Algebra and Functions

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Most tested topics

1. Functions

A function assigns exactly one output to every input in its domain. Different inputs may share an output, but one input cannot have two outputs.

Term	Meaning
domain	permitted input values
range (image set)	outputs actually produced
one-one	distinct inputs give distinct outputs
many-one	two or more inputs may give the same output
inverse	reverses a one-one function
composition	uses one function's output as another's input

A graph represents a function when every vertical line meets it at most once. It is one-one when every horizontal line also meets it at most once; a many-one function still passes the vertical-line test but fails the horizontal-line test.

‘Exactly one output’ is required only for inputs in the stated domain. Having several inputs map to one output does not stop a relation being a function; it stops that function being one-one.

A domain records allowed inputs and a range records the resulting outputs. Both must respect algebraic restrictions and any stated interval.

Expression feature	Domain condition
denominator $q(x)$	$q(x)\ne0$
square root $\sqrt{q(x)}$	$q(x)\geq0$
$\ln q(x)$ or $\lg q(x)$	$q(x)>0$

For an inverse, domain and range swap: $\operatorname{Dom}(f^{-1})=\operatorname{Range}(f)$ and $\operatorname{Range}(f^{-1})=\operatorname{Dom}(f)$ . Restrict a many-one formula to an interval on which it is one-one before defining its inverse.

For $gf(x)=g(f(x))$ , first require $x$ to be in the domain of $f$ , then require $f(x)$ to lie in the domain of $g$ . Therefore $\operatorname{Dom}(gf)\subseteq\operatorname{Dom}(f)$ and $\operatorname{Range}(gf)\subseteq\operatorname{Range}(g)$ .

If $f(x)=x^2$ is restricted to $x\geq0$ , then its range is $[0,\infty)$ and $f^{-1}(x)=\sqrt{x}$ has domain $x\geq0$ . Without the restriction, $x^2$ is many-one and has no inverse function on all real numbers.

State inequalities with the correct strictness. A forbidden denominator value is excluded, while a zero radicand is allowed for a square root.

Notation	Meaning
$f(x)$	output of $f$ at input $x$
$f:x\mapsto\lg x$	$f$ maps $x$ to $\lg x$
$f^{-1}(x)$	inverse-function output
$fg(x)$	$f(g(x))$ : apply $g$ first
$f^2(x)$	$f(f(x))$ : apply $f$ twice

If $f(x)=x^2+1$ and $g(x)=3x-2$ , then $fg(4)=f(g(4))=f(10)=101$ . For iteration, $f^2(2)=f(f(2))=f(5)=26$ .

$f^{-1}(x)$ is not $1/f(x)$ , and $f^2(x)$ is not generally $[f(x)]^2$ . Read adjacent function letters from right to left. In this syllabus, $f^2(x)$ notation is not used with trigonometric functions.

The graph of $y=|f(x)|$ keeps every part of $y=f(x)$ on or above the $x$ -axis and reflects every part below it across the $x$ -axis.

$(x,y)\mapsto(x,|y|)$

The domain and $x$ -intercepts stay unchanged, all output values become non-negative, and a crossing of the $x$ -axis usually becomes a sharp turning point. Apply the rule point by point to permitted linear, quadratic, cubic or trigonometric $f(x)$ .

For $f(x)=5x-7$ , the zero is $x=7/5$ . Thus $y=|5x-7|$ is V-shaped with vertex $(7/5,0)$ and $y$ -intercept $(0,7)$ ; the negative half of the original line is reflected upward.

Absolute value changes output signs, not input values: $|f(x)|$ is generally different from $f(|x|)$ . Do not reflect parts that are already above the axis.

An inverse function exists only when the original function is one-one on its stated domain.

Reversing a many-one function makes one former output point to several former inputs. That reversed relation fails the function rule because one input would have more than one output. Graphically, the original fails the horizontal-line test, so its reflection in $y=x$ fails the vertical-line test.

A suitable domain restriction can select one one-one branch. For example, $f(x)=x^2$ has no inverse on $\mathbb R$ , but on $x\geq0$ it has inverse $f^{-1}(x)=\sqrt{x}$ .

