P1.3 Polynomials, curve sketching and transformations
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This page covers expanding and factorising cubics, sketching cubic and reciprocal graphs, counting solutions from intersections, and the four standard transformations (Pure Mathematics 1, AS). Turning points by differentiation are in P1.6; the factor theorem and algebraic division are in P2; combined transformations and modulus graphs are in P3.
What this page covers
7 points
- Expanding products of brackets and factorising cubics, using \(f(x)\) notation
- Sketching cubic graphs, including repeated roots, with all axis intercepts
- Sketching \(y=\dfrac{k}{x}\) and \(y=\dfrac{k}{x^2}\) with their asymptotes, for positive and negative \(k\)
- Using the points where two graphs meet to state the number of solutions of an equation
- The transformations \(y=af(x)\), \(y=f(x)+a\), \(y=f(x+a)\) and \(y=f(ax)\)
- The effect of each transformation on intercepts, turning points and asymptotes
- Applying the same transformations to the graphs of \(\sin x\), \(\cos x\) and \(\tan x\)
In one minute
6 points
- Once a cubic is factorised you can sketch it in seconds, and once you know how each transformation acts you can move that sketch anywhere; a pair of sketches then tells you how many solutions an equation has without solving it.
- Factorise first: a cubic with no constant term has a common factor \(x\); take it out and factorise the quadratic that is left.
- Cubic shape: a positive \(x^3\) coefficient starts bottom left and ends top right; a negative one does the reverse. A single root crosses the axis; a squared (repeated) root touches it.
- \(y=\dfrac{k}{x}\) and \(y=\dfrac{k}{x^2}\) have asymptotes \(x=0\) and \(y=0\). For \(k>0\), \(\dfrac{k}{x}\) is in the 1st and 3rd quadrants; \(\dfrac{k}{x^2}\) is above the \(x\)-axis on both sides.
- The number of solutions of \(f(x)=g(x)\) equals the number of points where \(y=f(x)\) and \(y=g(x)\) meet.
- Outside the bracket acts on \(y\) as you expect: \(f(x)+a\) moves up \(a\); \(af(x)\) multiplies \(y\) by \(a\). Inside the bracket acts on \(x\) the opposite way: \(f(x+a)\) moves left \(a\); \(f(ax)\) divides \(x\) by \(a\).
Contents
13 sections
- How do you expand and factorise polynomials?
- How do you sketch a cubic graph?
- What do the graphs of y = k/x and y = k/x² look like?
- How can a sketch tell you the number of solutions?
- How do the four transformations change a graph?
- What happens to asymptotes and trig graphs?
- How do the ideas combine in a full question?
- Exam tips for P1.3
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
How do you expand and factorise polynomials?
Function notation \(f(x)\) names a polynomial so you can substitute into it. If \(f(x)=2x^3-3x^2-23x+12\), then \(f(1)=2-3-23+12=-12\). This notation is used throughout the course, and it is what makes transformations such as \(f(x+2)\) easy to write down.
Expanding three brackets
Multiply two of the brackets first, simplify, then multiply the result by the third. Every term of one bracket must meet every term of the other, so a bracket with two terms times a bracket with three terms gives six products before you collect like terms.
Factorising a cubic in P1
In this unit, cubics to factorise always have a common factor (usually \(x\)). Take it out, then factorise the quadratic that remains. Watch for two special patterns:
- a difference of two squares, \(x^2-a^2=(x-a)(x+a)\);
- a perfect square, such as \(x^2-6x+9=(x-3)^2\).
"Factorise completely" means that no bracket can be factorised any further.
Expand and simplify \((x+3)(2x-1)(x-4)\), giving your answer in the form \(ax^3+bx^2+cx+d\).
Multiply two brackets: \((x+3)(x-4)=x^2-4x+3x-12=x^2-x-12\).
Multiply by the third bracket, term by term: \((2x-1)(x^2-x-12)=2x^3-2x^2-24x-x^2+x+12\).
Collect like terms: \(2x^3-3x^2-23x+12\)
Check with \(x=1\): the brackets give \(4\times1\times(-3)=-12\), and the answer gives \(2-3-23+12=-12\) ✓. Calling this \(f(x)\), we also know \(f(-3)=f(4)=f(\tfrac12)=0\), because each value makes one bracket zero.
Factorise completely (a) \(3x^3-12x\) (b) \(x^3-6x^2+9x\) (c) \(2x^3+5x^2-3x\).
