Aqa a-level maths: core 1 revision notes

This is just my notes for core1 AQA Maths by chapter...I've not included methods of doing things, just the name of them as I cant yet format properly equations. Ive tried to explain my notations where they might be unclear.. As it is a maths exam, it is not a memory test, but a test of understanding. Therefore, just looking over the following notes the night before and having done nothing is futile. PRACTISE PAPERS!!

Hope it helps anyways with general understanding/recap/definitions, and good luck with ure maths exams. =]

Linear Graphs and Equations

For any straight line, the gradient (M) is: dy/dx or difference in y/difference in x which is (y2-y1)/(x2-x1)

Equation of a line: y=mx+c which is used when the gradient and intercept is known or y-y1=m(x-x1) when the gradient and the co-ordinates (x1,y1) of a single point that the line passes through is known. You'll need to learn this equation.

[The equation of the line can be kept in this form unless stated in the exam. (reduces error chance) Also, usually there is a working mark, so state the fact that the gradient is difference in y/difference in x]

The mid-point of two graphs is found by (x1+x2)/2 , (y1+y2)/2 in the form (x,y)

Lines with the same gradient are parallel, while lines with gradients that are negative reciprocals of each other is perpendicular to it. (perpendicular means at a right angle to)

[they usually want you to state that 'the perpendicular line's gradient is the negative reciprocal', so stick it in. It is also usefull to draw out diagrams if they ask about right angle triangles, (usually something to do with negative reciprocals rather than pythagoras.)]

In 2D Lines that are not parallel must intesect, the point of intersection can be found by simultaneous equations by:

  • equating coefficients of the two lines
  • substituing one equationinto the other
  • equating both equations for y.


surd form is exact. they involve irrational roots, which are roots that cannot be expressed as fractions as they are irrational for example: √5'(for clarification purposes marks the end of the root)

  • √a' x √b' = √a x b' = √ab'
  • √a' / √b' = √a/b'
  • √a+b' ≠ √a'+√b'

It is often more useful when denominator of a fraction is rationalised. This is done by multiplying the top and bottom by the conjugate, as the product of two conjugates is always rationalised because (a+b)(a-b)=(a^2)-(b^2) and a surd^2 is always rational. (^2 means squared)

Quadratic Graphs and Equations (and Further Equations)

The graph of a quadratic is a parabola, they can be factorised into two linear factors.

Parallel beams of light on a parabola reflector reflect through the same point.

Any moving object in gravity follow a parabolic curve. A plane that cuts a cone makes a parabola shape parallel to the cone.

A quadratic can be solved by:

  • factorising (finding y=(x+r1)(x+r2) where r=root)
  • completing the square (in the form y=k(x-p)^2 +q note the -ve infront of the p )
  • using the quadatic equation. (needs to be learnt)

the graph y=ax^2 +bx+c lets you find the y-intercept, c.

quadratics in completed square form have a translation of [p over q] (dont know how to format) with the vertex at (p,q) symmetry at x=p

[remember the word translation, question on it every exam without fail.]

quadratics completed in factorised form lets you find the root(s)/solution(s)/x-intercept(s)/when y is = 0

The discriminant b^2-4ac can be used to find the number of roots:

  • positive answer = 2 real roots
  • 0 = one root / repeated roots.
  • negative answer = no 'real' roots. (you cant square a negative number, so quadatic formula wouldnt work ∴ no real roots)

[if the exam doesnt say how many roots, just that they're real its ≥0, they also usually want a statement stating the bullet points above.]

If a pair of simultaneous equations lead to a quadratic, then the discriminant shows the relationship between the graphs. The same rules about the discriminant applies, ie. number of intersections.


finding solution set of an inequality means to solve an inequality.

linear equality = both expressions are linear (GCSE stuff)

remember that multiplying/dividing by negative numbers means that the inequality symbol is reversed.

The discriminant can also be turned into an inequality. > 0, = 0, < 0, ≥0

The solution can be represented on a number line, graphically or using sign diagrams showing critical values for both linear and quadratic inequalities.

[theres usually marks for showing this. even if you dont need to, draw it out. Also state what the critical values are]

[N.B] I dont know how to define a critical value properly. For a quadratic inequality its the -ve of the roots.


An index/indices is an expression like x^5 (x to the power 5)

Cubic expression = 3 linear factors

Coefficients = constant multiples 'k' eg. kx^3+kx^2+kx (the drawn graph would be stretched by k.)

y=x^3 translated by [3 over 1], replace x by (x-3) ; y by (y-1) so it should be y-1=(x-3)^3 = y=(x-3)^3 +1 (translations in more depth in core2)

polynomial expressions = a+bx+cx^2 +dx^3 +ex^4.......etc. where a,b,c,d,e.... are constants that are positive.

degree of a polynomial = highest index value (=number of roots)

function notation f(x) makes it easier to deal with/can be useful

A division results in a quotient and a remainder. Dividing polynomials can be done in three ways:

  • long division
  • grid method
  • inspection
  • sibi's method. (named after a fellow classmate whose method made it into our notes)

[with all the methods above, it might help to write in 0x^n when there is a missing factor. Feel free to ask about any of the methods and i'll try to help.]

Factor Theorem: if the polynomial p(x) is divided by (x-a) with no remainders, p(a) is a factor of p(x)

Remainder Theorem: when p(x) is divided by (x-a) the remainder is p(a)

[as always, the exam wants you to state the theorem before applying it]

Equation of a Circle

formula for a circle: (x-a)^2 +(y-b)^2 =r^2 with the centre at (a,b) comes from x^2 +y^2 =r^2 translated [a over b] there is a proof thing for this, that i might draw a picture out for another time.

