A Algebra · equations and reasoning — revision question pack

8 specification points · notes, questions, answers and worked methods

A6 · Know the difference between an equation and an identity; argue mathematically to show algebraic expressions are equivalent, and use algebra to support and construct arguments and proofs

Explanation

An equation is satisfied by particular values, whereas an identity states that two expressions are equivalent for every permitted value.
To show expressions are equivalent, expand or factorise one side until it matches the other.
Foundation tier: use algebra to support and construct an argument by defining quantities with variables, translating the claim and linking the resulting algebra back to it.
Higher tier: extend this to a proof by using a general form, such as $2n$ for an even integer or $2n+1$ for an odd integer, and reasoning that covers every permitted case.
Checking examples alone is not a proof.

Worked example

Higher tier: prove algebraically that the sum of two consecutive integers is odd.

1.Let the first integer be $n$ , so the next is $n+1$ .
2.Their sum is $n+(n+1)=2n+1$ .
3. $2n$ is even for every integer $n$ , so $2n+1$ is odd.

Answer: The sum has form $2n+1$ , so it is odd.

Common mistakes

Don't treat expressions that agree for one value as equivalent without simplifying them generally.
Don't make this mistake: Higher tier: checks several numerical cases and calls the pattern a proof.
Don't make this mistake: Higher tier: finishes with algebra but does not state why its form proves the claim.

Exam tip

Foundation tier: show each algebraic step and link the result to the argument. Higher tier: a “prove” question needs a general variable-based argument, not examples.

Tier 1 · Easy

1. Show that $4(n+2)-3(n-1)$ is equivalent to $n+11$ .[2 marks]

Tier 2 · Standard

1. Prove algebraically that the sum of two consecutive integers is odd.[3 marks]

Tier 3 · Hard

1. An odd number is written as $2n+1$ . Demonstrate algebraically that squaring it leaves remainder $1$ after division by $8$ .[4 marks]

A7 · Interpret simple expressions as functions with inputs and outputs; interpret the reverse as the 'inverse function' and two successive functions as a 'composite function' (formal notation expected)

Explanation

A function maps each allowed input to one output. To evaluate $f(a)$ , substitute the complete input $a$ into every occurrence of the variable and then simplify.
A function can be represented by a rule, mapping diagram or input-output table.
Higher tier: an inverse function reverses a one-to-one function, while a composite applies two functions successively; in $fg(x)=f(g(x))$ , $g$ acts first.
Formal function notation is expected for those extensions.
Examiners require careful brackets when the input is an expression, because the whole input replaces $x$ .

Worked example

Given $f(x)=3x-4$ , work out $f(6)$ and $f(2a)$ .

1.Substitute $6$ : $f(6)=3(6)-4=14$ .
2.Substitute the complete input $2a$ : $f(2a)=3(2a)-4$ .
3.Simplify to $f(2a)=6a-4$ .

Answer: $f(6)=14$ and $f(2a)=6a-4$ .

Common mistakes

Don't treat $f(x)$ as $f\times x$ instead of function notation.
Don't substitute only part of an expression supplied as the input.
Don't make this mistake: Higher tier: reads $f^{-1}(x)$ as $\dfrac{1}{f(x)}$ .

Exam tip

Put the complete input in brackets everywhere $x$ appears before simplifying.

Tier 1 · Easy

1. Given $f(x)=3x-4$ , work out $f(6)$ .[1 mark]

Tier 2 · Standard

1. Given $f(x)=5x+2$ , find $f^{-1}(x)$ and work out $f^{-1}(27)$ .[3 marks]

Tier 3 · Hard

1. Let $f(x)=2x+3$ and $g(x)=x^2-1$ . Solve $fg(x)=19$ , where $fg(x)=f(g(x))$ .[4 marks]

A8 · Work with coordinates in all four quadrants

Explanation

A coordinate $(x,y)$ gives horizontal position first and vertical position second. Positive $x$ is right, negative $x$ is left, positive $y$ is up and negative $y$ is down.
The signs identify the quadrant, numbered anticlockwise from the top right.
Find a displacement by subtracting starting coordinates from ending coordinates.
Find a midpoint by averaging the two $x$ -coordinates and separately averaging the two $y$ -coordinates.
Examiners expect coordinate order and negative signs to be preserved, with the two component calculations shown clearly.