(a) Common factor \(3x\), then a difference of two squares: \(3x(x^2-4)=\) \(3x(x-2)(x+2)\)
(b) Common factor \(x\), then a perfect square: \(x(x^2-6x+9)=\) \(x(x-3)^2\)
(c) Common factor \(x\). For \(2x^2+5x-3\), find two numbers with product \(2\times(-3)=-6\) and sum 5: they are 6 and \(-1\). So \(x(2x^2+6x-x-3)=x\bigl(2x(x+3)-(x+3)\bigr)=\) \(x(2x-1)(x+3)\)
Stopping at \(3x(x^2-4)\). The question says "completely", and \(x^2-4\) is a difference of two squares, so the answer is not finished until it is \(3x(x-2)(x+2)\).
How do you sketch a cubic graph?
A sketch is not a plot. It shows the correct shape and the key points, labelled with their coordinates, but it does not need to be to scale. For a cubic, the factorised form gives you almost everything at once.
What a sketch must show
- The right shape. For \(ax^3+\ldots\) with \(a>0\): as \(x\to\infty\), \(y\to\infty\), and as \(x\to-\infty\), \(y\to-\infty\). For \(a<0\) the ends are reversed. Find the sign of \(a\) by multiplying the \(x\)-terms of each bracket.
- Where it meets the \(x\)-axis: from the factorised form, set \(y=0\).
- Where it meets the \(y\)-axis: put \(x=0\). For \(y=(x-p)(x-q)(x-r)\) this is \(-pqr\).
- Repeated roots. A factor \((x-p)^2\) means the curve touches the \(x\)-axis at \(x=p\), with a turning point on the axis. A factor \((x-p)^3\), as in \(y=(x-1)^3\), means the curve flattens and crosses there.
The turning points do not need exact coordinates in a sketch (finding them needs differentiation, P1.6), but a maximum must sit above the axis and a minimum below it wherever the roots force that.
Sketch the curve with equation \(y=(3-x)(x+1)^2\), showing the coordinates of the points where it meets the axes.
Shape. The \(x^3\) term is \((-x)(x^2)=-x^3\): a negative cubic, starting top left and ending bottom right.
\(x\)-axis, \(y=0\). \(x=3\) (single root: crosses) and \(x=-1\), a repeated root (touches).
\(y\)-axis, \(x=0\). \(y=3\times1^2=3\), so \((0, 3)\).
Draw it. The curve comes down from the top left, touches the axis at \((-1, 0)\), rises through \((0, 3)\) to a maximum and comes down to cross at \((3, 0)\). Touches at \((-1, 0)\), crosses at \((3, 0)\), meets the \(y\)-axis at \((0, 3)\)
Expanded, \(y=-x^3+x^2+5x+3\): the constant term 3 confirms the \(y\)-intercept.
At a touching root, draw the curve genuinely turning on the axis, like the vertex of a parabola: not crossing it and not stopping short of it. Write the coordinates on the sketch itself, not only in your working.
What do the graphs of y = k/x and y = k/x² look like?
| \(y=\dfrac{k}{x}\) | \(y=\dfrac{k}{x^2}\) | |
|---|---|---|
| \(k>0\) | two branches, 1st and 3rd quadrants | two branches, both above the \(x\)-axis (1st and 2nd quadrants) |
| \(k<0\) | 2nd and 4th quadrants | both below the \(x\)-axis (3rd and 4th quadrants) |
| Asymptotes | \(x=0\) and \(y=0\) | \(x=0\) and \(y=0\) |
| Symmetry | rotational, order 2, about \(O\) | reflection in the \(y\)-axis |
| Meets the axes? | never | never |
Why \(x=0\) is an asymptote: as \(x\to0\), the fraction becomes enormous, and \(x=0\) itself gives no value. Why \(y=0\) is an asymptote: as \(x\to\pm\infty\) the fraction tends to 0 but never equals it, because \(k\ne0\).
Compare the two families: \(y=\dfrac{k}{x^2}\) falls away faster than \(y=\dfrac{k}{x}\) for \(|x|>1\) and climbs faster near the \(y\)-axis. Both pass through \((1, k)\).
How can a sketch tell you the number of solutions?
