Home / IGCSE Add Maths 0606 / Straight-line graphs / Reducing to straight-line form
Cambridge IGCSE · Topic 7.4
IGCSE Add Maths 0606 / Topic 7 / IGCSE

Bend the axes, straighten the curve.

Many relationships, like \(y = Ax^{n}\) or \(y = Ab^{x}\), give curves, and you cannot read constants off a curve. Choose new quantities \(X\) and \(Y\) to plot, such as \(\lg x\) and \(\lg y\), and the graph becomes a straight line \(Y = mX + c\). Its gradient and intercept then give the unknown constants, and a given straight line can be turned back into the relationship.

Topic 7All straight-line graphs lessons →
The key idea

Rewrite the relationship as \(Y = mX + c\), where \(X\) and \(Y\) contain only the variables \(x\) and \(y\) and the unknown constants sit in \(m\) and \(c\). For \(y = Ax^{n}\): \(\lg y = n\lg x + \lg A\), so plot \(\lg y\) against \(\lg x\): gradient \(n\), intercept \(\lg A\). For \(y = Ab^{x}\): \(\ln y = (\ln b)x + \ln A\), so plot \(\ln y\) against \(x\): gradient \(\ln b\), intercept \(\ln A\). From two points on the line, \(m = \dfrac{Y_2 - Y_1}{X_2 - X_1}\) and \(c = Y_1 - mX_1\).

lg x lg y O 0.5 1 1.5 2 1 2 3 1.6 − 0.4 = 1.2 2.7 − 0.9 = 1.8 (0.4, 0.9) (1.6, 2.7) 0.3 Plot lg y against lg x for y = Axⁿ: lg y = n lg x + lg A gradient = 1.8 ÷ 1.2 = 1.5 so n = 1.5 intercept c = 0.9 − 1.5 × 0.4 = 0.3 = lg A so A = 100.3 ≈ 2.00
Fig. 1: for y = Ax^n, the graph of lg y against lg x is a straight line. Through the points (0.4, 0.9) and (1.6, 2.7) the gradient is 1.8 / 1.2 = 1.5, so n = 1.5, and the intercept on the lg y axis is 0.3 = lg A, so A = 10^0.3, about 2.00. The intercept is where lg x = 0, that is where x = 1
Section 01

Choose what to plot to make a line.

A straight line has the form \(Y = mX + c\): gradient \(m\), intercept \(c\) on the \(Y\)-axis. To reduce a relationship to this form, rearrange it until one side is a quantity \(Y\) made only from \(x\) and \(y\), and the other side is (constant) \(\times\) (a quantity \(X\) made only from \(x\) and \(y\)) + (constant). For powers and exponentials, take logs first: the laws of logarithms turn \(y = Ax^{n}\) into \(\lg y = \lg A + n\lg x\), which is a straight line in \(\lg x\) and \(\lg y\). Natural logs or base 10 logs both work, as long as you use the same base on both sides.

RelationshipStraight-line formPlot \(Y\)against \(X\)GradientIntercept
\(y = Ax^{n}\)\(\lg y = n\lg x + \lg A\)\(\lg y\)\(\lg x\)\(n\)\(\lg A\)
\(y = Ab^{x}\)\(\ln y = (\ln b)x + \ln A\)\(\ln y\)\(x\)\(\ln b\)\(\ln A\)
\(y = Ae^{kx}\)\(\ln y = kx + \ln A\)\(\ln y\)\(x\)\(k\)\(\ln A\)
\(y = ax^{2} + bx\)\(\dfrac{y}{x} = ax + b\)\(\dfrac{y}{x}\)\(x\)\(a\)\(b\)
Worked example Powers and exponentials
The variables \(x\) and \(y\) are related by (a) \(y = Ax^{n}\) (b) \(y = Ab^{x}\), where \(A\), \(n\) and \(b\) are constants. In each case state what to plot to obtain a straight line, and give its gradient and intercept.
  1. (a) Take lg of both sides: \(\lg y = \lg (Ax^{n}) = \lg A + \lg x^{n} = \lg A + n\lg x\).
  2. Write it as \(Y = mX + c\): \(\lg y = n\lg x + \lg A\). Plot \(\lg y\) against \(\lg x\); the gradient is \(n\) and the intercept is \(\lg A\).
  3. (b) Take ln of both sides: \(\ln y = \ln A + \ln b^{x} = \ln A + x\ln b\).
  4. So \(\ln y = (\ln b)x + \ln A\). Plot \(\ln y\) against \(x\); the gradient is \(\ln b\) and the intercept is \(\ln A\). Here \(x\) itself is \(X\): it is already a power, so it is not logged.
Answer: (a) \(\lg y\) against \(\lg x\): gradient \(n\), intercept \(\lg A\)   (b) \(\ln y\) against \(x\): gradient \(\ln b\), intercept \(\ln A\)
Worked example No logs needed
The variables \(x\) and \(y\) satisfy \(y = ax^{2} + \dfrac{b}{x}\), where \(a\) and \(b\) are constants. Show that plotting \(xy\) against \(x^{3}\) gives a straight line, and state its gradient and intercept.
  1. The \(\dfrac{b}{x}\) term stops this being a line in \(x^{2}\), so clear the fraction: multiply every term by \(x\).
  2. \(xy = ax^{3} + b\).
  3. This is \(Y = mX + c\) with \(Y = xy\) and \(X = x^{3}\), both made only from the variables.
Answer: \(xy = ax^{3} + b\): a straight line with gradient \(a\) and intercept \(b\)
Your turn
The variables satisfy \(y = Ae^{-2x}\). What should be plotted to give a straight line, and what are its gradient and intercept?
Examiner trap