Do not say only ‘because it is quadratic’. The reason is that distinct permitted inputs give the same output; identify the many-one behaviour or a failed horizontal-line test.

To invert a one-one function: write $y=f(x)$ , rearrange to make $x$ the subject, then interchange the variable names and state the inverse domain. Any branch choice must match the original domain restriction.

$f(x)=(x-1)^2+3,\ x\geq1\quad\Longrightarrow\quad f^{-1}(x)=1+\sqrt{x-3},\ x\geq3$

The positive square-root branch is required because the original restriction gives $x-1\geq0$ . The inverse domain $x\geq3$ is the original range, while the inverse range $x\geq1$ is the original domain.

Check both compositions on their valid domains: $f^{-1}(f(x))=x$ and $f(f^{-1}(x))=x$ . Use $f^{-1}$ notation only for the inverse function, never for a reciprocal.

In $fg(x)=f(g(x))$ , the right-hand function acts first. Composition is substitution: replace every $x$ in the outer function by the complete inner expression.

$f(x)=x^2+1,\ g(x)=3x+2\quad\Rightarrow\quad fg(x)=(3x+2)^2+1,\quad gf(x)=3(x^2+1)+2$

For a value, follow the same inside-first order: $fg(1)=f(g(1))=f(5)=26$ . For an equation such as $fg(x)=5$ , first form the correct composite, then solve while enforcing the composite domain.

The composite exists only when the input is valid for the inner function and its output is valid for the outer function. A logarithm, square root or denominator can therefore remove inputs after substitution.

Usually $fg\ne gf$ ; switching the letters changes both the expression and possibly the domain. Keep brackets around the entire inner expression during substitution.

The graphs of a one-one function and its inverse are reflections of each other in the line $y=x$ .

$(a,b)\text{ on }y=f(x)\quad\Longleftrightarrow\quad(b,a)\text{ on }y=f^{-1}(x)$

Swap the coordinates of useful points, then reflect the whole shape: an $x$ -intercept $(a,0)$ becomes the inverse's $y$ -intercept $(0,a)$ ; horizontal and vertical asymptotes swap; domain and range swap. Points on $y=x$ remain fixed and satisfy $f(x)=x$ .

Sketch $y=x$ as a guide, mark reflected intercepts, endpoints and asymptotes, and preserve the restricted domains. Label both curves so their roles are unambiguous.

A many-one graph cannot be reflected into an inverse function until its domain is restricted to a one-one branch. Reflection does not mean changing every coordinate's sign.

2. Quadratic functions

For $f(x)=ax^2+bx+c$ with $a\ne0$ , the turning point gives the quadratic's minimum when $a>0$ and maximum when $a<0$ . Its $x$ -coordinate is the axis of symmetry, $x=-b/(2a)$ .

Completing the square writes the function as $f(x)=a(x-h)^2+k$ . Because $(x-h)^2\geq0$ , the turning point is $(h,k)$ : $k$ is the minimum if $a>0$ and the maximum if $a<0$ .

$2x^2+3x-4=2\left(x+\frac34\right)^2-\frac{41}{8}\quad\Rightarrow\quad\min f=-\frac{41}{8}\text{ at }x=-\frac34$

Alternatively, differentiate: $f'(x)=2ax+b$ . Set $f'(x)=0$ to obtain $x=-b/(2a)$ , substitute this value into $f$ , then use the sign of $a$ to classify the turning point. Both methods locate the same point.

Give both the extreme value and where it occurs when asked. The value is the $y$ -coordinate; it is not the same as the $x$ -coordinate of the turning point.

Vertex form $f(x)=a(x-h)^2+k$ reveals the turning point $(h,k)$ , axis $x=h$ , and opening direction. These features fix the basic sketch and the unrestricted range.

Situation	Range
$a>0$ , domain $\mathbb R$	$f(x)\geq k$
$a<0$ , domain $\mathbb R$	$f(x)\leq k$
restricted domain	compare the vertex, included endpoints and end behaviour that lie in the domain

For the sketch, mark the vertex and axis of symmetry, then find any real $x$ -intercepts and the $y$ -intercept. Draw a smooth symmetric parabola opening upward for $a>0$ or downward for $a<0$ .