The four coordinate quadrants and the signs of

(x,y)

in each.

Worked example

Find the midpoint of the segment joining $(-5,2)$ and $(3,-4)$ .

1.Average the $x$ -coordinates: $\dfrac{-5+3}{2}=-1$ .
2.Average the $y$ -coordinates: $\dfrac{2+(-4)}{2}=-1$ .
3.Write the coordinates in $(x,y)$ order.

Answer: $(-1,-1)$ .

Common mistakes

Don't write the vertical coordinate first and swaps $(x,y)$ .
Don't lose a negative sign when averaging coordinates.
Don't find half the coordinate differences but forgets to add them to an endpoint.

Exam tip

Write the midpoint as two separate averages before combining them into one ordered pair.

Tier 1 · Easy

1. State the quadrant containing the point $(-4,3)$ .[1 mark]

Tier 2 · Standard

1. Find the midpoint of the line segment joining $(-5,2)$ and $(3,-4)$ .[2 marks]

Tier 3 · Hard

1. The point $P$ divides the line segment from $A(-6,5)$ to $B(4,-5)$ in the ratio $AP:PB=3:2$ . Find the coordinates of $P$ .[3 marks]

A9 · Plot graphs of straight-line equations; use y = mx + c to identify parallel and perpendicular lines; find the equation of a line through two given points, or one point with a given gradient

Explanation

A non-vertical straight line has equation $y=mx+c$ , where $m$ is the gradient and $(0,c)$ is the $y$ -intercept. Plot a line by calculating at least two accurate coordinate pairs and joining them carefully with a ruler.
From two points, find $m=\dfrac{y_2-y_1}{x_2-x_1}$ , then substitute either point to find $c$ .
Parallel lines have equal gradients.
Higher tier: perpendicular non-vertical lines have gradients whose product is $-1$ .
Examiners expect the gradient calculation, substitution for the intercept and a final equation in a requested form.

A straight line

y=mx+c

crosses the vertical axis at

(0,c)

and has constant gradient.

Worked example

Find the equation of the line through $(-2,5)$ and $(4,-1)$ .

1. $m=\dfrac{-1-5}{4-(-2)}=\dfrac{-6}{6}=-1$ .
2.Use $y=-x+c$ and substitute $(-2,5)$ : $5=2+c$ .
3. $c=3$ , so the equation is $y=-x+3$ .

Answer: $y=-x+3$ .

Common mistakes

Don't subtract coordinates in different orders in the gradient numerator and denominator.
Don't use the $x$ -intercept as $c$ in $y=mx+c$ .
Don't make this mistake: Higher tier: changes only the sign of a gradient to make a perpendicular line.

Exam tip

For a line through two points, show the gradient first and then substitute one point to find $c$ .

Tier 1 · Easy

1. Write the equation of the line with gradient $4$ and $y$ -intercept $-7$ .[1 mark]

Tier 2 · Standard

1. Line $A$ has equation $4y=6x+7$ . Line $B$ passes through the points $(-2,4)$ and $(2,10)$ . Show that the two lines are parallel.[3 marks]

Tier 3 · Hard

1. Find the equation of the line through $(3,-5)$ that is perpendicular to $2x-3y=6$ . Give your answer in the form $y=mx+c$ .[4 marks]

A10 · Identify and interpret gradients and intercepts of linear functions graphically and algebraically

Explanation

The gradient of a linear function is the change in the vertical quantity per unit change in the horizontal quantity. Its sign shows whether the line rises or falls.
The $y$ -intercept is the output when $x=0$ ; the $x$ -intercept is where the output is zero.
Read axis scales and units before calculating or interpreting either feature.
In a context, a gradient is a rate and an intercept is often an initial value or fixed charge.
Examiners require a value, its unit and a sentence explaining what it represents.

The intercept is the output at zero; the gradient is rise divided by run.

Worked example

Water volume follows $V=120-6t$ , where $V$ is litres and $t$ is minutes. Interpret the gradient and intercept.

1.The coefficient of $t$ is $-6$ , so volume changes by $-6$ litres per minute.
2.Therefore the volume decreases by $6$ litres each minute.
3.When $t=0$ , $V=120$ , so the intercept is the initial volume.

