Sketch and interpret graphs of $y = a \sin(b x + c) + d$, $y = a \cos(b x + c) + d$ and $y = a \tan(b x + c) + d$, identifying amplitude, period, phase shift and vertical shift

What this dot point is asking

NESA wants you to sketch transformations of $\sin x$, $\cos x$ and $\tan x$ accurately, identify amplitude, period, phase shift and vertical shift from an equation, and read these off a sketch.

The deep idea is that every curve in this dot point is the same base wave seen through four independent dials: a vertical stretch ($a$), a horizontal squeeze ($b$), a horizontal slide ($h$ or $c$) and a vertical lift ($d$). If you can read those four numbers off the equation and know what each does to the base graph, you can sketch anything in the family without plotting a single point, and equally, read the four numbers back off a given sketch. The trap that costs marks is treating the four dials as if they interact: they do not, provided you apply them in the right grouping, and the only genuinely fiddly one is the phase shift, which must be read after factoring $b$ out of the bracket.

The answer

The base graphs

For $y = \sin x$:

• Domain $\mathbb{R}$, range $[-1, 1]$.
• Period $2 \pi$, amplitude $1$.
• Zeros at $x = k \pi$. Maxima at $x = \frac{\pi}{2} + 2 k \pi$. Minima at $x = -\frac{\pi}{2} + 2 k \pi$.
• Odd function: $\sin(-x) = -\sin x$.

For $y = \cos x$:

• Domain $\mathbb{R}$, range $[-1, 1]$.
• Period $2 \pi$, amplitude $1$.
• Zeros at $x = \frac{\pi}{2} + k \pi$. Maxima at $x = 2 k \pi$. Minima at $x = \pi + 2 k \pi$.
• Even function: $\cos(-x) = \cos x$.
• $\cos x = \sin\left(x + \frac{\pi}{2}\right)$: cosine is sine shifted left by $\frac{\pi}{2}$.

For $y = \tan x$:

• Domain $\mathbb{R} \setminus \left\{ \frac{\pi}{2} + k \pi \right\}$, range $\mathbb{R}$.
• Period $\pi$, no amplitude (unbounded).
• Zeros at $x = k \pi$. Vertical asymptotes at $x = \frac{\pi}{2} + k \pi$.
• Odd function: $\tan(-x) = -\tan x$.

Transformations of sine and cosine

For $y = a \sin(b(x - h)) + d$ or $y = a \cos(b(x - h)) + d$:

• Amplitude $= |a|$. The graph oscillates between $d - |a|$ and $d + |a|$.
• Period $= \frac{2 \pi}{|b|}$.
• Phase shift $= h$ (right if $h > 0$, left if $h < 0$).
• Vertical shift $= d$. The centre line is $y = d$.
• If $a < 0$, the graph is reflected in the centre line: a sine starts going down from the centre rather than up; a cosine starts at the minimum rather than the maximum.
• If $b < 0$, the graph is reflected in a vertical line. For sine, $\sin(-x) = -\sin x$, which is equivalent to flipping the sign of $a$. For cosine, $\cos(-x) = \cos x$, so a sign on $b$ has no effect.

If the equation is given as $y = a \sin(b x + c) + d$, factor: $b x + c = b\left(x + \frac{c}{b}\right)$. The phase shift is $-\frac{c}{b}$.

Transformations of tangent

For $y = a \tan(b(x - h)) + d$:

• Period $= \frac{\pi}{|b|}$ (note: tangent's period is $\pi$, not $2 \pi$).
• Asymptotes are at $b(x - h) = \frac{\pi}{2} + k \pi$, that is $x = h + \frac{\pi}{2 b} + \frac{k \pi}{b}$.
• Vertical shift by $d$ raises or lowers the graph but does not change the asymptotes.
• $a$ rescales the steepness but does not change the asymptotes or the period.

Reading features off the equation

Given any equation in the standard form, you can extract amplitude, period and shifts in seconds without sketching. Reverse process: given amplitude, period, centre line and a starting point, write the equation.

A sine function with period $T$ has $b = \frac{2 \pi}{T}$. A cosine function with amplitude $A$, centre line $y = c$, period $T$ and starting at the maximum at $x = h$ has equation

$y = A \cos\left( \frac{2 \pi}{T} (x - h) \right) + c.$

Building up a transformed curve, stage by stage

To sketch $y = 3\sin(2x) - 1$ on $0 \le x \le 2\pi$, do not plot points. Start from the base curve and turn the four dials one at a time, redrawing after each.

Stage 1, draw the base curve. Sketch $y = \sin x$ over $[0, 2\pi]$: through the origin, up to $+1$ at $\frac{\pi}{2}$, back to $0$ at $\pi$, down to $-1$ at $\frac{3\pi}{2}$, and back to $0$ at $2\pi$. Amplitude $1$, period $2\pi$. Everything else is a controlled distortion of this shape.

Stage 2, stretch the amplitude. The coefficient $a = 3$ multiplies every $y$-value by $3$, so the curve now peaks at $+3$ and dips to $-3$. The $x$-positions of the peaks, troughs and zeros do not move; only the height changes. This gives $y = 3\sin x$, amplitude $3$, period still $2\pi$.

Stage 3, squeeze the period. The coefficient $b = 2$ inside the bracket squeezes the curve horizontally: the period becomes $\frac{2\pi}{b} = \frac{2\pi}{2} = \pi$, so two complete cycles now fit between $0$ and $2\pi$. The amplitude is unchanged at $3$. This gives $y = 3\sin(2x)$.

Stage 4, lift (here, lower) the centre line. The constant $d = -1$ slides the whole curve down by $1$, so it now oscillates about the dashed centre line $y = -1$ instead of $y = 0$. The new maximum is $d + |a| = -1 + 3 = 2$ and the new minimum is $d - |a| = -1 - 3 = -4$. The finished curve is $y = 3\sin(2x) - 1$.

(This example has no phase shift. A phase shift $h$ would slide every feature horizontally by $h$ as a final stage, after factoring $b$ out of the bracket so that the slide is measured in $x$, not in the argument.)

How exam questions ask about trig graphs

The phrasing points straight at which features to extract:

• "Sketch $y = \dots$ for $0 \le x \le 2\pi$, marking the amplitude, period and centre line." A construction task. Build it up dial by dial as above, and label the centre line, the max and min values, and the $x$-positions of at least one full cycle.
• "State / find the period (and amplitude)." A read-off. Period $= \frac{2\pi}{|b|}$ (or $\frac{\pi}{|b|}$ for tan); amplitude $= |a|$. No sketch needed.
• "Find the equation of the centre line" or "the maximum / minimum value." Centre line $y = d$; max $= d + |a|$; min $= d - |a|$.
• "Find the phase shift" or "... in the form $a\sin(b(x - h)) + d$." Factor $b$ out of the bracket first; the shift is $h = -\frac{c}{b}$.
• "Write an equation for the graph shown." The reverse task. Read amplitude (half the min-to-max gap), centre line (midline), period (one full cycle width, then $b = \frac{2\pi}{\text{period}}$), and a convenient start point to fix the phase; choose sine or cosine to make the phase simplest.
• "For what values of $x$ is the curve increasing / equal to its maximum?" Read intervals and points off the sketched shape; the maximum occurs once per period.
• "State the number of solutions of $f(x) = k$ in the interval" (graph-and-line reasoning). Draw the horizontal line $y = k$ and count crossings; this is the graphical face of solving a trig equation.