A restricted domain can remove the vertex. For $f(x)=(x-2)^2-5$ with domain $x\geq4$ , the permitted branch starts at $(4,-1)$ and rises, so the range is $f(x)\geq-1$ , not $f(x)\geq-5$ .

A domain restricts $x$ ; a range restricts $f(x)$ or $y$ . Check whether an endpoint is included before choosing $\leq$ or $<$ .

For $ax^2+bx+c=0$ , the discriminant $D=b^2-4ac$ tells how many real roots exist without solving the equation.

Discriminant	Roots of the equation	Line and curve
$D>0$	two distinct real roots	two intersections
$D=0$	two equal real roots	tangent; one point of contact
$D<0$	no real roots	no intersection

For a line and a curve, first equate their $y$ -expressions and rearrange to one quadratic in $x$ . Only then identify $a$ , $b$ and $c$ and apply the required condition to $D$ . If a parameter is present, solving the resulting equation or inequality gives its allowed values.

$y=x^2+1,\ y=kx\quad\Rightarrow\quad x^2-kx+1=0,\quad D=k^2-4$

Thus the line cuts the curve twice when $|k|>2$ , is tangent when $k=\pm2$ , and does not meet it when $|k|<2$ .

Use $D\geq0$ for ‘real roots’, but $D>0$ for ‘two different real roots’. Tangency requires equality, not merely a non-negative discriminant.

First rearrange the equation to $ax^2+bx+c=0$ . Then choose the method that exposes the roots with the least unnecessary work.

Method	Best choice when	Essential move
factorisation	factors are visible	use the zero-product rule
quadratic formula	factors are awkward	substitute $a,b,c$ with their signs
completing the square	vertex form is useful	isolate the square, then square-root both sides

$x^2-5x+4=0\Rightarrow(x-1)(x-4)=0\Rightarrow x=1\text{ or }x=4$

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

For $x^2-6x+1=0$ , write $(x-3)^2=8$ , so $x=3\pm2\sqrt2$ . The $\pm$ is required because both numbers have the same square.

Substitute each proposed root into the original equation. If $D<0$ , there are no real roots in this syllabus context; do not replace $\sqrt{D}$ with a real number.

A quadratic inequality asks where the graph lies above, on or below the $x$ -axis. Its real roots are critical values where the sign may change.

Rearrange so one side is zero, solve the corresponding quadratic equation, then split the number line at its real roots. Determine the sign in each interval by a test value or by the parabola's opening direction. Include a root only for $\leq$ or $\geq$ .

$(x+2)(x-3)\leq0\quad\Rightarrow\quad-2\leq x\leq3$

For $x^2-x-6\geq0$ , the roots are $-2$ and $3$ and the upward-opening graph is on or above the axis outside them. Hence $x\leq-2$ or $x\geq3$ .

Use ‘and’ for one interval such as $-2<x<3$ , but ‘or’ for two separate outer intervals. If there are no real roots, the quadratic keeps the sign of its leading coefficient for every real $x$ .

3. Factors of polynomials

The remainder theorem replaces polynomial division by one substitution. When $p(x)$ is divided by $x-a$ , the constant remainder is $p(a)$ .

Divisor	Value to substitute	Conclusion
$x-a$	$x=a$	remainder $p(a)$
$mx-n$	$x=n/m$	remainder $p(n/m)$
either linear divisor	its zero gives $p(x)=0$	the divisor is a factor

$p(x)=6x^3+x^2-12x+5,\quad p\left(\frac12\right)=0\quad\Rightarrow\quad2x-1\text{ is a factor}$

If coefficients are unknown, translate each statement into an equation: ‘ $x+2$ is a factor’ gives $p(-2)=0$ , while ‘remainder 40 on division by $x-3$ ’ gives $p(3)=40$ . Solve the resulting simultaneous equations.

For the divisor $mx-n$ , substitute $n/m$ , not $n$ . A zero remainder proves a factor; any non-zero value is only the remainder and does not prove a factor.

Once one linear factor of a cubic is known, dividing by it leaves a quadratic. Factorising that quadratic can reveal the complete product of linear factors.