Every point where \(y=f(x)\) and \(y=g(x)\) meet has an \(x\)-coordinate that satisfies \(f(x)=g(x)\). So the number of real solutions of \(f(x)=g(x)\) equals the number of intersection points, and you can count them without solving the equation.
Questions often hide this link. An equation such as \(x^4-4x^2-3=0\) is really "cubic \(=\) reciprocal" after rearranging, and you are expected to show that rearrangement line by line. When you multiply through by \(x\) or \(x^2\), note that \(x\ne0\): neither reciprocal curve exists at \(x=0\), so no solutions are lost or gained.
(a) On the same axes, sketch \(y=x(x^2-4)\) and \(y=\dfrac{3}{x}\).
(b) Show that the \(x\)-coordinates of the points of intersection satisfy \(x^4-4x^2-3=0\).
(c) State the number of real solutions of this equation and find them exactly.
(a) Cubic: \(x(x-2)(x+2)\), positive, roots \(-2\), 0, 2. Reciprocal: 1st and 3rd quadrants, asymptotes the axes. See the figure.
(b) Set the two expressions equal and multiply by \(x\) (\(x\ne0\)): \(x(x^2-4)=\dfrac{3}{x}\Rightarrow x^4-4x^2=3\Rightarrow x^4-4x^2-3=0\) as required.
(c) The curves meet twice, so there are 2 real solutions. Solve as a quadratic in \(x^2\): \(x^2=\dfrac{4\pm\sqrt{16+12}}{2}=2\pm\sqrt7\).
\(2-\sqrt7<0\) is impossible for \(x^2\), so \(x=\pm\sqrt{2+\sqrt7}\) (\(\approx\pm2.16\))
The algebra confirms the sketch: one positive and one negative solution, matching \(Q\) and \(P\).
Giving four solutions because a quartic "has four roots". A quartic has at most four real roots; here two values of \(x^2\) are found, but one is negative, so it gives no real \(x\).
How do the four transformations change a graph?
| New graph | Transformation | Point \((x, y)\) moves to |
|---|---|---|
| \(y=f(x)+a\) | translation by \(a\) units in the positive \(y\)-direction (up if \(a>0\)) | \((x, y+a)\) |
| \(y=f(x+a)\) | translation by \(a\) units in the negative \(x\)-direction (left if \(a>0\)) | \((x-a, y)\) |
| \(y=af(x)\) | stretch parallel to the \(y\)-axis, scale factor \(a\) | \((x, ay)\) |
| \(y=f(ax)\) | stretch parallel to the \(x\)-axis, scale factor \(\tfrac1a\) | \(\left(\tfrac{x}{a}, y\right)\) |
Outside the function (\(f(x)+a\), \(af(x)\)) the change happens to \(y\), exactly as written. Inside the function (\(f(x+a)\), \(f(ax)\)) the change happens to \(x\), and it is the inverse: \(+a\) moves left, \(\times a\) divides by \(a\). The reason: the new graph reaches the old height \(f(p)\) when \(x+a=p\), that is at \(x=p-a\), which is \(a\) units to the left.
- \(a=-1\) gives reflections: \(-f(x)\) reflects in the \(x\)-axis; \(f(-x)\) reflects in the \(y\)-axis.
- Points on the axis that is not being stretched stay put: \(x\)-intercepts are unchanged by \(af(x)\); the \(y\)-intercept is unchanged by \(f(ax)\).
- A turning point stays a turning point (a maximum stays a maximum unless \(a<0\) flips it).
The curve \(y=f(x)\), where \(f(x)=x^3-3x^2-9x\), passes through the origin, has a maximum at \((-1, 5)\) and a minimum at \((3, -27)\). Write down the coordinates of the turning points of (a) \(y=f(x-2)\) (b) \(y=3f(x)\) (c) \(y=f(\tfrac12x)\). (d) Find the number of distinct real roots of \(f(x)+27=0\).
(a) Translation 2 units in the positive \(x\)-direction: maximum \((1, 5)\), minimum \((5, -27)\)
(b) Stretch parallel to the \(y\)-axis, scale factor 3: maximum \((-1, 15)\), minimum \((3, -81)\)
(c) Stretch parallel to the \(x\)-axis, scale factor \(1\div\tfrac12=2\): maximum \((-2, 5)\), minimum \((6, -27)\)
(d) \(y=f(x)+27\) moves the minimum to \((3, 0)\), on the axis, and the maximum to \((-1, 32)\). The curve touches the axis at \(x=3\) and crosses once to the left: 2 distinct roots
Check: \(x^3-3x^2-9x+27=(x-3)^2(x+3)\), with roots 3 (repeated) and \(-3\).