Answer: The tank starts with $120$ litres and loses $6$ litres per minute.

Common mistakes

Don't calculate run divided by rise for the gradient.
Don't read the wrong intercept because the axes have been confused.
Don't give a contextual gradient as a bare number without units or meaning.

Exam tip

For “interpret”, state what happens per horizontal-axis unit and what the intercept means at zero.

Tier 1 · Easy

1. State the gradient and $y$ -intercept of $y=-3x+8$ .[2 marks]

Tier 2 · Standard

1. A straight line crosses the axes at $(0,12)$ and $(6,0)$ . Find its gradient and both intercepts.[3 marks]

Tier 3 · Hard

1. A straight-line graph of water volume $V$ litres against time $t$ minutes passes through $(0,120)$ and $(8,72)$ . Find and interpret its gradient and $V$ -intercept, then write $V$ in terms of $t$ .[4 marks]

A11 · Identify and interpret roots, intercepts, turning points of quadratic functions graphically; deduce roots algebraically and turning points by completing the square

Explanation

A root of a quadratic is an $x$ -coordinate where its graph meets the $x$ -axis, so $y=0$ . The $y$ -intercept is found by setting $x=0$ .
A turning point is the maximum or minimum point, and its vertical line is the axis of symmetry.
Factorising can reveal roots algebraically and the graph confirms their positions.
Higher tier: completing the square into $a(x-h)^2+k$ reveals turning point $(h,k)$ .
Examiners expect coordinates for points, equations for axes, and algebraic working when roots are to be deduced.

A quadratic's roots are its

x

-axis intersections, and its turning point lies on the axis of symmetry.

Worked example

Find the roots and $y$ -intercept of $y=x^2-5x+6$ .

1.Factorise: $x^2-5x+6=(x-2)(x-3)$ .
2.Set $y=0$ : each factor can be zero, giving $x=2$ or $x=3$ .
3.Set $x=0$ : $y=6$ , so the $y$ -intercept is $(0,6)$ .

Answer: Roots $x=2$ and $x=3$ ; $y$ -intercept $(0,6)$ .

Common mistakes

Don't report roots as $y$ -values instead of $x$ -coordinates.
Don't find the $y$ -intercept by setting $y=0$ .
Don't make this mistake: Higher tier: reads $(x-h)^2+k$ as having turning point $(-h,k)$ .

Exam tip

State roots as $x$ -values, intercepts as coordinates, and the symmetry line as an equation.

Tier 1 · Easy

1. Find the roots and the $y$ -intercept of $y=x^2-5x+6$ .[2 marks]

Tier 2 · Standard

1. For $y=(x-4)^2-9$ , state the turning point and axis of symmetry, and find the roots.[3 marks]

Tier 3 · Hard

1. Complete the square for $y=2x^2+8x-10$ . Hence state the turning point and find the roots.[4 marks]

A12 · Recognise, sketch and interpret graphs of linear, quadratic and simple cubic functions, the reciprocal y = 1/x (x ≠ 0), exponential y = k^x (k > 0), and y = sin x, cos x, tan x for angles of any size

Explanation

Recognise graph families from their defining shapes and features: linear graphs have constant gradient, quadratics are symmetric parabolas, simple cubics have an S-shape, and $y=\dfrac{1}{x}$ has two reciprocal branches with asymptotes $x=0$ and $y=0$ .
Sketch by marking intercepts, roots, turning points, asymptotes and representative values.
Higher tier: also recognise exponentials and sine, cosine and tangent graphs for angles of any size, using their periods and standard values.
Examiners expect a sketch to show correct shape and position, not merely a collection of plotted points.

The reciprocal graph

y=1/x

has branches in quadrants I and III and approaches, but never reaches, both axes.

Worked example

For $y=\dfrac{6}{x}$ , state both asymptotes and the quadrants containing its branches.

1. $x=0$ is excluded, so the vertical asymptote is $x=0$ .
2.As $|x|$ grows, $\dfrac{6}{x}$ approaches $0$ , so the horizontal asymptote is $y=0$ .
3. $x$ and $y$ have the same sign, so the branches lie in quadrants I and III.