Write the cubic in descending powers, inserting a zero coefficient for any missing term. Divide by the known factor using algebraic long division, synthetic division when appropriate, or coefficient comparison. Multiply the divisor by the quotient to check that every coefficient is recovered.

$2x^3-3x^2-8x-3=(x-3)(2x^2+3x+1)$

$2x^2+3x+1=(2x+1)(x+1)\quad\Rightarrow\quad p(x)=(x-3)(2x+1)(x+1)$

A known factor is only the first step. Do not stop at ‘linear factor × quadratic’ when the quadratic can be factorised further, and do not lose a missing-power zero during division.

To solve $p(x)=0$ , express the cubic as a product and use the zero-product rule: if a product is zero, at least one factor is zero.

$p(x)=(x-2)(3x+2)(2x-5)=0\quad\Rightarrow\quad x=2,-\frac23,\frac52$

Remaining factor	Real solutions contributed
two distinct linear factors	two real roots
a repeated linear factor	one repeated real root
quadratic with $D<0$	no additional real roots

For $(3x-1)(2x^2+3x+5)=0$ , the quadratic has $D=3^2-4(2)(5)=-31<0$ . It contributes no real roots, so the only real solution is $x=1/3$ .

List roots from every factor and verify them in the original cubic. A cubic need not have three distinct real roots: repeated factors or an irreducible quadratic change the number of distinct real solutions.

4. Equations, inequalities and graphs

An equation $|A(x)|=c$ with $c\geq0$ means $A(x)=c$ or $A(x)=-c$ . Solve both branches and check every result in the original equation.

Form	Algebraic branches	Extra condition
$\|A\|=c$	$A=c$ or $A=-c$	no solution if $c<0$
$\|A\|=B$	$A=B$ or $A=-B$	retained roots must have $B\geq0$
$\|A\|=\|B\|$	$A=B$ or $A=-B$	verify both branches
$\|Q(x)\|=d$	$Q(x)=d$ or $Q(x)=-d$	solve both resulting quadratics

$5|2x-1|+8=23\Rightarrow|2x-1|=3\Rightarrow2x-1=\pm3\Rightarrow x=2\text{ or }x=-1$

Graphically, solutions are the $x$ -coordinates where $y=|A(x)|$ meets the other side. An accurate graph must show all intersections, including tangencies that produce one distinct solution.

Do not write $|A|=c\Rightarrow A=c$ only. When the other side contains $x$ , solving both algebraic branches is not enough: reject any value for which that side is negative.

A modulus measures distance from zero. For $c>0$ , $|A|<c$ places $A$ between $-c$ and $c$ , while $|A|>c$ places it outside those limits; include endpoints for $\leq$ or $\geq$ .

$|A|<c\Longleftrightarrow-c<A<c,\qquad |A|>c\Longleftrightarrow A<-c\text{ or }A>c$

For $|A(x)|$ compared with a linear expression, use piecewise branches or graph $y=|A(x)|$ and the line, then choose where one graph is above or below the other. For $|A|$ compared with $|B|$ , both sides are non-negative, so squaring preserves the comparison.

$4|x-1|\leq|3x+2|\Rightarrow16(x-1)^2\leq(3x+2)^2\Rightarrow7x^2-44x+12\leq0$

The critical values are $2/7$ and $6$ . The upward-opening quadratic is non-positive between them, so $2/7\leq x\leq6$ .

Do not square an inequality against an expression whose sign is unknown: squaring can create false values. On a graph, ‘greater than’ means above, and strict inequalities exclude intersection points.

When an equation repeats one expression and its square, replace that expression by a single variable. Solve the quadratic in the substitute, then return to the original variable and apply its domain conditions.

Repeated expressions	Useful substitute	Condition
$a^{2x}$ and $a^x$	$u=a^x$	$u>0$ for $a>0$
$e^{2x}$ and $e^{-2x}$	multiply to clear the negative power, then $u=e^{2x}$	$u>0$
reciprocal logarithms	$u=\log_a x$	valid log base and argument
$x^{1/3}$ and $x^{1/6}$	$u=x^{1/6}$	respect the real-domain restriction

$3^{2x+1}+8(3^x)-3=0,\quad u=3^x\Rightarrow3u^2+8u-3=(3u-1)(u+3)=0$

Because $u=3^x>0$ , reject $u=-3$ and keep $u=1/3$ . Therefore $3^x=3^{-1}$ and $x=-1$ .

