Graph simple interest as a straight line of interest I (or amount A) against time n, interpret the gradient as the interest earned each period, compare two rates on the same axes by their gradients, and read interest, amount or time off the line

What this dot point is asking

NESA wants you to graph simple interest and read meaning from the graph. Simple
interest adds the same fixed amount every period. So plotting the interest $I$ (or the
total amount $A$) against time $n$ gives a straight line. The whole dot point is
about that line. You draw it from a table of values, and you learn what its gradient
(its steepness) means. You put two rates on the same axes to compare them. And you
read values off the line both ways: the interest or amount at a given time, and the
time to reach a target.

This builds directly on simple interest,
where you met $I = Prn$ and $A = P + I$ and saw why the amount grows in equal steps.
Here we put that line to work. It is strictly Year 11 Money Matters (MS-F1); the curve
of compound interest, where interest earns interest, is a Year 12 topic and stays out
of scope.

The answer

A simple-interest graph is a straight line, and three facts about that line answer
almost every question.

• Interest against time is a line through the origin. Since $I = Prn$, for a fixed
  principal and rate the interest is just a constant times $n$:

At $n = 0$ no time has passed, so $I = 0$: the line starts at the origin.

• The gradient is the interest earned each period. The number multiplying $n$ is
  $Pr$, so the line rises by $Pr$ dollars for every $1$ period across. That gradient
  is the interest per period:

• A steeper line means a higher rate. With the same principal, a bigger rate $r$
  gives a bigger gradient $Pr$, so its line climbs faster and sits above the others.
  Compare two deposits by comparing the steepness of their lines.

If instead you plot the amount $A = P + Prn$, the line has the same gradient $Pr$
but starts at the principal $P$ (its vertical intercept) rather than at the origin,
because at $n = 0$ the account already holds $P$.

A $10,000 term deposit at $6\%$ per annum has gradient $10\,000 \times 0.06 = 600$,
so its interest line is $I = 600n$ and it earns $600.00 each year. The same
$10,000 at $4\%$ has gradient $10\,000 \times 0.04 = 400$ and the flatter line
$I = 400n$.

Drawing the line from a table of values

To draw the graph, first simplify the formula for your principal and rate so only
$n$ is left, then make a short table of values, plot the points, and join them with
a ruled straight line. For $10,000 at $4\%$ per annum, $I = 10,000 \times 0.04
\times n = 400n$, giving the table

Every interest value is $400 above the one before it, the signature of a straight
line. You only ever need two points to fix a straight line, but plotting three or four
lets you check they line up, and the origin $(0, 0)$ is always one of them for an
interest graph.

Comparing two rates on the same axes

The most common exam task is comparing rates. It is built up in stages below: draw the
axes, plot the lower rate, add the higher rate, then read off. Both deposits below are
$10,000. The only difference is the rate, $4\%$ versus $6\%$. So the only difference
in the picture is the steepness of the two lines.

Stage 1, draw and label the axes. Put time $n$ across the bottom (the
horizontal axis) and interest $I$ up the side (the vertical axis), the standard
convention for an interest graph. Scale the time axis to the years asked for and the
interest axis high enough to fit the steeper line.

Stage 2, plot the lower rate. For $10,000 at $4\%$ the line is $I = 400n$.
Plot $(0, 0)$, $(1, 400)$, $(2, 800)$ and so on, and rule a straight line through them.
It passes through the origin because no time means no interest, and rises by
$400 each year, its gradient.

Stage 3, add the higher rate. On the same axes, plot the $10,000 at $6\%$
deposit, $I = 600n$. It also starts at the origin, but rises by $600 each year, so
it is steeper and sits above the $4\%$ line at every time after the start. That gap
is the whole point of the comparison: the steeper line is the better rate.

Stage 4, read the graph. To read the interest at a chosen time, go up from that
time to a line and across to the interest axis. At $3$ years the $6\%$ line reads
$1800 and the $4\%$ line reads $1200, a difference of $600. To go the other
way and find a time, start at a value on the interest axis, read across to a line,
then down to the time axis.

