Turning Curves into Straight Lines

Rewrite curved relationships like y = Axⁿ and y = Abˣ as straight lines, find the constants from the gradient and intercept, and go back again.

In the exam for IGCSE Additional Maths 0606 · 8 real past-paper questions below

اقرأ هذا الدرس مشروحاً بالعربي ←

Why straight lines?

Scientists collect data that makes a curve. A curve is hard to read, but a straight line is easy: you can measure its gradient and where it crosses the axis. So we change what we plot until the points lie on a straight line.

The trick: the axes do not have to be x and y. You can plot $y$ against $x^{2}$ , or $ln y$ against $x$ . If the points make a straight line, that line still has the form you know.

Formula

Y = m X + c

$Y$ — whatever is on the vertical axis, such as $y$ , $ln y$ or $x y$ .
$X$ — whatever is on the horizontal axis, such as $x^{2}$ or $ln x$ .
$m$ — the gradient: change in Y divided by change in X.
$c$ — the value of Y where X = 0.

Worked example

1
When y is plotted against $x^{2}$ , a straight line passes through (1, 7) and (4, 16). Find y in terms of x.
2
$y = m x^{2} + c$ — Write the line with the real axes: Y is y, X is $x^{2}$ .
3
$m = \frac{16 - 7}{4 - 1} = 3$ — Gradient from the two points.
4
$7 = 3 (1) + c \Rightarrow c = 4$ — Put one point into the line.
5
$y = 3 x^{2} + 4$ — Replace m and c.

Rule

Start every answer by writing the line in its real variables, such as $y = m x^{2} + c$ or $ln y = m x + c$ . Mark schemes give a mark for this line.

Common mistake

Treating (4, 16) as x = 4. The 4 is the value of $x^{2}$ , the thing on the horizontal axis. Here x would be 2.

Key idea

Name the axes first. Then it is just the straight line $Y = m X + c$ .

Back from a straight line

Often the vertical axis has a log or an e in it. Find m and c as before, then undo the log at the end.

Worked example

1
When $ln y$ is plotted against x, a straight line passes through (2, ln 18) and (4, ln 162). Show that $y = A \times b^{x}$ and find A and b.
2
$ln y = m x + c$ — Write the line in its real variables.
3
$m = \frac{l n 162 - l n 18}{4 - 2} = \frac{l n 9}{2} = ln 3$ — Quotient law: $162 \div 18 = 9$ . Then power law: $\frac{1}{2} ln 9 = ln 3$ .
4
$ln 18 = 2 ln 3 + c \Rightarrow c = ln 2$ — $c = ln 18 - ln 9 = ln 2$ .
5
$ln y = x ln 3 + ln 2 = ln (2 \times 3^{x})$ — Power law, then product law: one single log.
6
$y = 2 \times 3^{x}$ — Remove the ln. So A = 2 and b = 3.

Worked example

1
When $e^{y}$ is plotted against $x^{2}$ , a straight line passes through (1, 9) and (3, 1). Find y in terms of x, and the values of x for which y exists.
2
$e^{y} = m x^{2} + c$ — Write the line in its real variables.
3
$m = \frac{1 - 9}{3 - 1} = - 4$ — Gradient.
4
$9 = - 4 (1) + c \Rightarrow c = 13$ — Put in the point (1, 9).
5
$y = ln (13 - 4 x^{2})$ — Take ln of both sides. ln undoes e.
6
$13 - 4 x^{2} > 0$ — You can only take the log of a positive number.
7
$- \frac{13}{2} < x < \frac{13}{2}$ — $x^{2} < \frac{13}{4}$ , so x lies between the two square roots.

Rule

No calculator on 0606 Paper 1: keep values like ln 3 and $13$ exact. On Paper 2, give 3 significant figures.

Common mistake

From $ln y = x ln 3 + ln 2$ , writing $y = 3^{x} + 2$ . A plus between logs means times inside: $y = 2 \times 3^{x}$ .

Key idea

Find m and c, then combine into a single log and remove it. Check where the inside must be positive.

Curves into lines

Now go the other way. You are given a curve with unknown constants, and you must choose what to plot to get a straight line. Logs do the work, because they turn powers into multiplying.

$y = A x^{n}$ — the curve, with unknowns A and n.

$ln y = ln A + ln x^{n}$ — take ln of both sides, then the product law.

$ln y = n ln x + ln A$ — power law, and put the terms in line order.

Formula

l n y = n l n x + l n A

$ln y$ — Y: plot it up the vertical axis.
$ln x$ — X: plot it along the horizontal axis.
$n$ — the gradient.
$ln A$ — the intercept. Undo it at the end: $A = e^{c}$ .

$y = A b^{x}$ — the curve, with unknowns A and b.

$ln y = ln A + x ln b$ — take ln of both sides, then product and power laws.