Solving the quadratic is only the middle step. Always translate each admissible $u$ back, and reject roots that violate positivity, logarithm or fractional-power conditions.

For $f(x)=a(x-r_1)(x-r_2)(x-r_3)$ , the factors give the $x$ -intercepts and $f(0)$ gives the $y$ -intercept. These points, root multiplicities and the sign of $a$ determine the sketch.

Feature	Effect on the graph
simple root	crosses the $x$ -axis
repeated root	touches and turns at the $x$ -axis
$a>0$	falls to the left and rises to the right
$a<0$	rises to the left and falls to the right

$f(x)=3(x-3)(x-1)(x+2)\Rightarrow x=-2,1,3,\quad f(0)=18$

To sketch $y=|f(x)|$ , keep every section on or above the $x$ -axis and reflect each negative section upward. The intercepts stay fixed; a simple crossing becomes a sharp turn, while a section already above the axis is unchanged.

Do not reflect the whole cubic or change the $x$ -coordinates. Absolute value acts on the output only, and all axis intersections must be labelled.

To solve $f(x)\square d$ graphically, compare the cubic $y=f(x)$ with the horizontal line $y=d$ . Their intersection $x$ -coordinates divide the graph into intervals with a consistent above/below relationship.

Draw or use the accurate graph, mark every intersection with $y=d$ , then read where the cubic is above the line for $>$ or $\geq$ and below it for $<$ or $\leq$ . Include intersection values only for non-strict inequalities.

$f(x)=-\frac15(x+2)(2x-1)(x+5)\geq0$

The zero-level intersections are $x=-5,-2,1/2$ . Reading the sign of the cubic gives $x\leq-5$ or $-2\leq x\leq1/2$ .

Do not assume the sign switches at every contact: at a repeated root the graph touches the level and turns back. For $d\ne0$ , use the actual intersections with $y=d$ , not the polynomial's roots.

5. Simultaneous equations

A solution of two simultaneous equations is an ordered pair $(x,y)$ that satisfies both equations. Eliminate one unknown or express it in terms of the other, solve the resulting one-variable equation, then reconstruct and verify each pair.

Structure	Strong first move	Main check
two linear equations	eliminate matching coefficients, or substitute from the simpler equation	one pair unless equations coincide or conflict
line with a quadratic relation	rearrange the line and substitute	each quadratic root needs its matching second coordinate
an equation involving $xy$	factor or isolate one variable, such as $y=k/x$	record $x\ne0$ when division by $x$ occurs
fractional equations	state forbidden denominator values, then clear denominators or substitute	reject any forbidden or extraneous value

$xy+x^2=15,\quad y+3x=11\Rightarrow y=11-3x\Rightarrow2x^2-11x+15=0$

Factorising gives $(2x-5)(x-3)=0$ , so $x=5/2$ or $x=3$ . Back-substitution into $y=11-3x$ gives the paired solutions $(5/2,7/2)$ and $(3,2)$ . The two $x$ -values and two $y$ -values are not independent lists: keep their correspondence.

Substitution may produce a quadratic, so there can be several pairs. Check every pair in both original equations and discard values introduced by multiplying through a zero denominator, squaring, or using a rearrangement outside its valid domain.

6. Logarithmic and exponential functions

$y=e^x$ and $y=\ln x$ are inverse functions. Their graphs are reflections in $y=x$ : swapping every point $(p,q)$ to $(q,p)$ also swaps domain with range and horizontal features with vertical ones.

