Further Graph Transformations and Modelling

This focus area takes the transformation rules from Year 11 (Graph transformations) and applies them to the trigonometric functions, then uses transformed functions and logarithmic scales to model real situations. It was worth 8 marks on NESA's sample paper for the 2024 syllabus.

• Graph $y=kf(a(x+b))+c$ when $f$ is $\sin x$, $\cos x$ or $\tan x$, and describe each transformation in words.
• State the domain, range, amplitude, period, phase shift and vertical shift of a transformed trig graph.
• Solve equations such as $k\sin (a(x+b))+c=d$ in a given domain, algebraically or from a graph.
• Model periodic phenomena (tides, daylight hours, Ferris wheels, temperature) with transformed trig functions. No calculus is needed here.
• Model practical problems with any function from the course and its transformations.
• Explain when a logarithmic scale is useful, read one, and solve problems involving decibels, the Richter scale, star magnitudes and pH.

All angles in this topic are in radians unless a question says otherwise.

Start from the general form

and read off each parameter (the same rules hold for $\cos$):

For $y=k\tan (a(x+b))+c$ the period is $\pi /a$, there is no amplitude, and the vertical asymptotes move with the graph: they sit where $a(x+b)=\frac{\pi }{2}+n\pi$.

The range of a transformed sine or cosine graph is $c-|k|\le y\le c+|k|$.

Factorise before you translate

The most common error in this topic is reading the phase shift from an unfactorised bracket. In $y=\sin (2x-\frac{\pi }{3})$ the shift is not $\frac{\pi }{3}$. Factorise first:

so the graph is $y=\sin 2x$ moved $\frac{\pi }{6}$ to the right.

Order of transformations

When you build a graph step by step, do dilations before translations in each direction. A safe order is: horizontal dilation, then horizontal translation (using the factorised form), then vertical dilation or reflection, then vertical translation. Describe the transformations in this order when a question asks you to.

The method is the same as in Year 11 but with a transformed argument:

1. Rearrange so the trig function is on its own.
2. Find the related (reference) angle.
3. Use the quadrants (ASTC) to find every angle in the right quadrants.
4. Adjust the domain for the argument: if $0\le x\le \pi$, then $0\le 2x\le 2\pi$.
5. Solve for $x$ and check each answer lies in the original domain.

Worked example 1

Consider $y=3\cos \left(2\left(x-\frac{\pi }{4}\right)\right)-1$.

(a) Features. Amplitude 3, period $\frac{2\pi }{2}=\pi$, phase shift $\frac{\pi }{4}$ to the right, centre line $y=-1$. The range is $-4\le y\le 2$.

(b) Solve $y=-2.5$ for $0\le x\le \pi$.

Let $u=2x-\frac{\pi }{2}$. Since $0\le x\le \pi$, the new domain is $-\frac{\pi }{2}\le u\le \frac{3\pi }{2}$. The reference angle is $\frac{\pi }{3}$ and cosine is negative in quadrants 2 and 3:

Neither negative angle $-\frac{2\pi }{3}$ nor $-\frac{4\pi }{3}$ lies in the domain. Then $2x=u+\frac{\pi }{2}$:

A quick sketch confirms two intersections in one period. When a question says "how many solutions", a sketch of both sides is often the fastest method.

To build a model $y=k\sin (a(t+b))+c$ or a cosine version from a description:

• Centre line: $c=\frac{\text{max}+\text{min}}{2}$.
• Amplitude: $k=\frac{\text{max}-\text{min}}{2}$.
• Period $T$ gives $a=\frac{2\pi }{T}$.
• Choose cosine if the cycle starts at a maximum, $-\cos$ if it starts at a minimum, and sine if it starts on the centre line going up.

Worked example 2

A Ferris wheel has diameter 30 m and its centre is 17 m above the ground. It turns once every 12 minutes, and a rider boards at the lowest point at $t=0$. Find a model for the rider's height $h$ metres after $t$ minutes, and find how long during the first turn the rider is more than 25 m above the ground.