The numbers agree with the formula exactly: at $n = 3$, the $6\%$ deposit has
$I = 600 \times 3 = 1800$, i.e. $1800.00, and the $4\%$ deposit has
$I = 400 \times 3 = 1200$, i.e. $1200.00. Because the lines are straight, a read-off
between plotted points is exact, not an estimate, which is what makes these graphs so
quick to use.

Why the gradient is the interest each period

The gradient deserves a second look, because reading it off a graph is a favourite exam
move. The gradient is rise over run: pick any two points on the line and divide the rise
in interest by the run in time. On the $6\%$ line, going from $1$ year to $3$ years is a
run of $2$ years, and the interest rises from $600 to $1800, a rise of $1200.

That is exactly the $600 of interest earned each year, and it equals $Pr = 10,000
\times 0.06$. So if a graph gives you the line but not the rate, find the gradient from
two points and divide by the principal to recover the rate: $r = \dfrac{\text{gradient}}{P}$.

Graphing the amount instead of the interest

Some questions plot the amount $A$ (the value of the investment, or total owed),
not the interest. The shape is the same straight line with the same gradient $Pr$, but
it is lifted up by the principal: at $n = 0$ the account already holds $P$, so the
line starts at $A = P$ on the vertical axis, not at the origin. For $8000 at $5\%$
per annum the gradient is $8000 \times 0.05 = 400$ and the intercept is $8000, so
$A = 8000 + 400n$. Reading up at $4$ years gives $9600.

The check is exact: $A = 8000 + 400 \times 4 = 9600$, i.e. $9600.00. Read the
question carefully to see whether the vertical axis is interest (line from the
origin) or amount (line from the principal); it changes the intercept but never the
gradient.

How exam questions ask about this

Each wording points at one move on the graph. Learn the translation:

• "Draw / construct the graph of the interest (or amount) against time." Simplify
  $I = Prn$ (or $A = P + Prn$) for your $P$ and $r$, make a table of values, plot, and
  rule a straight line. The interest line starts at the origin; the amount line starts
  at $P$.
• "What is the gradient / slope of the line, and what does it represent?" The
  gradient is $Pr$, the interest earned each period. Find it as rise over run from two
  points.
• "Find the interest rate from the graph." Read the gradient, then $r =
  \dfrac{\text{gradient}}{P}$, and write it as a percentage.
• "Which investment has the higher rate?" The one with the steeper line (for
  the same principal).
• "Use the graph to find the interest (or amount) after $n$ years." Read up from
  the time on the horizontal axis to the line, then across to the vertical axis.
• "How long until the interest (or amount) reaches $X?" Read across from
  $X to the line, then down to the time axis.
• "How much more does the higher rate earn after $n$ years?" Read both lines at that
  time and subtract.
• "Extend the line to estimate ... after $n$ years" (beyond the plotted range).
  Because the line is straight, continue it with a ruler at the same gradient and read
  off, or just substitute into the equation.

Edge case: same rate but different principals, and crossing lines

Two subtleties round out the topic. First, "steeper means higher rate" only holds when
the principal is the same. This is because the gradient is $Pr$, not $r$ alone. A
$5000 deposit at $8\%$ has gradient $5000 \times 0.08 = 400$, while a $20,000
deposit at $4\%$ has gradient $20\,000 \times 0.04 = 800$. The lower-rate deposit is
steeper here because its principal is far larger. So compare gradients to rank the
dollars earned per year, but read the actual rates before claiming one rate is
higher. Second, two interest lines both start at the origin, so they never cross
again. The steeper one is above for all time after the start. Two amount lines start
at their own principals. So a deposit that starts lower but climbs faster can overtake
another later, and the crossing point is where their amounts are equal. Both behaviours
follow from the lines being straight, which is the defining feature of simple interest.