This note covers finding the rule for a number sequence (linear, quadratic and geometric) and recognising, plotting and using the graphs of common functions. Sequences model growth and patterns, and graphs let you read off solutions and spot features such as turning points and asymptotes. Quadratic and geometric nth terms, and estimating gradients from tangents, are Extended content.
Key ideas
Linear sequences. A sequence with a constant difference \(d\) between terms has nth term \(dn+c\), where \(c\) is found by comparing: \(c=\) first term \(-\,d\). For \(7,11,15,19,\ldots\), the difference is 4, so the nth term is \(4n+3\). To test whether a number is in the sequence, set the nth term equal to it and check that \(n\) is a positive whole number.
Other sequences. Terms can also be found by a rule from the previous term, and some sequences use special numbers such as squares (\(n^2\)), cubes (\(n^3\)) or triangular numbers. If the multiplier between terms is constant, the sequence is geometric (Extended), and the nth term is \(ar^{n-1}\) with first term \(a\) and common ratio \(r\).
Quadratic sequences (Extended). If the first differences are not constant but the second differences are, the nth term is quadratic, \(an^2+bn+c\). The coefficient \(a\) is half of the second difference. Subtract \(an^2\) from each term. What is left is a linear sequence whose nth term gives \(bn+c\).
Graphs of functions. Know the shape and key features of each family.
- Linear \(y=mx+c\): a straight line with gradient \(m\) and \(y\)-intercept \(c\).
- Quadratic \(y=ax^2+bx+c\): a parabola, "U" shaped for \(a\gt0\) and "n" shaped for \(a\lt0\), with one turning point and a vertical line of symmetry through it.
- Cubic \(y=ax^3+\ldots\): an S-shaped curve that can have up to two turning points.
- Reciprocal \(y=\dfrac{a}{x}\): two separate branches that approach the axes. The axes are asymptotes, lines the curve gets closer to but never reaches.
- Exponential \(y=a^x\) (Extended): passes through \((0,1)\), grows quickly for \(a\gt1\), and has the \(x\)-axis as an asymptote.
Using graphs. The roots (solutions of \(y=0\)) are where the curve meets the \(x\)-axis. To solve an equation such as \(f(x)=k\), draw the line \(y=k\) and read the \(x\)-values where it meets the curve. To find the gradient of a curve at a point (Extended), draw a tangent, a straight line that just touches the curve there, then calculate rise \(\div\) run for the tangent. To plot a curve, make a table of values, mark points accurately, and join them with a smooth curve, not with ruled lines.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| Linear nth term \(dn+c\) | \(d\) is the common difference, \(c=\) first term \(-\,d\). |
| Quadratic nth term \(an^2+bn+c\) | \(a=\tfrac12\times\) second difference (Extended). |
| Geometric nth term \(ar^{n-1}\) | First term \(a\), common ratio \(r\) (Extended). |
| Root (zero) | An \(x\)-value where \(y=0\); the graph crosses the \(x\)-axis there. |
| Turning point | A maximum or minimum point where the curve changes direction. |
| Asymptote | A line that a curve approaches without reaching. |
| \(x=-\dfrac{b}{2a}\) | The line of symmetry of \(y=ax^2+bx+c\); the \(x\)-coordinate of its turning point. |
Worked examples
Example 1 A savings plan has monthly balances (in pounds) of 7, 11, 15, 19, \(\ldots\) Find the nth term, and decide whether 203 and 100 appear in the sequence.
The difference is \(11-7=4\), so the nth term starts with \(4n\).
When \(n=1\), \(4n=4\), but the first term is 7, so add 3. The nth term is \(4n+3\).
Check: \(n=4\) gives \(16+3=19\). Correct.
For 203: \(4n+3=203\), so \(4n=200\) and \(n=50\). This is a whole number, so 203 is the 50th term.
For 100: \(4n+3=100\), so \(4n=97\) and \(n=24.25\). This is not a whole number, so 100 is not in the sequence.
Answer: \(4n+3\); 203 is in the sequence (50th term), 100 is not
Example 2 (Extended) Find the nth term of the sequence \(5,12,23,38,57,\ldots\)
First differences: \(7,11,15,19\). Second differences: \(4,4,4\). These are constant, so the sequence is quadratic.
Half of the second difference is 2, so the nth term begins \(2n^2\).
The values of \(2n^2\) are \(2,8,18,32,50\). Subtract these from the sequence: \(3,4,5,6,7\).
This remainder is linear with difference 1 and first term 3, so it is \(n+2\).
So the nth term is \(2n^2+n+2\). Check \(n=3\): \(18+3+2=23\). Correct. Check \(n=5\): \(50+5+2=57\). Correct.
Answer: \(2n^2+n+2\)
Example 3 Consider the graph of \(y=x^2-2x-3\). (a) Find the \(y\)-intercept, the roots and the turning point. (b) Use the graph to solve \(x^2-2x-3=2\), giving answers to 3 significant figures.
(a) The \(y\)-intercept is where \(x=0\): \(y=-3\).
The roots are where \(y=0\): \(x^2-2x-3=(x-3)(x+1)=0\), so \(x=3\) and \(x=-1\).
The line of symmetry is halfway between the roots, \(x=\dfrac{-1+3}{2}=1\). At \(x=1\): \(y=1-2-3=-4\). The turning point is \((1,-4)\), a minimum, because the curve is "U" shaped.
(b) Draw the horizontal line \(y=2\) on the same axes. Its crossings with the curve give the solutions.
Algebraically, \(x^2-2x-3=2\) becomes \(x^2-2x-5=0\). Then \(x=\dfrac{2\pm\sqrt{4+20}}{2}=1\pm\sqrt{6}\).
\(1+\sqrt6=3.45\) and \(1-\sqrt6=-1.45\) (3 s.f.). A graph read carefully should give values close to these.
Answer: (a) \(y\)-intercept \(-3\), roots \(-1\) and \(3\), minimum \((1,-4)\); (b) \(x\approx3.45\) or \(x\approx-1.45\)
Common errors
- Finding the nth term as \(n+d\) instead of \(dn+c\), or forgetting to subtract from the first term to get \(c\).
- Writing "\(n=24.25\)" and still saying the number is in the sequence. The position must be a positive whole number.
- For a quadratic sequence, using the second difference as \(a\) instead of half of it.
- Joining plotted points with a ruler, or with a curve that has flat spikes at the turning point. Use a smooth curve.
- Letting a reciprocal graph touch or cross an asymptote, or joining its two branches together.
Exam tip: When you find an nth term, substitute \(n=1\) and \(n=2\) back in. If either term is wrong, you know to look for the error before moving on.
Check yourself
- Find the nth term of \(10,7,4,1,\ldots\)
Answer
The difference is \(-3\), so the nth term starts \(-3n\). At \(n=1\): \(-3\), but the first term is 10, so add 13. The nth term is \(-3n+13\). - (Extended) A geometric sequence begins \(3,6,12,\ldots\) Find the next term and the nth term.
Answer
The ratio is 2, so the next term is 24. The nth term is \(3\times2^{n-1}\). - (Extended) Find the nth term of \(3,9,19,33,\ldots\)
Answer
First differences: \(6,10,14\). Second differences: \(4,4\), so \(a=2\). Subtract \(2n^2=2,8,18,32\) from the terms to get \(1,1,1,1\), which is constant. So the nth term is \(2n^2+1\). - Find the turning point of \(y=x^2+4x+1\).
Answer
The line of symmetry is \(x=-\dfrac{4}{2(1)}=-2\). Then \(y=4-8+1=-3\). The turning point is \((-2,-3)\), a minimum.