To check an "inside the bracket" answer, substitute back: for \(y=f(x-2)\) at \(x=1\), you get \(f(1-2)=f(-1)=5\), the old maximum. If the substitution does not land on the old point, the shift went the wrong way.
What happens to asymptotes and trig graphs?
Asymptotes move with the graph
- A vertical asymptote \(x=c\) is moved by \(f(x+a)\) to \(x=c-a\), and by \(f(ax)\) to \(x=\tfrac{c}{a}\).
- A horizontal asymptote \(y=d\) is moved by \(f(x)+a\) to \(y=d+a\), and by \(af(x)\) to \(y=ad\).
- An asymptote along an axis that is being stretched towards stays where it is: \(y=0\) is still an asymptote of \(\dfrac{5}{x}\), and \(x=0\) is still one of \(\dfrac{1}{2x}\).
- After moving the asymptotes, find any new intercepts by substitution: a translated reciprocal curve can now cross an axis.
For each curve, state the transformation of \(y=\dfrac{k}{x}\) or \(y=\dfrac{k}{x^2}\), the equations of its asymptotes and the coordinates of any points where it meets the axes. (a) \(y=\dfrac{3}{x+2}\) (b) \(y=\dfrac{4}{x^2}-1\)
(a) \(f(x+2)\) with \(f(x)=\dfrac{3}{x}\): a translation 2 units in the negative \(x\)-direction. Asymptotes \(x=-2\) and \(y=0\). \(x=0\) gives \(y=\tfrac32\). It never meets the \(x\)-axis, since 3 divided by anything is not 0. Asymptotes \(x=-2\), \(y=0\); meets the axes at \((0, 1.5)\) only
(b) \(g(x)-1\) with \(g(x)=\dfrac{4}{x^2}\): a translation 1 unit in the negative \(y\)-direction. Asymptotes \(x=0\) and \(y=-1\). \(y=0\) gives \(x^2=4\). It never meets the \(y\)-axis, since \(x=0\) is excluded. Asymptotes \(x=0\), \(y=-1\); meets the axes at \((-2, 0)\) and \((2, 0)\)
Trig graphs
The same four rules apply to \(y=\sin x\), \(\cos x\) and \(\tan x\) (in degrees here; radians work in the same way).
- \(a\sin x\): the amplitude becomes \(|a|\).
- \(\sin ax\): the period 360° is divided by \(a\), so \(\sin 2x\) has period 180°.
- \(\sin(x+a)\): the whole wave moves \(a\)° to the left; each zero, maximum and minimum moves with it.
- The asymptotes of \(\tan x\) at 90° and 270° move too: \(\tan 2x\) has asymptotes every 90°, starting at 45°.
(a) For \(0\le x\le360°\), find the coordinates of the points where \(y=\cos(x+60°)\) meets the axes, and of its maximum and minimum points.
(b) State the period of \(y=\tan 2x\) and the equations of its asymptotes for \(0\le x\le360°\).
(a) \(y\)-axis: \(x=0\) gives \(\cos60°=\tfrac12\), so \((0, \tfrac12)\).
\(x\)-axis: \(\cos(x+60°)=0\) when \(x+60°=90°\) or \(270°\) (with \(60°\le x+60°\le420°\)), so \((30°, 0)\) and \((210°, 0)\).
Minimum where \(x+60°=180°\); maximum where \(x+60°=360°\). Minimum \((120°, -1)\), maximum \((300°, 1)\)
(b) A stretch parallel to the \(x\)-axis, scale factor \(\tfrac12\): the period is \(180°\div2\). Period 90°; asymptotes \(x=45°, 135°, 225°, 315°\)
Moving \(\cos(x+60°)\) 60° to the right, or giving \(\sin 2x\) a period of 720°. Both come from treating changes inside the bracket like changes outside it.
How do the ideas combine in a full question?
A typical long question walks through the whole topic: factorise, sketch, add a second graph to count solutions, then transform. Each part feeds the next, so a correct sketch early on is worth the time.
\(f(x)=x^3-2x^2-3x\).
(a) Factorise \(f(x)\) completely.
(b) Sketch \(y=f(x)\), showing the coordinates of the points where the curve meets the axes.