Answer: Asymptotes $x=0$ and $y=0$ ; branches in quadrants I and III.

Common mistakes

Don't draw a reciprocal branch touching or crossing an axis.
Don't sketche a cubic as a parabola rather than an S-shaped curve.
Don't make this mistake: Higher tier: treats exponential growth as a straight line.

Exam tip

Before sketching, list the intercepts, turning points and asymptotes that fix the graph's shape.

Tier 1 · Easy

1. A graph passes through $(0,1)$ and its $y$ -value doubles whenever $x$ increases by $1$ . Name the function $y=2^x$ as linear, quadratic, cubic, reciprocal or exponential.[1 mark]

Tier 2 · Standard

1. For the graph $y=\frac6x$ , state both asymptotes and the two quadrants containing its branches.[3 marks]

Tier 3 · Hard

1. For $y=\cos x$ on $-180^\circ\leq x\leq360^\circ$ , list the $x$ -intercepts and the coordinates of every maximum and minimum needed for an accurate sketch.[4 marks]

A13 · Sketch translations and reflections of a given function [Higher only]

Explanation

For $y=f(x)+a$ , translate the graph vertically by vector $\begin{pmatrix}0\\a\end{pmatrix}$ . For $y=f(x-a)$ , translate it horizontally by $\begin{pmatrix}a\\0\end{pmatrix}$ ; the sign inside the function appears opposite to the movement.
The graph $y=-f(x)$ is the reflection of $y=f(x)$ in the $x$ -axis, while $y=f(-x)$ is its reflection in the $y$ -axis.
Track distinctive points, including intercepts and turning points, and preserve the graph's exact shape and scale.
A point $(p,q)$ provides a reliable coordinate check after transforming.
Examiners expect a fully described transformation, including the correct axis or translation vector.

A translation moves every point by the same vector while preserving the graph's shape.

Worked example

The point $(p,q)$ lies on $y=f(x)$ . Find its image on $y=f(x-3)-2$ .

1. $f(x-3)$ translates the graph $3$ units right.
2.Subtracting $2$ outside the function translates it $2$ units down.
3.Therefore $(p,q)$ maps to $(p+3,q-2)$ .

Answer: The image is $(p+3,q-2)$ , under translation by $\begin{pmatrix}3\\-2\end{pmatrix}$ .

Common mistakes

Don't move $f(x-3)$ three units left instead of right.
Don't reflect $y=f(-x)$ in the $x$ -axis rather than the $y$ -axis.
Don't move only selected points and changes the graph's shape.

Exam tip

For a translation, state the vector; for a reflection, name the mirror axis.

Tier 1 · Easy

1. Describe fully the transformation from $y=f(x)$ to $y=f(x)+4$ .[2 marks]

Tier 2 · Standard

1. The point $(3,-2)$ lies on $y=f(x)$ . Find the corresponding point on $y=f(-x)$ and name the transformation.[3 marks]

Tier 3 · Hard

1. The point $(-1,5)$ lies on $y=f(x)$ . Find the corresponding point on $y=f(2-x)-3$ , and describe the reflection and translations that produce the new graph.[4 marks]

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

A6 · Know the difference between an equation and an identity; argue mathematically to show algebraic expressions are equivalent, and use algebra to support and construct arguments and proofs

Tier 1 · Easy

1. Answer
- $4(n+2)-3(n-1)\equiv n+11$
Method: Expand both brackets, remembering that the subtraction acts on both terms: $4n+8-3n+3=n+11$ . Therefore the two expressions are equivalent.

Tier 2 · Standard

1. Answer
- $n+(n+1)=2n+1$ , so the sum is odd
Method: Let the first integer be $n$ , so the next is $n+1$ . Their sum is $n+(n+1)=2n+1$ . Since $2n$ is even for every integer $n$ , $2n+1$ is odd, proving the claim.

Tier 3 · Hard

1. Answer
- $(2n+1)^2=8k+1$ for an integer $k$
Method: Write an odd integer as $2n+1$ . Then $(2n+1)^2=4n^2+4n+1=4n(n+1)+1$ . One of the consecutive integers $n$ and $n+1$ is even, so $n(n+1)=2k$ for some integer $k$ . Therefore the square is $4(2k)+1=8k+1$ .

