PremiumBrad  D. A Level Maths tutor, GCSE Maths tutor, Mentoring Maths tutor...
£30 /hr

Brad D.

Degree: Natural Sciences (Masters) - Durham University

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About me

Hey everyone, I'm Brad! I'm studying for a Masters degree in Natural Sciences at Durham, specializing in Maths and Physics. Even while studying at school, I loved the sciences, and I loved being able to help people fully come to grips with a topic that they struggled with. I've been a tutor for around 4 years, and had a good amount of previous tutoring experience across a wide range of ages, subjects and exam boards. I hope that through my tutorial sessions, I can help students feel confident and comfortable in tackling any question that the exam boards can throw at you. Above all, I intend the tutorials to be enjoyable and fun, but still productive. Science is absolutely fascinating, and hopefully once you fully understand it, you'll agree with me as well! Outside of tutoring times, I welcome any emails with questions/problems you might have.

About my sessions

Prior to the session, I'll ask to you to pick up on one/two major areas that you particularly want to work on, rather than wasting your time and money going through general areas which you already feel comfortable with. This ensures that we focus on and fix specific problems, and allows me to prepare material in advance!

I'll usually prepare about 20-30 minutes of "teaching", which includes going through some worked examples while covering the key concepts and explanations of all the syllabus points related to your chosen topic. I'll then formulate a "summary" section, including key formulas and critical processes. With this to hand, we will then work through some exam problems together, and analyse the solutions to see exactly what the exam board wants. Obviously, each student is different, so I can amend the plan to your individual needs; this is my only general plan which I find maximizes learning and understanding!

Subjects offered

SubjectLevelMy prices
Maths A Level £30 /hr
Physics A Level £30 /hr
Maths GCSE £30 /hr
Physics GCSE £30 /hr


Further Maths (AS)A-LevelA
Disclosure and Barring Service

CRB/DBS Standard


CRB/DBS Enhanced


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Ratings and reviews

4.8from 37 customer reviews

Sogil (Student) March 19 2016

His lessons are very enjoyable.

Sogil (Student) March 4 2016

very good

Sogil (Student) February 27 2016

Very helpful and informative.

Khadija (Student) February 19 2016

very helpful lesson on logarithms with clear explanations, thanks!
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Questions Brad has answered

How do I draw and sketch an equation?

Graph sketching can look like an intimidating process at times, but is easily broken down into a collection of fairly easy stages.  In summary: Factorise the equation - Find the roots/intersections of the equation - Find the turning points - Find the nature of the turning points - Find the a...

Graph sketching can look like an intimidating process at times, but is easily broken down into a collection of fairly easy stages. 

In summary:

Factorise the equation - Find the roots/intersections of the equation - Find the turning points - Find the nature of the turning points - Find the asymptotes.

0) Factorise the equation as much as possible! It always helps to see a simpler version of the equation.

1) The first step to solving a graphing problem is to figure out where it crosses both of the axis. The easiest way to find these is to substitute the values of y=0 and x=0 seperately into your given equation. These will give you coordinates of some points where the graph passes through.

Take an example of y=x3 - 2x. By substituting x=0, we find that y=0, and so we know the graph passes through the origin. When we sub y=0, we can solve the equation by factorising, and find that x is equal to 0, and +/- the square root of 2. We now know 3 points where the graph crosses the axis.

2) Turning points (or stationary points) are where the graph 'changes direction'. If you imagine a smiling face, like the letter U, the point where it is at the bottom and turns from moving downwards to moving upwards is the turning point. This is where the gradient is 0, since a line which is horizontal and flat has a gradient of 0. Using this fact, we can find the stationary points by differentiating using basic rules and setting this value to 0.

Using the previous example, y= x3 - 2x, the differential (dy/dx) = 3x2 - 2. To find where the gradient is 0, we set it to 0 and solve for x, giving us solutions of +/- the square root of 2/3. We can then substitute these back into the original equation to find coordinates. 

3) The nature of the turning point can be described as whether it is a normal U or an updside-down U, like an arch. If it is a normal U, it is a minimum, and if it's an arch, or upside down U, it's a maximum. The nature can be found by differentiating a second time, and substituting in the found stationary points. If the second derivative is positive, it is a minimum, and if it's a negative, then it's a maximum. 

In the previous example the second derivative (d2y/dx2) is 6x. The positive stationary point gives a positive value for d2y/dx2, and so is a minimum. The negative root gives a negative value, and so is a maximum. 

4) Some equations which have fractions in them may result in equations which may seem impossible to do; these are called asymptotes.

Take the equation y=1/x. If we follow our previous steps, and substitute y=0 and x=0, we have some problems. For x=0, we find that we need to evaluate 1/0, which is not possible to do. We can then deduce that this must be an asymptote, which means that the graph approaches the line x=0; it will get closer and closer and closer to 0 but never actually touches it. This means that no matter what value of y, we can never find a solution that crosses the x axis, since x can never be 0. The same situation happens with y=0, since the equation becomes 0=1, which of course is not true. Therefore, there is another asymptote where y=0, and the graph never touches this line. 

Following these steps means that you should be able to graph any given function!

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2 years ago

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