\(X\) and \(Y\) must not contain the unknown constants: for \(y = Ax^{n}\), plotting \(y\) against \(x^{n}\) is no use because you do not know \(n\). And the intercept is \(\lg A\), not \(A\): \(\lg (Ax^{n})\) is \(\lg A + n\lg x\), never \(A + n\lg x\) or \(A\lg x^{n}\).

Section 02

Two points give the constants.

Exam questions do not ask you to measure a gradient from a drawn graph; they give two points that the transformed line passes through, as values of \(X\) and \(Y\). Find the gradient \(m = \dfrac{Y_2 - Y_1}{X_2 - X_1}\), then the intercept from \(c = Y_1 - mX_1\). Finally match \(m\) and \(c\) to the straight-line form and undo the logs: if \(\lg A = c\) then \(A = 10^{c}\); if \(\ln A = c\) then \(A = e^{c}\). Keep the exact value of \(c\) for later working and only round the final answer to 3 significant figures.

Worked example lg y against lg x
The variables \(x\) and \(y\) are related by \(y = Ax^{n}\). The graph of \(\lg y\) against \(\lg x\) is a straight line passing through the points \((0.4,\ 0.9)\) and \((1.6,\ 2.7)\), as in Fig. 1. Find \(n\) and \(A\), and hence find \(y\) when \(x = 5\).
  1. Straight-line form: \(\lg y = n\lg x + \lg A\), so the gradient is \(n\) and the intercept is \(\lg A\).
  2. Gradient: \(n = \dfrac{2.7 - 0.9}{1.6 - 0.4} = \dfrac{1.8}{1.2} = 1.5\).
  3. Intercept: \(\lg A = 0.9 - 1.5 \times 0.4 = 0.3\), so \(A = 10^{0.3} = 1.995\ldots \approx 2.00\).
  4. When \(x = 5\): \(\lg y = 1.5\lg 5 + 0.3 = 1.348\ldots\), so \(y = 10^{1.348\ldots} = 22.3\) (3 s.f.). Using the rounded \(A = 2\) would give \(22.4\), so keep \(10^{0.3}\) exact.
Answer: \(n = 1.5\), \(A = 10^{0.3} \approx 2.00\); \(y = 22.3\) when \(x = 5\)
Worked example ln y against x
The variables \(x\) and \(y\) are related by \(y = Ab^{x}\). The graph of \(\ln y\) against \(x\) is a straight line passing through \((2,\ 3.4)\) and \((5,\ 4.6)\). Find \(A\) and \(b\).
  1. Straight-line form: \(\ln y = (\ln b)x + \ln A\).
  2. Gradient: \(\ln b = \dfrac{4.6 - 3.4}{5 - 2} = \dfrac{1.2}{3} = 0.4\), so \(b = e^{0.4} = 1.49\) (3 s.f.).
  3. Intercept: \(\ln A = 3.4 - 0.4 \times 2 = 2.6\), so \(A = e^{2.6} = 13.5\) (3 s.f.).
  4. So \(y = 13.5 \times 1.49^{x}\), or exactly \(y = e^{2.6} \times \left(e^{0.4}\right)^{x} = e^{2.6 + 0.4x}\).
Answer: \(A = e^{2.6} \approx 13.5\), \(b = e^{0.4} \approx 1.49\)
Your turn
For \(y = Ax^{n}\), the graph of \(\lg y\) against \(\lg x\) passes through \((1,\ 2)\) and \((3,\ 8)\). Find \(n\) and \(A\).
Examiner trap

The intercept \(c\) is \(\lg A\), not \(A\): answering \(A = 0.3\) in Fig. 1 loses the mark, since \(A = 10^{0.3}\). Undo the log in the SAME base you took: a line in \(\ln y\) needs \(e^{c}\), not \(10^{c}\). And gradient is rise over run, \(\dfrac{\Delta Y}{\Delta X}\); \(\dfrac{1.2}{1.8}\) instead of \(\dfrac{1.8}{1.2}\) gives the reciprocal.

Stage 1 · Learn

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Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

For \(y = Ax^{n}\), the graph of \(\lg y\) against \(\lg x\) is a straight line. What is its gradient?

