Hello! I'm currently a 2nd year Mathematics student at the University of Edinburgh with a sturdy passion for all things Mathematics. Not only do I love Maths, I love teaching Maths and helping people with problems. In particular, I enjoy finding the best way to explain tricky concepts so that they suddenly become easily understandable.
In terms of previous experience, I helped GCSE Maths students in the run up to their exams. I also volunteered at a local primary school helping the kids out during their lessons.
I am also a fully-trained MathPALS (Peer Assisted Learning Scheme) leader at the University of Edinburgh, which means I organise and run informal interactive study sessions for 1st year Maths students. Through my PALS training, I have learned the importance of facilitation whereby I guide and prompt the student to the correct solution rather than just revealing what it is straight-away. This has proven to be much more beneficial to the student’s understanding as it requires more independent thought.
In my lessons, what we cover is completely up to you. I am able to aid you with whatever you’re struggling with, and can teach you whatever you would like to learn. I ensure that my students really understand the material so that they can apply it to any situation they encounter in Maths.
Additionally, I can offer advice on UCAS applications for Maths courses, such as how to perfect that personal statement or how to prepare for admissions tests, having been there myself. And of course I can also give first-hand experience of what it’s actually like studying Maths at university.
|Further Mathematics||A Level||£20 /hr|
|Maths||A Level||£20 /hr|
|Further Mathematics||GCSE||£18 /hr|
|Further Mathematics||IB||£20 /hr|
|Maths||13 Plus||£18 /hr|
|Maths||11 Plus||£18 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
First, we must find the value of y when x = 2.
y = x3+4x+5 = (2)3+4(2)+5 = 21
Then we must find the gradient of the tangent line. This can be done by differentiating y with respect to x and substituting x = 2.
dy/dx = 3x2+4 = 3(2)2+4 = 16
Now that we have a point (2,21) and the gradient (m = 16) of our tangent line, we can find the equation of the tangent using the formula:
y-y1 = m(x-x1)
y-21 = 16(x-2)
y = 16x-32+21
Thus y = 16x-11 is the equation of the tangentsee more