Linear Law | Sec 3 & 4 A Maths Study Guide

Secondary 3 & 4 Additional Mathematics · Study guide

Choose what to plot, interpret gradient and intercept, and return to the original variables. Connect algebraic, power and exponential models to straight-line graphs.

Concept → worked examples → practice23 min read · practice linked below

A straight line in new variables

A curved relationship can become a straight line when we change what goes on the axes. In $Y=mX+c$ , the capital letters represent the quantities plotted; they need not be the original $x$ and $y$ . A graph of $Y$ against $X$ has $X$ horizontally and $Y$ vertically.

This chapter is taught in Sec 3 and revisited in Sec 4. Start with algebraic rearrangement, then use logarithms for power and exponential models. Revise indices and logarithms if the logarithm laws are unfamiliar.

Why it works. After an equivalent rearrangement, every valid original pair $(x,y)$ gives a transformed pair $(X,Y)$ satisfying the same straight-line equation. Thus its gradient and intercept encode the original constants. The transformation must be defined: denominators cannot be zero and real logarithms require positive arguments.

Rearrange to $Y=mX+c$ , where $m$ and $c$ contain only constants.
State both axes and match $m,c$ to the constants.
Read or calculate the gradient using transformed coordinates: $m=(Y_2-Y_1)/(X_2-X_1)$ .
Solve for the original constants; undo any logarithm or reciprocal.
Check by substituting into the original model, retaining its restrictions.

Useful linear forms and their conditions
Relationship	Vertical $Y$ , horizontal $X$	Gradient and intercept
$y=ax^2+bx$ , $x\ne0$	$Y=y/x,\ X=x$	$m=a,\ c=b$
$y=a/(x+b)$ , $a\ne0,\ x\ne-b$	$Y=1/y,\ X=x$	$m=1/a,\ c=b/a$
$y=ax^n$ , $a>0,\ x>0$	$Y=\lg y,\ X=\lg x$	$m=n,\ c=\lg a$
$y=Ab^x$ , $A>0,\ b>0$	$Y=\lg y,\ X=x$	$m=\lg b,\ c=\lg A$
$y=d+Ae^{kx}$ , $A>0$	$Y=\ln(y-d),\ X=x$	$m=k,\ c=\ln A$

Here $\lg$ means logarithm to base 10. Natural logarithms also work for power and exponential models, but use the same base consistently and undo $\ln$ with $e^{(\cdot)}$ . The constant exponential case $b=1$ gives a horizontal line.

Algebraic rearrangements and restrictions

Example 1: dividing by a variable

For $y=ax^2+bx$ , the graph of $y/x$ against $x$ passes through $(1,7)$ and $(4,13)$ . Find $a,b$ and explain what happens at $x=0$ .

\frac yx=ax+b\qquad(x\ne0).

The given points are $(x,y/x)$ , so

a=\frac{13-7}{4-1}=2,\qquad 7=2(1)+b\implies b=5.

Therefore $y=2x^2+5x$ . At $x=4$ , $y=52$ and $y/x=13$ , agreeing with the graph. The original formula gives $y=0$ at $x=0$ , but $y/x$ is undefined there. The extended straight line has intercept 5; this is not an observed transformed point at $x=0$ .

Example 2: reciprocal axes

Let $y=a/(x+b)$ , where $a\ne0$ . The graph of $1/y$ against $x$ passes through $(0,1/2)$ and $(6,2)$ . Find the model.

\frac1y=\frac{x+b}{a}=\frac1a x+\frac ba.

\frac1a=\frac{2-1/2}{6-0}=\frac14\implies a=4,\qquad \frac b4=\frac12\implies b=2.

Hence $y=4/(x+2)$ , with $x\ne-2$ . At $x=6$ , $y=1/2$ , so $1/y=2$ . Although the extended line meets the horizontal axis at $(-2,0)$ , there is no original finite value of $y$ there: $1/y=0$ has no finite solution.

A transformation can exclude values. Dividing $xy^2=ax+by$ by $x$ gives $y^2=b(y/x)+a$ , so one possible graph is $y^2$ against $y/x$ , for $x\ne0$ . Do not divide by a variable without stating its restriction. If $x=0$ matters, handle it in the original equation.