The expanded form x^2 +cx+y^2 +dy+e=0 can be sorted by completing the square for the first part x^2+cx then the y^2+dy+e

A tangent to the circle is a straight line that touches it once.

The radius is perpendicular to the tangent.

The normal is a line at right angles to the tangent, and as it will go through the centre of the circle the diameter would be a part of it.

The gradient of the normal is therefore the gradient of the radius and the tangent the negative reciprocal.

These gradients can all be worked out (linear graphs and equations section)

The discriminant can be used to show if the line:

  • intersects the circle (chord) >0 (crosses twice)
  • is at a tangant to the circle =0 (touches once)
  • doesnt meet the circle <0

The line segment between the two intersections of the circle is known as the chord. The perpendicular bisector goes through the centre of the circle.

Rates of Change, Diffrentiation

rate of change = rate of change in y with respect to x = gradient = dy/dx

The gradient of a curve is constantly changing so the tangent is taken at a point, the gradient of that tangent is the gradient at that particular point.

Take δy/δx as the gradient of the line joining two points on a graph, and dy/dx as the gradient of the tangent; as the distance between δx and δy get closer, δy/δx gets closer to dy/dx

finding dy/dx for y is called diffrentiating with respect to x.

the gradient function dy/dx is called the derivative also written as f'(x)

To diffrentiate a polynomial, look at it term by term (as the sum of functions = sum of seperate derivatives) and apply the rule for y=x^n ,dy/dx= nx^n-1

The normal to the tangent on the curve is the negative reciprocal of M.

When the gradient of the tangent to the curve/derivative of a constant/y=k/gradient is 0, it is at a stationary/turning point.

The stationary point is either a maximum or minimum point. A maximum is the highest part of the curve while the minimum the lowest.

Using Diffrentiation

Although the gradient of all the stationary points are 0, you can tell the maximum and minimum apart by looking either side.

A maximum has positive gradients on the left, and a negative gradient on the right, as the curve now goes down after meaching the maximum.

A minimum has negative gradients on the left and a positive on the right.

To work out if the f(x) is increasing or decreasing, work out the derivative,f'(x), and see if its +ve or -ve. (+ve=increasing, -ve=decreasing)

The second derivative f(x) also written as d^2 y/dx^2 (derivative of the derivative; done in the same way) can be used to see if a stationary point is at maximum or minumum.

If f(x) = +ve then minimum, if f(x) = -ve then its a maximum

Optimization means finding the maximum/minimum depending on the situation.

[and here to, statement about why its increasing/decreasing/max/min]


Integration is the reverse of diffrentiation.

When integrating there is always an indefinate integral which includes a constant c, as 'c is indefinate. c is called the constant of integration.

The notation for indefinate integral is: ∫(......)dx

so if dy/dx=x^n (finding y in terms of x:) y is the indefinate integral; y=∫(x^n) = x^n +1/n+1 +c

You can find the constant if you substitue the (x,y) values with a co-ordinate on the graph.

Area Under a Graph

The area under the graph y=x^n between b and c is shown by ∫c^b(x^n )dx = [(x^n+1)/(n+1) ]c^b = b^n-1 / n+1 - c^n+1 / n+1(c^b means c in subscript and b in superscript ontop)

 is a long 's' for sum. It was thought of as summing up lots of strips of very small widths of δx

This expression is a definate integral. It works out to a numerical value. The constant of integration is not included as +c and -c cancel out.

Areas below the x-axis would have a negative value of integration, so we take the positive (modulus) value of the negative answer of integration.

The fundamental theorem of calculus says that the area under the graph f(x) is found by integrating f(x)

If A=area under the graph y=f(x) then dA/dx=f(x)

[usually they try to make this look hard, by sticking a triangle or something in it. Work out the area of triangle too, and take them away from each other etc. show what your doing clearly.]

Exam Tips

A few more random tips:

  • Whenever they ask you to prove something, they need words and numbers - include a statement. For example, therefore as f(3)=0, (x-3) is a factor of f(x).
  • You're not usually finished after factorising. Either you need to find the y-coordinate for both of them by substitution or you need to consider the critical values.
  • Keep it clear: re-write your answer somewhere obvious/box it/underline it if it's not obvious.
  • Practise papers allow you to see the types of questions asked and effectively 'play the examination game'. For your first paper or two, study the mark scheme carefully. Ensure you note where you lose marks not from error but from 'silly mistakes' like forgetting a statement on a factor theorem question.
  • Diagrams and brief sketches may help you visualise the question and even award you marks.
  • If you're stuck and unsure of where to begin, you dig right in (maybe in the pencil if you're completely uncertain). You may find it is easier than it appears - they can't examiner you on anything outside of the specification.
  • Normally, answers aren't typically 'nasty' - aka horrible fractions like 92/8 (e.g. 23/2). However, if you are continually figuring a 'nasty' answer, trust your gut.
  • Maths required a rested mind: sleep > cramming.
  • Use your common sense: for example, integrating x^2+4x+7 between the limits of 1 and 3 will not be negative. If your answer is illogical, plug your answers back in to check or consider a diagram that might assist you.
  • Assume the examiner is stupid. If you feel your working if a little unclear, annotate. If you feel you've gotten the wrong answer, annotate. Doing this will maximise your marks.