, so for the graph to be unchanged we need $f(-x) = f(x)$ for every $x$ in the domain. The powers $y = x^2, x^4, x^6$ are even, and so is $y = |x|$; their graphs fold exactly onto themselves across the $y$-axis. A function is odd if its graph has point symmetry in the origin, meaning a $180^\circ$ rotation about $O$ maps it onto itself (equivalently, reflect in both axes). That double flip sends $y = f(x)$ to $y = -f(-x)$, so for no change we need $-f(-x) = f(x)$, that is $f(-x) = -f(x)$ for every $x$. The powers $y = x^3, x^5$ are odd, and so are $y = x$ and $y = \dfrac{1}{x}$.

The test is the same one calculation for both, and it is the marker's preferred method because it needs no graph: simplify $f(-x)$ and compare it with $f(x)$ and with $-f(x)$. If $f(-x) = f(x)$ the function is even; if $f(-x) = -f(x)$ it is odd; if it matches neither it is neither (most functions are neither). Take $f(x) = x^3 - x$: then $f(-x) = (-x)^3 - (-x) = -x^3 + x = -(x^3 - x) = -f(x)$, so it is odd; a numeric check agrees, with $f(2) = 8 - 2 = 6$ and $f(-2) = -8 + 2 = -6 = -f(2)$. By contrast $f(x) = x^2 + x$ has $f(-x) = x^2 - x$, which is neither $f(x)$ nor $-f(x)$, so it is neither even nor odd. The diagram contrasts an even curve with an odd one.

Two facts are worth banking. An odd function defined at $x = 0$ must pass through the origin, because $f(-0) = -f(0)$ forces $f(0) = -f(0)$, so $f(0) = 0$. And the symmetry doubles your information: once you know an even function's graph for $x \ge 0$ you know it for $x \le 0$ by mirroring, and for an odd function you know the left half by rotating the right half $180^\circ$.

Absolute-value graphs by transformation

The absolute value $|x|$ is the size of $x$ ignoring its sign, that is, its distance from $0$ on the number line: $|3| = 3$ and $|-3| = 3$. Written in cases, $|x| = x$ for $x \ge 0$ and $|x| = -x$ for $x < 0$, two straight lines of gradient $+1$ and $-1$ that meet at the origin. So the graph of $y = |x|$ is a V-shape with its sharp corner (vertex) at $(0, 0)$, sitting on or above the $x$-axis; it is even, with the $y$-axis as its mirror line.

Because $y = |x|$ is a single known shape, every $y = |x - h| + k$ is just that V translated: replace $x$ by $x - h$ to shift right $h$, and add $k$ to shift up $k$, exactly the rules from the first two sections. The corner moves from $(0, 0)$ to $(h, k)$, the arms keep their gradients of $\pm 1$, and you finish by marking the intercepts. Take $y = |x - 2| - 1$: the V is shifted right $2$ and down $1$, so the corner sits at $(2, -1)$. The $x$-intercepts come from $|x - 2| = 1$, giving $x - 2 = \pm 1$, so $x = 1$ or $x = 3$; the $y$-intercept is $y = |0 - 2| - 1 = 2 - 1 = 1$, the point $(0, 1)$. If you ever need the two straight branches explicitly, split at the corner: $y = (x - 2) - 1 = x - 3$ for $x \ge 2$, and $y = -(x - 2) - 1 = -x + 1$ for $x < 2$. The four-panel build shows the transformation stage by stage.

Stage 1, the basic V

Start from $y = |x|$, the V-shape with its corner at the origin and arms of gradient $\pm 1$. This is the known graph every absolute-value sketch is built from.

Stage 2, shift right $2$

Replacing $x$ by $x - 2$ slides the whole V right $2$ units, carrying the corner from $(0, 0)$ to $(2, 0)$. The shape and the arm gradients are unchanged.

Stage 3, shift down $1$

Subtracting $1$ from the function lowers the V by $1$, so the corner drops from $(2, 0)$ to $(2, -1)$. Now part of the graph lies below the $x$-axis.