(c) On the same axes, sketch \(y=\dfrac{2}{x^2}\). Hence state, with a reason, the number of real solutions of \(x^5-2x^4-3x^3-2=0\).
(d) Sketch \(y=f(x+1)\), showing the coordinates of the points where it meets the axes.
(a) Common factor, then the quadratic: two numbers with product \(-3\) and sum \(-2\) are \(-3\) and 1. \(f(x)=x(x-3)(x+1)\)
(b) Positive cubic; roots \(-1\), 0, 3; it passes through the origin, so the \(y\)-intercept is \((0, 0)\). Crosses at \((-1, 0)\), \((0, 0)\), \((3, 0)\)
(c) \(\dfrac{2}{x^2}\): two branches above the \(x\)-axis, asymptotes \(x=0\) and \(y=0\). Link the equation to the curves (\(x\ne0\)): \(x^3-2x^2-3x=\dfrac{2}{x^2}\Leftrightarrow x^5-2x^4-3x^3-2=0\).
Count the intersections. For \(x<-1\) and \(0<x<3\) the cubic is negative and the reciprocal positive: no meeting. For \(-1<x<0\) the cubic stays below 1 (its maximum is about 0.88) while \(\dfrac{2}{x^2}>2\): no meeting. For \(x>3\) the cubic rises from 0 to \(\infty\) while \(\dfrac{2}{x^2}\) falls towards 0: exactly one meeting. The curves meet once, so there is 1 real solution
(d) Translation 1 unit in the negative \(x\)-direction: subtract 1 from each root; the \(y\)-intercept is \(f(0+1)=f(1)=1-2-3=-4\). Crosses at \((-2, 0)\), \((-1, 0)\), \((2, 0)\); meets the \(y\)-axis at \((0, -4)\)
In a "hence" counting question, the reason is part of the answer. Write a sentence such as "the curves meet at one point, so the equation has one real solution", and show the algebra that turns the given equation into "curve 1 \(=\) curve 2".
Exam tips for P1.3
Write the coordinates of every axis crossing next to the point, and the equation of every asymptote next to its dashed line. A correct shape with no labels shows only part of the answer.
Before plotting any roots, decide the sign of the \(x^3\) term (multiply the \(x\)-terms of each bracket). It fixes which corner the curve starts in, and the roots then decide the rest.
Changes inside the bracket act on \(x\) the opposite way: \(f(x+3)\) moves left 3, \(f(4x)\) divides \(x\)-coordinates by 4. Outside the bracket, the change to \(y\) is exactly what you see.
After expanding, test \(x=1\) in both forms. After transforming, put your new point into the new equation and confirm it gives the old \(y\)-value. Each check takes seconds.
"Sketch" means the correct shape with key points labelled, not plotted to scale. "Show that" needs every line of algebra leading to the given result. "Hence" means use the previous part, which here is usually the sketch.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| \(f(x+2)\) moves the graph 2 units right. | The old value \(f(p)\) is now reached at \(x=p-2\). | \(f(x+a)\) is a translation \(a\) units to the left. |
| \(f(2x)\) doubles every \(x\)-coordinate. | The old value \(f(p)\) is reached when \(2x=p\). | \(f(ax)\) divides \(x\)-coordinates by \(a\): scale factor \(\tfrac1a\). |
| A repeated root is where the curve crosses twice. | A squared factor does not change sign either side of the root. | The curve touches the axis there; it is a turning point. |
| \(y=\dfrac{3}{x^2}\) has a branch below the \(x\)-axis, like \(\dfrac{3}{x}\). | \(x^2>0\) for every \(x\ne0\). | Both branches are on the same side of the axis as the sign of \(k\). |
| Asymptotes stay on the axes after a transformation. | The asymptote is part of the picture and moves with it. | Apply the same transformation to the asymptote's equation. |
| A stretch \(af(x)\) moves the roots. | \(0\times a=0\), so points on the \(x\)-axis stay there. | Roots are unchanged by \(af(x)\); only the \(y\)-coordinates are scaled. |
How this topic connects
P1.1 Quadratics and P1.6 Differentiation
The discriminant from P1.1 decides whether a line meets a curve; differentiation finds the exact turning points that a sketch here only shows roughly.