A7 · Interpret simple expressions as functions with inputs and outputs; interpret the reverse as the 'inverse function' and two successive functions as a 'composite function' (formal notation expected)

Tier 1 · Easy

1. Answer
- $f(6)=14$
Method: Substitute $x=6$ : $f(6)=3(6)-4=18-4=14$ .

Tier 2 · Standard

1. Answer
- $f^{-1}(x)=\frac{x-2}{5}$
- $f^{-1}(27)=5$
Method: Write $y=5x+2$ and rearrange: $x=(y-2)/5$ . Hence $f^{-1}(x)=(x-2)/5$ . Substituting $27$ gives $f^{-1}(27)=(27-2)/5=5$ .

Tier 3 · Hard

1. Answer
- $x=-3$ or $x=3$
Method: Form the composite by substituting $g(x)$ into $f$ : $fg(x)=2(x^2-1)+3=2x^2+1$ . Hence $2x^2+1=19$ , so $2x^2=18$ and $x^2=9$ . Therefore $x=-3$ or $x=3$ .

A8 · Work with coordinates in all four quadrants

Tier 1 · Easy

1. Answer
- Quadrant II
Method: The $x$ -coordinate is negative and the $y$ -coordinate is positive, which places the point in quadrant II.

Tier 2 · Standard

1. Answer
- $(-1,-1)$
Method: Average the $x$ -coordinates and the $y$ -coordinates: $((-5+3)/2,(2-4)/2)=(-2/2,-2/2)=(-1,-1)$ .

Tier 3 · Hard

1. Answer
- $P=(0,-1)$
Method: The vector from $A$ to $B$ is $(10,-10)$ . Since $AP$ is $3/5$ of the whole segment, $AP=(6,-6)$ . Adding this to $A$ gives $P=(-6,5)+(6,-6)=(0,-1)$ .

A9 · Plot graphs of straight-line equations; use y = mx + c to identify parallel and perpendicular lines; find the equation of a line through two given points, or one point with a given gradient

Tier 1 · Easy

1. Answer
- $y=4x-7$
Method: In $y=mx+c$ , use $m=4$ and $c=-7$ . This gives $y=4x-7$ .

Tier 2 · Standard

1. Answer
- Each line has gradient $\dfrac{3}{2}$ , and their $y$ -intercepts differ ( $\tfrac74$ and $7$ ), so the lines are parallel and distinct.
Method: Rearranging line $A$ gives $y=\frac32x+\frac74$ , so its gradient is $\frac32$ and its $y$ -intercept is $\frac74$ . The gradient of line $B$ is $(10-4)/(2-(-2))=6/4=\frac32$ , and substituting $(2,10)$ gives $y=\frac32x+7$ . The gradients are equal, so the lines never converge; the $y$ -intercepts differ, so the lines are not the same line. Hence they are parallel.

Tier 3 · Hard

1. Answer
- $y=-\frac32x-\frac12$
Method: Rearrange the given line: $-3y=6-2x$ , so $y=(2/3)x-2$ and its gradient is $2/3$ . A perpendicular line has gradient $-3/2$ . Write $y=-(3/2)x+c$ and substitute $(3,-5)$ : $-5=-9/2+c$ , so $c=-1/2$ . Hence $y=-(3/2)x-1/2$ .

A10 · Identify and interpret gradients and intercepts of linear functions graphically and algebraically

Tier 1 · Easy

1. Answer
- Gradient $-3$
- $y$ -intercept $8$
Method: Compare with $y=mx+c$ . The coefficient of $x$ is $m=-3$ , and the constant is $c=8$ .

Tier 2 · Standard

1. Answer
- Gradient $-2$
- $y$ -intercept $12$
- $x$ -intercept $6$
Method: Using the two points, $m=(0-12)/(6-0)=-12/6=-2$ . The point with $x=0$ gives the $y$ -intercept $12$ , and the point with $y=0$ gives the $x$ -intercept $6$ .

Tier 3 · Hard

1. Answer
- Gradient $-6$ litres per minute
- $V$ -intercept $120$ litres, the initial volume
- $V=120-6t$
Method: The gradient is $(72-120)/(8-0)=-48/8=-6$ , meaning the volume decreases by $6$ litres each minute. At $t=0$ , $V=120$ , so the initial volume is $120$ litres. Therefore $V=-6t+120$ .