Quick check+10 XP

For \(y = Ab^{x}\), the graph of \(\ln y\) against \(x\) is a straight line. What is its intercept on the \(\ln y\)-axis?

Quick check+10 XP

For \(y = Ax^{n}\), the line of \(\lg y\) against \(\lg x\) has gradient 2 and intercept 1. Find \(A\).

Quick check+10 XP

When \(y^{2}\) is plotted against \(x\), the straight line has gradient 3 and passes through \((0,\ 4)\). Which equation connects \(x\) and \(y\)?

Section 03

From a line back to the equation.

Sometimes the question goes the other way: it tells you which quantities were plotted and gives two points on the line, and you must find the relationship between \(x\) and \(y\). Work out \(m\) and \(c\) exactly as before, then write \(Y = mX + c\) with the plotted quantities put back in for \(X\) and \(Y\). For example, if \(y^{2}\) is plotted against \(x^{3}\), the line is \(y^{2} = mx^{3} + c\). Then use this equation to find \(y\) for a given \(x\), or \(x\) for a given \(y\).

Worked example y squared against x cubed
The variables \(x\) and \(y\) are such that when \(y^{2}\) is plotted against \(x^{3}\) a straight line is obtained, passing through \((2,\ 5)\) and \((6,\ 13)\). Find \(y^{2}\) in terms of \(x\), and hence find the positive value of \(y\) when \(x = 1.5\).
  1. Here \(X = x^{3}\) and \(Y = y^{2}\), so the points mean \(x^{3} = 2,\ y^{2} = 5\) and \(x^{3} = 6,\ y^{2} = 13\).
  2. Gradient: \(m = \dfrac{13 - 5}{6 - 2} = 2\). Intercept: \(c = 5 - 2 \times 2 = 1\).
  3. So \(y^{2} = 2x^{3} + 1\).
  4. When \(x = 1.5\): \(y^{2} = 2 \times 3.375 + 1 = 7.75\), so \(y = \sqrt{7.75} = \dfrac{\sqrt{31}}{2} = 2.78\) (3 s.f.).
Answer: \(y^{2} = 2x^{3} + 1\); \(y = \frac{\sqrt{31}}{2} \approx 2.78\)
Worked example An exponential of y
When \(e^{2y}\) is plotted against \(x^{2}\), a straight line is obtained through \((1,\ 7)\) and \((4,\ 16)\). Express \(e^{2y}\) in terms of \(x\) and find the exact value of \(y\) when \(x = 2\).
  1. Gradient: \(m = \dfrac{16 - 7}{4 - 1} = 3\). Intercept: \(c = 7 - 3 \times 1 = 4\).
  2. So \(e^{2y} = 3x^{2} + 4\).
  3. When \(x = 2\): \(e^{2y} = 3 \times 4 + 4 = 16\).
  4. Take ln: \(2y = \ln 16\), so \(y = \tfrac{1}{2}\ln 16 = \ln 4\) (\(\approx 1.39\)).
Answer: \(e^{2y} = 3x^{2} + 4\); \(y = \ln 4\)
Your turn
When \(y^{3}\) is plotted against \(\ln x\), a straight line is obtained through \((1,\ 5)\) and \((3,\ 11)\). Find \(y^{3}\) in terms of \(\ln x\), and find the exact value of \(x\) when \(y = 2\).
Examiner trap

Put the plotted quantities back, not plain \(x\) and \(y\): the line for \(y^{2}\) against \(x^{3}\) gives \(y^{2} = 2x^{3} + 1\), not \(y = 2x + 1\). The given points are values of \(X\) and \(Y\): the point \((2,\ 5)\) means \(x^{3} = 2\), not \(x = 2\). And finish the job: if the line gives \(e^{2y} = 16\), the question wants \(y\), so take ln and halve.

Stage 2 · Exam

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The variables \(x\) and \(y\) are related by \(y = Ax^{n}\). The graph of \(\lg y\) against \(\lg x\) is a straight line through \((0.2,\ 1.1)\) and \((0.8,\ 2.9)\). Find \(n\) and \(A\).

Exam style+20 XP

The variables \(x\) and \(y\) are related by \(y = Ab^{x}\). The graph of \(\ln y\) against \(x\) is a straight line through \((1,\ 2.5)\) and \((3,\ 3.7)\). Find \(A\) and \(b\), each to 3 significant figures.

Exam style+20 XP

The variables satisfy \(y = ax^{2} + bx\), where \(a\) and \(b\) are constants. Which graph is a straight line with gradient \(a\) and intercept \(b\)?

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When \(y^{3}\) is plotted against \(\ln x\), a straight line is obtained through \((2,\ 9)\) and \((4,\ 15)\). Find the value of \(y\) when \(x = 10\).

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Curves straightened, constants found.

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Six original Cambridge-style questions
Choosing what to plot, constants from two points on \(\lg y\) against \(\lg x\) and \(\lg y\) against \(x\), a line back to an equation, a quadratic relationship through \(\frac{y}{x}\), and a pendulum model.