Stage 4, mark the intercepts

Solve $|x - 2| - 1 = 0$ for the $x$-intercepts $(1, 0)$ and $(3, 0)$, and put $x = 0$ for the $y$-intercept $(0, 1)$. The finished graph is a V with corner $(2, -1)$ through those three points.

Composite functions and their domain

A composite function chains two functions so the output of one becomes the input of the other. Writing $f(g(x))$ means "do $g$ first, then $f$": feed $x$ into $g$, then feed the result into $f$. It is built by substitution, putting the whole rule for $g(x)$ wherever $x$ appears in $f$. If $f(x) = \sqrt{x}$ and $g(x) = x - 3$, then $f(g(x)) = \sqrt{g(x)} = \sqrt{x - 3}$. Order matters: the other composite $g(f(x)) = f(x) - 3 = \sqrt{x} - 3$ is a different function, so $f(g(x))$ and $g(f(x))$ are generally not the same.

The part the exam really tests is the domain. A value $x$ is allowed into $f(g(x))$ only if two things hold: $x$ must be in the domain of the inner function $g$, and the output $g(x)$ must then be a legal input for the outer function $f$. In practice it is almost always enough to write down the equation of the composite and read its domain off as a single function, applying the usual rules (no division by zero, no square root of a negative). For $f(g(x)) = \sqrt{x - 3}$ the square root needs $x - 3 \ge 0$, so the domain is $x \ge 3$; a check confirms it, since $f(g(3)) = \sqrt{0} = 0$ and $f(g(7)) = \sqrt{4} = 2$ are fine but $f(g(2)) = \sqrt{-1}$ is not. Switching the order, $g(f(x)) = \sqrt{x} - 3$ instead needs $x \ge 0$, a different domain again, which is the clearest sign that the order of composition genuinely changes the function. As a reciprocal example, with $f(x) = \dfrac{1}{x}$ and $g(x) = x - 1$ the composite $f(g(x)) = \dfrac{1}{x - 1}$ is undefined where the denominator is $0$, so its domain is $x \ne 1$.

How exam questions ask about transformations and symmetry

The wording points straight to the move or the test:

• "Sketch $y = f(x) + k$ / $y = f(x - h)$" or "describe the transformation that maps ... onto ..." Name the shift precisely: $+k$ is up $k$, $f(x - h)$ is right $h$ (remember the horizontal one runs backwards to the sign). Track the vertex/key point.
• "Sketch $y = -f(x)$ / $y = f(-x)$." $-f(x)$ is a reflection in the $x$-axis; $f(-x)$ is a reflection in the $y$-axis. State which axis.
• "Show that $f(x)$ is even / odd," or "test $f(x)$ for even or odd symmetry." Compute $f(-x)$, simplify, and write the conclusion: $f(-x) = f(x)$ (even) or $f(-x) = -f(x)$ (odd). "Show that" wants the algebra, not a graph.
• "What symmetry does the graph have?" Even means line symmetry in the $y$-axis; odd means point symmetry about the origin ($180^\circ$ rotation).
• "Sketch $y = |x - h| + k$." Transform $y = |x|$: corner at $(h, k)$, arms of gradient $\pm 1$, then mark intercepts.
• "Solve $|x - h| + k = 0$" or "find the $x$-intercepts." Set $|x - h| = -k$ and use $x - h = \pm(-k)$ (two answers, provided $k \le 0$).
• "Find $f(g(x))$ / find $g(f(x))$." Substitute the inner rule into the outer one and simplify. Watch the order.
• "State the domain of $f(g(x))$." Read it from the composite's equation: denominator $\ne 0$, square-root argument $\ge 0$. Give it in $a < x < b$ or $x \ge a$ form.
• "Hence find $f(-a)$" after proving oddness/evenness. Use $f(-a) = -f(a)$ (odd) or $f(-a) = f(a)$ (even) rather than recomputing.