Graph	Domain	Range	Intercept	Asymptote
$y=e^x$	all real $x$	$y>0$	$(0,1)$	horizontal $y=0$
$y=\ln x$	$x>0$	all real $y$	$(1,0)$	vertical $x=0$

For $y=ke^{nx}+a$ with $k\ne0$ and $n\ne0$ , the horizontal asymptote is $y=a$ and the $y$ -intercept is $(0,k+a)$ ; its range is $y>a$ if $k>0$ and $y<a$ if $k<0$ . For $y=k\ln(ax+b)$ with non-zero $a,k$ , require $ax+b>0$ , the vertical asymptote is $x=-b/a$ , and the $x$ -intercept satisfies $ax+b=1$ .

$y=4e^x+3$ has horizontal asymptote $y=3$ and $y$ -intercept $(0,7)$ . For $y=\ln(4x-3)$ , the domain is $x>3/4$ , the vertical asymptote is $x=3/4$ , and the $x$ -intercept is $(1,0)$ .

An asymptote is found from the transformed input or output, not copied unchanged. The exponential graph never reaches $y=a$ , and the logarithmic graph is undefined where $ax+b\le0$ . Series expansions are outside this syllabus objective.

For a real logarithm, the base must satisfy $a>0$ and $a\ne1$ , and every logarithm argument must be positive. Logarithm laws rewrite multiplication, division and powers; they do not distribute across addition.

Structure	Logarithm form
$uv$	$\log_a(uv)=\log_a u+\log_a v$
$u/v$	$\log_a(u/v)=\log_a u-\log_a v$
$u^r$	$\log_a(u^r)=r\log_a u$
change of base	$\log_a u=\dfrac{\ln u}{\ln a}=\dfrac{\log_b u}{\log_b a}$
reciprocal bases	$\log_a b=1/\log_b a$

$3\lg x-2\lg(y^2)-3=\lg\left(\frac{x^3}{1000y^4}\right),\qquad x>0,\ y\ne0$

The coefficients become powers, and $3=\lg1000$ supplies the constant inside the quotient. Because the original expression contains $\lg(y^2)$ , its real-domain condition is $y\ne0$ ; the final logarithm has the same condition.

Never write $\log(u+v)=\log u+\log v$ . When solving, record the original positive-argument restrictions before combining logs, then reject any candidate that violates them.

Equation after isolation	Best move
$a^{f(x)}=a^{g(x)}$	equate exponents: $f(x)=g(x)$
$a^{f(x)}=b$ with no useful common base	take logarithms: $f(x)\ln a=\ln b$

$a^x=b,\quad a>0,\ a\ne1,\ b>0\quad\Longrightarrow\quad x=\frac{\ln b}{\ln a}$

Solve $4e^{2x-3}=7e^{5-x}$ . Divide first: $e^{3x-8}=7/4$ . Taking natural logarithms gives $3x-8=\ln(7/4)$ , so $x=\dfrac{8+\ln(7/4)}{3}\approx2.85$ .

Isolate the exponential expression before taking logs. Keep the exact logarithmic form until the final line, then round only if requested. Any logarithm base gives the same solution when used consistently.

For a positive base, $a^x$ is always positive, so $a^x=b$ has no real solution when $b\le0$ . Do not equate exponents unless both sides have first been expressed with the same valid base.

Algebra and Functions

Exam analysis

Most tested topics

1. Functions

Understand functions and their key terms

Find domains and ranges, including restrictions

Read and use function notation

Transform y=f(x) into y=|f(x)|

Explain when an inverse function does not exist

Find an inverse function with the correct domain

Form and use composite functions in the right order

Sketch a function and its inverse

2. Quadratic functions

Find the turning value of a quadratic

Use the turning point to sketch and find a range

Use the discriminant to count roots and intersections

Choose a method to solve a quadratic equation

Solve a quadratic inequality by interval signs

3. Factors of polynomials

Use evaluation to find remainders and test factors

Reduce a cubic to find all its factors

Solve a cubic from its factors

4. Equations, inequalities and graphs

Solve modulus equations by controlling both branches

Solve modulus inequalities with valid sign logic

Turn a related equation into a quadratic

Sketch a factored cubic and its modulus

Read a cubic inequality from the graph

5. Simultaneous equations

Choose a route through simultaneous equations

6. Logarithmic and exponential functions

Connect exponential and logarithmic graphs

Use logarithm laws with valid arguments

Solve exponential equations efficiently