Factor theorem and algebraic division (P2); functions (P3)
When a cubic has no common factor of \(x\), you find one root by trying values in \(f(x)\) and divide, then sketch exactly as here. Two or more transformations in sequence, modulus graphs \(|f(x)|\) and \(f(|x|)\), and inverse functions in P3 all build on the four rules on this page.
IAL Physics and Mechanics
Inverse-square laws (gravitational and electric field strength) follow \(y=\dfrac{k}{x^2}\), and Boyle's law, \(p=\dfrac{k}{V}\), is a \(y=\dfrac{k}{x}\) graph.
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(x^2-a^2=(x-a)(x+a)\) | difference of two squares when factorising | No – learn it |
| \(y=(x-p)(x-q)(x-r)\) | cubic with roots \(p\), \(q\), \(r\); \(y\)-intercept \(-pqr\) | No – learn it |
| \(y=\dfrac{k}{x}\), \(y=\dfrac{k}{x^2}\) | reciprocal graphs; asymptotes \(x=0\) and \(y=0\) | No – learn it |
| \(y=f(x)+a\) | translation \(a\) units in the positive \(y\)-direction | No – learn it |
| \(y=f(x+a)\) | translation \(a\) units in the negative \(x\)-direction | No – learn it |
| \(y=af(x)\) | stretch parallel to the \(y\)-axis, scale factor \(a\) | No – learn it |
| \(y=f(ax)\) | stretch parallel to the \(x\)-axis, scale factor \(\tfrac1a\) | No – learn it |
Check your understanding
1. Expand \((x-2)^3\).
\((x-2)^2=x^2-4x+4\); multiplying by \((x-2)\) gives \(x^3-4x^2+4x-2x^2+8x-8=\) \(x^3-6x^2+12x-8\).
2. State the equations of the asymptotes of \(y=\dfrac{5}{x-4}\).
It is \(\dfrac{5}{x}\) translated 4 units in the positive \(x\)-direction: \(x=4\) and \(y=0\).
3. The point \((6, -2)\) lies on \(y=f(x)\). Write down its image on \(y=f(3x)\) and on \(y=f(x)-5\).
\(f(3x)\): divide the \(x\)-coordinate by 3, giving \((2, -2)\). \(f(x)-5\): subtract 5 from the \(y\)-coordinate, giving \((6, -7)\).
4. How many times does \(y=x^3\) meet \(y=-\dfrac{1}{x}\)?
Never. For \(x>0\), \(x^3\) is positive and \(-\dfrac1x\) negative; for \(x<0\) it is the other way round. The algebra agrees: \(x^4=-1\) has no real solution.
5. Write down the coordinates of the maximum and minimum points of \(y=4\sin x\) for \(0\le x\le360°\).
A stretch parallel to the \(y\)-axis, scale factor 4: maximum \((90°, 4)\), minimum \((270°, -4)\).
6. Factorise \(x^3-x^2-12x\) completely and state where the curve \(y=x^3-x^2-12x\) meets the \(x\)-axis.
\(x(x^2-x-12)\); two numbers with product \(-12\) and sum \(-1\) are \(-4\) and 3, so \(x(x-4)(x+3)\). The curve crosses the \(x\)-axis at \((-3, 0)\), \((0, 0)\) and \((4, 0)\).
Specification checklist
Tick every box before the exam. The wording is ours; the numbers refer to the P1 specification sections.
- (P1 1.10) I can expand products of two or three brackets and simplify, and use \(f(x)\) notation to substitute values.
- (P1 1.10) I can factorise a cubic completely by taking out a common factor and factorising the quadratic left, including differences of two squares and perfect squares.
- (P1 1.11) I can sketch a cubic from its factorised form, with the correct shape for a positive or negative \(x^3\) term, all axis intercepts, and a touching point at a repeated root.
- (P1 1.11) I can sketch \(y=\dfrac{k}{x}\) and \(y=\dfrac{k}{x^2}\) for positive and negative \(k\), and state their asymptotes.
- (P1 1.11) I can sketch two graphs on the same axes, use their points of intersection to state the number of solutions of an equation, and show the algebra that links the equation to the graphs.
- (P1 1.12) I can sketch \(y=af(x)\), \(y=f(x)+a\), \(y=f(x+a)\) and \(y=f(ax)\), and give the new intercepts, turning points and asymptotes.
- (P1 1.12) I can apply the same transformations to the graphs of \(\sin x\), \(\cos x\) and \(\tan x\), including changes of period, amplitude and asymptotes.