$ln y = (ln b) x + ln A$ — put the terms in line order.

Formula

l n y = (l n b) x + l n A

$ln y$ — Y: plot it up the vertical axis.
$x$ — X: plain x along the horizontal axis.
$ln b$ — the gradient. Undo it at the end: $b = e^{m}$ .
$ln A$ — the intercept, so $A = e^{c}$ .

You may use lg instead of ln. Then undo with powers of 10: $A = 1 0^{c}$ . Nothing here is printed on the formula list, so learn both forms.

Worked example

1
Variables are related by $y = A x^{2} + \frac{B}{x}$ . Write this as a straight line and say what to plot.
2
$x y = A x^{3} + B$ — Multiply every term by x to clear the fraction.
3
$Y = x y, X = x^{3}$ — Each axis holds only x and y. The unknowns A and B sit where m and c go.
4
$m = A, c = B$ — The gradient gives A and the intercept gives B.

Rule

For a straight-line form, the axes X and Y may contain only x and y. Every unknown constant must end up as the gradient or the intercept.

Key idea

In 1619 Johannes Kepler found a rule linking each planet's year to its distance from the Sun. Plot lg of the year against lg of the distance for the planets, and they fall on a straight line with gradient 1.5.

Key idea

A power of x: plot ln y against ln x. A number to the power x: plot ln y against plain x.

Constants from a graph

Calculator questions give a table of values. Change each pair into (X, Y), plot the points, and draw ONE ruled line of best fit. Then read the gradient and intercept from your line.

Worked example

1
Variables x and y are related by $y = A x^{n}$ . The table gives x = 1, 2, 4, 8 and y = 2.5, 7.07, 20.0, 56.57. Find A and n.
2
$ln x : 0, 0.693, 1.386, 2.079$ — Work out X for each point.
3
$ln y : 0.916, 1.956, 2.996, 4.035$ — Work out Y for each point, then plot and draw the line.
4
$n = \frac{4.035 - 0.916}{2.079 - 0} = 1.50$ — Gradient from two points far apart on your line.
5
$ln A = 0.916$ — The line crosses the vertical axis, X = 0, at 0.916.
6
$A = e^{0.916} = 2.50$ — Undo the ln.

Worked example

1
Using $y = 2.50 x^{1.5}$ from your line, estimate y when x = 3.
2
$ln 3 = 1.099$ — Find X for x = 3.
3
$ln y = 1.5 (1.099) + 0.916 = 2.564$ — Use the line, or read Y from your graph.
4
$y = e^{2.564} = 13.0$ — Undo the ln.

Worked example

1
Variables are related by $y = A b^{x}$ . When lg y is plotted against x, the line passes through (1, 0.8) and (5, 2.0). Find A and b.
2
$l g y = (l g b) x + l g A$ — The straight-line form, using lg.
3
$l g b = \frac{2.0 - 0.8}{5 - 1} = 0.3$ — The gradient is lg b.
4
$b = 1 0^{0.3} = 2.00$ — Undo lg with a power of 10.
5
$0.8 = 0.3 (1) + l g A \Rightarrow l g A = 0.5$ — Put in a point to find the intercept.
6
$A = 1 0^{0.5} = 3.16$ — Undo lg.

Rule

Plotting marks need every point in the right place and a single ruled straight line. Take the gradient from two points on your line that are far apart.

Common mistake

Reading c off the graph when the horizontal axis does not start at 0. Then the line does not show c. Work it out from a point on the line instead.

Key idea

Change to X and Y, find m and c, then undo the logs to get the constants.

Which method? Read the question

Before you start, read which way the question goes: to a straight line, or back from one.

You see	You do
“Y is plotted against X … find y in terms of x”	Write $Y = m X + c$ in real variables, find m and c, rearrange
ln y or lg y on the vertical axis	Find m and c, combine into one log, remove it
$e^{y}$ on the vertical axis	Take ln at the end; the inside must be positive
$y = A x^{n}$	Plot ln y against ln x: gradient n, intercept ln A
$y = A b^{x}$	Plot ln y against x: gradient ln b, intercept ln A
A table of values	Make X and Y, plot, ruled line, read m and c
“For which x is y valid?”	Inside of the log > 0, then solve

Common mistake

Forgetting to undo the log. If the intercept is ln A = 0.916, then A is $e^{0.916}$ , not 0.916.

Key idea

Every question is the same straight line, $Y = m X + c$ , dressed up. Find what X and Y are, and the rest follows.

Practise turning curves into straight lines now

Interactive visual aids, then real past-paper questions marked on your working like the exam. Free to start.

Practise free on Lemmara

Real past-paper questions on turning curves into straight lines

Each with a full worked solution.

Lemmara is an independent study tool, not affiliated with or endorsed by Cambridge International Education or Pearson Edexcel. The lessons and worked examples are Lemmara’s own.