Power and exponential models

Example 3: a power law

For $y=ax^n$ , $a>0,x>0$ , a graph of $\lg y$ against $\lg x$ passes through $(0,1)$ and $(2,4)$ . Find $a,n$ , then $y$ when $x=4$ .

Taking logarithms uses the product and power laws:

\lg y=\lg(ax^n)=\lg a+n\lg x.

n=\frac{4-1}{2-0}=\frac32,\qquad\lg a=1\implies a=10.

The model is $y=10x^{3/2}$ on the stated domain $x>0$ . Thus $y=10(\sqrt4)^3=80$ . The transformed point $(2,4)$ means $x=100,y=10\,000$ , and $10(100)^{3/2}=10\,000$ checks it.

Example 4: exponential decay

The graph of $\lg y$ against $x$ for $y=Ab^x$ passes through $(0,2)$ and $(4,1)$ . Find the model and $y$ at $x=2$ .

\lg y=(\lg b)x+\lg A,\qquad\lg b=\frac{1-2}{4}=-\frac14,\qquad\lg A=2.

Therefore $A=100$ , $b=10^{-1/4}$ , and

y=100\left(10^{-1/4}\right)^x=100\,10^{-x/4}.

Since $0<b<1$ , the model decreases as $x$ increases but stays positive. At $x=2$ , $y=100/\sqrt{10}=10\sqrt{10}\approx31.6$ . At $x=4$ , $y=10$ , whose base-10 logarithm is 1 as required. A negative logarithmic gradient does not mean a negative exponential base.

Example 5: subtract the background before taking logs

A cooling model is $T=20+Ce^{-kt}$ , with $C,k>0$ and $t\ge0$ minutes. On the graph of $\ln(T-20)$ against $t$ , the line passes through $(0,\ln80)$ and $(10,\ln40)$ . Find the initial temperature and the time until it halves.

\ln(T-20)=\ln C-kt,\qquad C=80,\qquad-k=\frac{\ln40-\ln80}{10}=-\frac{\ln2}{10}.

Thus $T=20+80\,2^{-t/10}$ , and $T(0)=100$ . Half the initial temperature is 50, so

50=20+80\,2^{-t/10}\implies 2^{-t/10}=\frac38\implies t=\frac{10\ln(8/3)}{\ln2}\approx14.2\text{ minutes}.

Substitution gives $T=20+80(3/8)=50$ . The excess above 20 halves every 10 minutes, but the temperature itself does not. For every finite $t\ge0$ , $T>20$ , so the logarithm is defined; the model approaches 20 without reaching it.

Explore the meaning of a logarithmic gradient

For $y=Ae^{kx}$ , the transformed graph is $\ln y=kx+\ln A$ . In the default case, $A=e$ and $k=1/2$ : increasing $x$ by one adds $1/2$ to $\ln y$ , and multiplies $y$ by $e^{1/2}$ . Changing the intercept changes the starting scale; changing the gradient changes the growth factor.

The axes are x and ln y. The highlighted point has x = 2; its vertical coordinate is not y itself.

Gradient k = 0.50; intercept ln A = 1.00. A = e ≈ 2.718. Each increase of 1 in x multiplies y by e^0.5 ≈ 1.649. At x = 2, ln y = 2 and y = e² ≈ 7.389. The model grows.

From a table to a useful line of best fit

Calculate the transformed coordinates first. Use clear scales and axis labels, plot every point, then draw one straight line that represents the trend. For measurement data, do not join points one by one or force the line through the origin. Retain extra digits in calculations; round only the plotted table as needed.

Choose two well-separated points on your drawn line to estimate the gradient. They need not be data points. Graph readings are estimates, so use $\approx$ and a precision supported by the scale. A prediction within the measured range is interpolation; outside it is extrapolation and depends more strongly on whether the model remains appropriate.