A11 · Identify and interpret roots, intercepts, turning points of quadratic functions graphically; deduce roots algebraically and turning points by completing the square

Tier 1 · Easy

1. Answer
- Roots $x=2$ and $x=3$
- $y$ -intercept $6$
Method: Factorise: $x^2-5x+6=(x-2)(x-3)$ , so $y=0$ at $x=2$ and $x=3$ . Setting $x=0$ gives $y=6$ , so the $y$ -intercept is $6$ .

Tier 2 · Standard

1. Answer
- Turning point $(4,-9)$
- Axis $x=4$
- Roots $x=1$ and $x=7$
Method: The completed-square form gives turning point $(4,-9)$ and axis $x=4$ . For the roots, set $y=0$ : $(x-4)^2=9$ , so $x-4=\pm3$ and $x=1$ or $x=7$ .

Tier 3 · Hard

1. Answer
- $y=2(x+2)^2-18$
- Turning point $(-2,-18)$
- Roots $x=-5$ and $x=1$
Method: Factor $2$ from the quadratic and linear terms: $y=2(x^2+4x)-10=2((x+2)^2-4)-10=2(x+2)^2-18$ . The turning point is therefore $(-2,-18)$ . Setting $y=0$ gives $(x+2)^2=9$ , so $x+2=\pm3$ and the roots are $-5$ and $1$ .

A12 · Recognise, sketch and interpret graphs of linear, quadratic and simple cubic functions, the reciprocal y = 1/x (x ≠ 0), exponential y = k^x (k > 0), and y = sin x, cos x, tan x for angles of any size

Tier 1 · Easy

1. Answer
- Exponential
Method: Equal increases in $x$ multiply the output by a constant factor, so $y=2^x$ is an exponential function.

Tier 2 · Standard

1. Answer
- Asymptotes $x=0$ and $y=0$
- Branches in quadrants I and III
Method: The expression is undefined at $x=0$ , giving vertical asymptote $x=0$ . As $|x|$ grows, $6/x$ approaches $0$ , giving horizontal asymptote $y=0$ . Since $6/x$ has the same sign as $x$ , the branches lie in quadrants I and III.

Tier 3 · Hard

1. Answer
- $x$ -intercepts at $x=-90^\circ,90^\circ,270^\circ$
- Maxima $(0^\circ,1)$ and $(360^\circ,1)$
- Minima $(-180^\circ,-1)$ and $(180^\circ,-1)$
Method: Cosine is zero at odd multiples of $90^\circ$ , giving $-90^\circ$ , $90^\circ$ and $270^\circ$ in the interval. It reaches $1$ at multiples of $360^\circ$ , here $0^\circ$ and $360^\circ$ , and reaches $-1$ at odd multiples of $180^\circ$ , here $-180^\circ$ and $180^\circ$ .

A13 · Sketch translations and reflections of a given function [Higher only]

Tier 1 · Easy

1. Answer
- Translation by vector $\begin{pmatrix}0\\4\end{pmatrix}$
Method: Adding $4$ outside the function increases every $y$ -coordinate by $4$ and leaves every $x$ -coordinate unchanged. This is translation by vector $(0,4)$ .

Tier 2 · Standard

1. Answer
- $(-3,-2)$
- Reflection in the $y$ -axis
Method: Replacing $x$ by $-x$ reverses every $x$ -coordinate and leaves every $y$ -coordinate unchanged. Thus $(3,-2)$ maps to $(-3,-2)$ , a reflection in the $y$ -axis.

Tier 3 · Hard

1. Answer
- Corresponding point $(3,2)$
- Reflect in the $y$ -axis, translate $2$ units right, then translate $3$ units down
Method: If $(a,b)$ lies on $y=f(x)$ , then $f(2-x)=b$ when $2-x=a$ , so the new $x$ -coordinate is $2-a$ and the new $y$ -coordinate is $b-3$ . With $(a,b)=(-1,5)$ this gives $(3,2)$ . Since $f(2-x)=f(-(x-2))$ , the graph is reflected in the $y$ -axis, moved $2$ units right, and then moved $3$ units down.