A12, A13, A14 - Exponential, trigonometric and transformed graphs

Exponential growth and decay (Higher)

A savings balance starts at £100 and is multiplied by 1.1 at each year end. When does it first exceed £125? Adding £10 each year would predict a different answer from multiplying the latest balance.

An exponential function has its input in the exponent. For $y=2^x$, equal increases of 1 in the input double the output. At inputs $-2,-1,0,1,2,3$, the outputs are $\frac14,\frac12,1,2,4,8$. The differences are not constant, so this is not a straight line.

For $y=k^x$, the base $k$ is positive. If $k>1$, the graph increases. If $0<k<1$, it decreases. For $k=1$, it is simply the horizontal line $y=1$. Every such graph passes through $(0,1)$. A growth or decay curve stays positive, approaching the asymptote $y=0$ in one direction without reaching it.

[DIAGRAM: asset_name: Exponential growth and decay; asset_slug: 1ma1_m26_exponentials; caption: Growth doubles and decay halves the output for each unit increase in input.; description: The graph plots y equals 2 to the power x in blue and y equals one half to the power x in sage, with their shared intercept (0,1), representative integer-input points and horizontal asymptote y equals 0.]

For the decay graph $y=(\frac12)^x$, the same inputs give $4,2,1,\frac12,\frac14,\frac18$. Notice that $(\frac12)^x=2^{-x}$: these two curves are mirror images in the vertical axis.

Change only the yearly multiplier, from 1.1 to 1.2. Keep the opening balance, target and payment dates.

If a context explicitly defines continuous growth between dates, a smooth graph can estimate a fractional time. For the first rule, $100(1.1)^t=125$ crosses near $t=2.34$. That is different from asking for the first whole-year payment date.

To sketch a pure exponential, plot a table including negative, zero and positive inputs, then draw a smooth curve. Preserve the positive outputs and the intercept $(0,1)$. Do not join a small output to the origin or confuse $2^x$ with $x^2$.

A coordinate can also determine an unknown base. If $y=k^x$ passes through $(-1,\frac15)$, then $k^{-1}=\frac15$, so $1/k=1/5$ and $k=5$. Substitute both coordinates before rearranging.

Keep the target word “exceeds” in mind in the next savings model.

Sine, cosine and tangent in degrees (Higher)

Trigonometric graphs repeat because a full turn returns to the same direction. Use degrees, including degree mode on your calculator.

Sine and cosine stay between $-1$ and 1. These quarter-turn values anchor their smooth waves.

Sine starts at zero and rises; cosine starts at its maximum. Repetition works for any angle: $\sin450^\circ=1$, $\cos(-360^\circ)=1$, and $\sin(-90^\circ)=-1$. A negative input does not force a negative output; $\cos(-90^\circ)=0$.

Tangent is $\tan x=\sin x/\cos x$ where $\cos x\ne0$. It is undefined at $90^\circ+180^\circ n$, for integer $n$, giving vertical asymptotes. It crosses zero at multiples of $180^\circ$ and increases along each separate branch. Useful points are $(-45^\circ,-1),(0^\circ,0),(45^\circ,1)$. Tangent has no upper or lower output bound.

[DIAGRAM: asset_name: Degree-based trigonometric graphs; asset_slug: 1ma1_m26_trigonometry; caption: Sine and cosine repeat every 360 degrees, while tangent repeats every 180 degrees with vertical asymptotes.; description: Three aligned panels show sine and cosine from minus 180 to 540 degrees with key quarter-turn values, and tangent on separate intervals with dashed asymptotes at minus 90, 90, 270 and 450 degrees. Tangent branches are clipped to the displayed vertical range, not joined across asymptotes.]

To solve an equation, imagine drawing its horizontal target level. Find every intersection inside the given interval, including endpoints if they work.

Change only the sign of the target. The reference angle stays the same, but the intersections move to the negative half of the wave.

For a longer or negative interval, first find solutions in a full turn, then add or subtract periods and retain only values in range.