Maths › Algebra and functions › Graphs, proportion and transformations
Graphs, proportion and transformations
Curve sketching is reading, not drawing. Factors name the roots, squared factors touch instead of crossing, reciprocal curves settle onto lines without ever arriving, and four standard transformations let one known graph stand in for a whole family of relatives.
Builds on Polynomials and the factor theorem.
IN THIS TOPIC
- Sketch a cubic or quartic from its factors, with crossings and touches in the right places.
- Sketch y = a/x and y = a/x2 and state their asymptotes.
- Set up direct, inverse and inverse-square proportion, and fit the constant from one data point.
- Apply the four transformations of y = f(x), singly and in combination.
COMMON MISCONCEPTION
Replacing x by (x + a) moves the graph a units to the right.
Sketching from the factors
A polynomial in factorised form gives you its sketch directly. The leading term fixes the ends, so a positive cubic runs from bottom-left to top-right and a positive quartic from top-left to top-right, and each factor plants a root. The power on the factor is the instruction. A simple factor crosses the axis. A squared factor touches it and turns back, as (x + 2)2 did in the last lesson's cubic.
WORKED EXAMPLE
A cubic read straight off its factors
Sketch y = (x + 1)(x − 2)2, marking every axis meeting.
Roots first. A crossing at x = −1 from the simple factor, a touch at x = 2 from the squared one.
The y-intercept comes from x = 0, namely 1 × 4 = 4.
A positive cubic with those features rises through −1, peaks, dips to kiss the axis at 2, then climbs away. No table of values needed anywhere.
Intersections keep the meaning they had in the simultaneous equations lesson. Where two sketches cross, the pair of equations has a solution, so counting crossings on a rough sketch counts solutions in advance.
Reciprocal curves and proportion
The graphs of y = a/x and y = a/x2 bring in a new feature, the asymptote, a line the curve settles onto: the gap between them shrinks to nothing as you go out far enough. Both axes play that role here. Near x = 0 the curves blow up, and for large x they flatten towards the horizontal axis.
Say it that way round, because "a line the curve never touches" is the version most people carry out of GCSE, and it is not true. Nothing stops a curve crossing its own asymptote. The curve y = (sin x)/x cuts through y = 0 again and again on its way out and still settles onto it, because the crossings shrink for ever. What makes a line an asymptote is where the curve ends up, not whether it keeps its distance on the way there. These two reciprocal curves happen never to meet their axes, which makes them a good first example and a bad definition.
Sign is what separates the two, and a carries a sign of its own. For y = a/x the sign of y is the sign of a times the sign of x, so a positive a puts the branches in the first and third quadrants and a negative a puts them in the second and fourth. Either way there is one branch in each of two opposite quadrants. For y = a/x2 the square makes the denominator positive, so y takes the sign of a alone. Both branches then sit on one side of the horizontal axis, above it when a is positive and below it when a is negative.
These are also the curves of proportion. Direct proportion, y = kx, is a straight line through the origin. Inverse proportion, y = k/x, is the reciprocal curve. Inverse square, y = k/x2, is its one-sided cousin. One data point fixes k and the whole curve follows.
WORKED EXAMPLE
Inverse square proportion, pinned by one point
y is inversely proportional to x2, and y = 4 when x = 3. Find y when x = 2.
Write the relationship with a constant, y = k/x2, then substitute the known pair. k = 4 × 9 = 36.
So y = 36/x2, and at x = 2, y = 36/4 = 9.
A modest change in x produced a large change in y, because the square in the denominator bites twice. Halve x and y quadruples.
The four transformations
Every change to a graph's equation moves its picture in a predictable way, and four moves cover the whole specification. The two stretches are stated below for a > 0, which is the case exam questions almost always set.
| Equation | Effect on the graph of y = f(x), for a > 0 |
|---|---|
| y = f(x) + a | translate a units up |
| y = f(x + a) | translate a units to the left |
| y = af(x) | stretch ×a vertically |
| y = f(ax) | stretch ×1/a horizontally |
A negative a needs one extra word. y = af(x) is a stretch of scale factor |a| vertically and a reflection in the x-axis, and y = f(ax) is a stretch of scale factor 1/|a| horizontally and a reflection in the y-axis. So y = −2f(x) doubles every height and then flips the picture upside down. A stretch of scale factor −2 is not a thing to write. The translations need no such caution, since a negative a there simply translates the other way. Zero is the dead end. y = 0f(x) flattens the whole curve onto the x-axis, and y = f(0x) is the constant f(0), so neither is a stretch and neither can be undone.