Model. Max height $17+15=32$, min $17-15=2$, so $c=17$ and amplitude 15. The period is 12, so $a=\frac{2\pi }{12}=\frac{\pi }{6}$. Starting at the minimum gives $-\cos$:

Check: $h(0)=2$ and $h(6)=32$.

Height above 25 m.

Solve the equality first. With $\theta =\frac{\pi t}{6}$ in $[0,2\pi ]$, $\cos \theta =-\frac{8}{15}$ gives $\theta =2.1333$ or $\theta =2\pi -2.1333$. So

The rider is above 25 m from about 4.07 to 7.93 minutes, which is about 3.85 minutes per turn. Notice the symmetry about $t=6$, the time of maximum height.

The syllabus also asks you to model practical problems with any function in the course (linear, quadratic, cubic, exponential, logarithmic, reciprocal, trig) and its transformations. Typical tasks: choose a function type that fits a graph or table, find parameters by substituting known points, then interpret a translation or dilation in context (for example, $c$ as a starting value or $a$ as a growth rate). Always state the domain that makes sense in context, such as $t\ge 0$.

A logarithmic scale places values so that equal distances mean equal ratios, not equal differences. Each step of 1 on a base-10 scale multiplies the quantity by 10. Use one when data cover several orders of magnitude (from 0.001 to 1 000 000, say), because a linear scale would squash most of the data near zero.

The four contexts named in the syllabus:

A question normally gives you the formula. The key skill is turning a difference on the scale into a ratio of the underlying quantity: subtract the two scale readings and use the index law $\log A-\log B=\log \frac{A}{B}$.

Worked example 3

(a) Reading a log axis. On a base-10 logarithmic axis, the gridlines are labelled 100 and 1000. A point lies 0.3 of the way from the 100 line to the 1000 line. What value does it represent?

The 100 line is at position 2 and the 1000 line at position 3, so the point is at $2.3$:

Not 370, which you would get by reading the axis as if it were linear.

(b) Comparing pH. Lemon juice has pH 2.4 and a sample of rainwater has pH 6.1. Each solution satisfies $[{\text{H}}^{+}]=10^{-\text{pH}}$, so

The lemon juice has about 5000 times the hydrogen ion concentration. Lower pH means more acidic, so watch the negative sign.

• Reading the phase shift from $\sin (2x-\frac{\pi }{3})$ as $\frac{\pi }{3}$ instead of factorising to get $\frac{\pi }{6}$.
• Forgetting to stretch the domain when the argument is $ax$: solutions of $\sin 2x=c$ in $[0,2\pi ]$ come from $[0,4\pi ]$.
• Giving answers in degrees, or using a calculator in degree mode.
• Using period $\frac{2\pi }{a}$ for tangent (it is $\frac{\pi }{a}$).
• Giving the amplitude as a negative number when $k<0$.
• Treating a log scale difference as a difference in the quantity (a 2-unit difference means a ratio of 100, not 2).
• Losing the minus sign in the pH formula, so a higher pH is wrongly read as more acidic.

• Sketch: show the shape, the key points (intercepts, maxima and minima, asymptotes for tan) and label the axes with scale values in radians. One period is not enough if the question's domain covers more.
• Describe the transformations: name each one (dilation, reflection, translation), give the direction and the factor or distance, in a sensible order.
• Solve: show the change of domain for the argument, list every solution, and give exact values unless told otherwise.
• Model: state what each parameter represents in context and give the domain. Check your model against one or two given data points before using it.
• Periodic modelling questions often have several parts that build on each other; if you can't get part (a), use the given model in later parts.

• MA11-GT Graph transformations: the transformation rules here are the same ones applied to new functions.
• MA11-TM and MA11-TI: exact values, the unit circle, radians and the ASTC rule are assumed.
• MA11-EL Exponential and logarithmic functions: log laws are what make logarithmic scales work.
• MA12-DC and MA12-IC: once a trig model is set up, Year 12 calculus finds rates of change and totals for it.