Example 6: fit data and undo the logarithm

Measurements approximately follow $y=Ab^x$ :

Original and transformed readings
x	0	1	2	3	4
y	3.02	4.45	6.78	10.03	15.20
lg y (3 d.p.)	0.480	0.648	0.831	1.001	1.182

Plot $Y=\lg y$ against $X=x$ . A suitable line of best fit passes approximately through $(0,0.478)$ and $(4,1.182)$ . Thus

m\approx\frac{1.182-0.478}{4}=0.176,\qquad c\approx0.478.

A\approx10^{0.478}=3.01,\qquad b\approx10^{0.176}=1.50.

To estimate $y$ at $x=1.5$ , read $Y\approx0.176(1.5)+0.478=0.742$ . Hence $y\approx10^{0.742}=5.52$ . The answer is not 0.742. As a check, the line predicts $y\approx6.76$ at $x=2$ , close to the measurement 6.78; it need not reproduce every measurement exactly.

Crosses are measurements, while the solid line represents the overall trend. Read a gradient triangle from the line.

Example 7: recognise an anomalous reading

Positive data are expected to follow $y=ax^n$ . The readings are $(1,12)$ , $(2,3)$ , $(3,2)$ , $(4,0.75)$ and $(6,1/3)$ . Four lie on the line $\lg y=-2\lg x+\lg12$ . Identify the inconsistent reading and estimate its corrected $y$ , assuming $x$ is correct.

Undoing logarithms gives $y=12x^{-2}=12/x^2$ . The predictions at $x=1,2,4,6$ are respectively $12,3,3/4,1/3$ , which match. At $x=3$ , the recorded value 2 differs from

y=\frac{12}{3^2}=\frac43\approx1.33.

Equivalently, the line predicts $\lg y=\lg12-2\lg3=\lg(4/3)$ , not $\lg2$ . Thus $(3,2)$ is anomalous, and the model suggests $y\approx1.33$ . A data point away from a fitted line should be investigated; the graph alone does not establish which measurement caused the discrepancy unless the question states an assumption.

Solve a new equation on the same axes

Example 8: translate the second condition too

For $x>0$ , measurements follow $y=5/x^2+2/x$ . A graph plots $Y=xy$ against $X=1/x$ . Use a second straight line to find where $x=3/y$ .

Multiplying the model by $x$ gives

xy=5\left(\frac1x\right)+2\implies Y=5X+2.

The new condition $x=3/y$ is equivalent to $xy=3$ , since here $y>0$ . Draw the horizontal line $Y=3$ . At its intersection with the model,

3=5X+2\implies X=\frac15,\qquad x=\frac1X=5,\qquad y=\frac3x=\frac35.

The exact illustrative model gives $(x,y)=(5,3/5)$ ; on a measured graph these would be estimates. Check: $5/25+2/5=3/5$ . The horizontal coordinate read from the transformed graph is $1/x$ , not $x$ .

For a logarithmic graph, transform the extra condition with the same logarithm and the same axes. For example, if $Y=\lg(y-10)$ , $X=x$ , then $y-10=10^{2x}$ becomes $Y=2X$ . Draw that line and recover the original variable from the intersection.

Common mistakes and a final check

Using original coordinates for a transformed gradient. Convert both axes first; a point $(\lg x,\lg y)=(1,2)$ means $(x,y)=(10,100)$ .
Confusing the model types. For $ax^n$ , log both variables. For $Ab^x$ , log only the vertical variable when the horizontal variable is $x$ .
Reading the constant straight from the intercept. If $c=\lg A$ , recover $A=10^c$ ; if $c=\ln A$ , recover $A=e^c$ .
Taking logarithms of a sum term by term. $\ln(T-20)\ne\ln T-\ln20$ . First isolate the exponential part.
Ignoring missing points. A transformed straight line can extend through coordinates that do not correspond to valid original values.
Overstating accuracy. Use a sensible line of best fit, well-separated line points and approximate answers for graph estimates.

Before finishing, write the original model with all constants recovered, check one original data pair or stated transformed point, and state any restriction introduced by division or logarithms.

Continue with topic practice

Choose the bank for your current year: Sec 3 Linear Law questions and worked solutions or Sec 4 Linear Law questions and worked solutions. Work on paper before opening a solution.