The second row is the one to argue with. Adding inside the bracket feeds the function an x from further to the right, so the picture slides left, not right. Changes outside the bracket act on y and do what they say. Changes inside act on x and do the opposite.
WORKED EXAMPLE
Tracking a single point
The curve y = f(x) passes through (2, 5). Find the corresponding point on y = f(x − 1) + 3.
Inside the bracket, x − 1 must equal 2, so x = 3. The picture has moved right by 1.
Outside, 3 is added to the output, lifting 5 to 8. The point is (3, 8).
Tracking one point through the brackets is more reliable than memorising rules, and it is how the mark scheme checks a sketch anyway.
Combinations, and the order they happen in
Two transformations at once cause trouble only when both act inside the bracket. Outside, a stretch and a translation behave themselves. Inside they do not commute, and f(2x + 6) is the standard test of whether you know that.
GUIDED PRACTICE
A transformed reciprocal
Sketch y = 1/(x − 2) + 1, stating the equations of its asymptotes and where it meets the axes, before opening the working.
Show the working
Start from y = 1/x and read the moves. Right 2, up 1, so the asymptotes travel too, to x = 2 and y = 1.
Axis meetings come from the equation. At x = 0, y = −1/2 + 1 = 1/2. Setting y = 0 gives 1/(x − 2) = −1, so x = 1.
The curve meets the axes at (0, 1/2) and (1, 0), with one branch below-left of the asymptote crossing and one above-right.
Transformed asymptotes are the skeleton of the sketch. Draw them dashed first and the branches hang off them naturally.
INDEPENDENT PRACTICE
A combination, in the right order
The curve y = f(x) has a root at x = 0. Describe two correct sequences of transformations taking y = f(x) to y = f(2x + 6), and state where the root ends up.
Show the working
Read f(2x + 6) as f(2(x + 3)). One route translates 6 left, then stretches horizontally ×1/2. The other stretches ×1/2 first, then translates only 3 left.
Either way the root lands where 2x + 6 = 0, at x = −3.
Stretching first and then translating 6 left would park the root at −6, which is wrong. Checking against 2x + 6 = 0 exposes that in seconds.
ASSESSMENT FOCUS
- Sketch from factors in three moves. Ends from the leading term, roots from the brackets, y-intercept from x = 0. Label all three or the accuracy marks go.
- A squared factor touches and a simple factor crosses. Examiners choose curves like x2(2x − 1)2 precisely to test that reading.
- Draw asymptotes dashed and state their equations in full, x = 2 and y = 1, never the bare numbers on their own.
- Proportion questions want the constant found first. Write y = k/x2, fix k from the given pair, then answer everything else from the fitted equation.
- Inside the bracket means horizontal and opposite; outside means vertical and literal. When a combination looks ambiguous, factorise the inside and track one point through it.
- Two rough sketches answer a “how many solutions” question faster than any algebra. Crossings are solutions, and a sketch is allowed to be rough as long as the features are right.
CHECK YOURSELF
The curve y = 3/x is transformed to y = 3/x − 2. State the equations of the asymptotes of the new curve, and find where it crosses the x-axis.
Show a hint
The whole curve moves down 2, asymptotes included; then set y = 0.
Show the answer
Subtracting 2 outside translates the curve down 2, so the asymptotes are x = 0 and y = −2.
Setting y = 0 gives 3/x = 2, so x = 3/2, and the curve crosses at (3/2, 0).
The vertical asymptote never moved, because nothing happened inside the function. That one line of reasoning is worth writing down in the answer.
The factors are the sketch. Simple roots cross, squared roots touch, and an asymptote is what the curve settles onto, which is not the same as a line it never touches.
Outside the bracket acts on y and does what it says. Inside acts on x and does the opposite.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the graphs, proportion and transformations questions page.
CHECK YOUR PROGRESS
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- Sketch a cubic or quartic from its factors, with crossings and touches in the right places.
- Sketch y = a/x and y = a/x2 and state their asymptotes.
- Set up direct, inverse and inverse-square proportion, and fit the constant from one data point.
- Apply the four transformations of y = f(x), singly and in combination.
Open the full revision checklist to see every objective in the course in one place.