Barnum S. A Level Maths tutor, A Level Further Mathematics  tutor, A ...

Barnum S.

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Degree: Mathematics (Bachelors) - Cambridge University

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About me

My name is Barnum and I am a second year maths undergraduate at Cambridge University. I have A-Levels in Maths, Further Maths, Physics, Chemistry and German, all at grade A*

I am patient, friendly and passionate about maths and science, and enjoy sharing this passion with others. I can help you to understand the key concepts, develop problem solving skills and ultimately succeed in examinations — and beyond!

I have many years of experience tutoring maths and science from KS2 up to A-Level, both in person and online. Additionally, I have worked within schools as a Maths Intervention Tutor, helping GCSE and A-Level students with exam preparation, as well as supporting talented mathematicians with their Oxbridge applications.

Please feel free to get in touch if you have any questions or would like arrange a free ‘Meet the Tutor Session’. I look forward to meeting you.

Subjects offered

SubjectLevelMy prices
Further Mathematics A Level £20 /hr
Maths A Level £20 /hr
Physics A Level £20 /hr
Maths GCSE £18 /hr
Science GCSE £18 /hr
.STEP. Uni Admissions Test £25 /hr

Qualifications

QualificationLevelGrade
MathsA-LevelA*
Further MathsA-LevelA*
PhysicsA-LevelA*
ChemistryA-LevelA*
GermanA-LevelA*
STEPUni Admissions Test1
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

24/11/2014

Currently unavailable: for new students

Ratings and reviews

4.7from 40 customer reviews

Nancy (Parent) May 5 2016

Whole hour only done two questions. Couldn't answer other questions so had to ask my son to figure out them himself. Not good value for the price!

Jahnavi (Student) June 23 2016

Good lesson

Dom (Parent) June 4 2016

Barnum is a very precise tutor, and keen to carry his explanations over the online interface. We appreciated.

Miriam (Student) June 15 2016

Extremely helpful and easy to work with!
See all reviews

Questions Barnum has answered

Find the coordinates of the stationary points of the curve y = 2x^3 + 3x^2 - 12x + 1

This is a typical A/S level question, which can be broken down into three key steps. 1) Differentiation: Remember that at a stationary point, the gradient of the curve is zero. Therefore, we first have to find the gradient, which means differentiating. (Multiply the coefficient of x by the po...

This is a typical A/S level question, which can be broken down into three key steps.

1) Differentiation: Remember that at a stationary point, the gradient of the curve is zero. Therefore, we first have to find the gradient, which means differentiating. (Multiply the coefficient of x by the power, and then reduce the power by 1)

y = 2x3 + 3x2 - 12x + 1

dy/dx = 6x2 + 6x - 12

2) Solve Quadratic: To find where the gradient is zero (dy/dx = 0), we must solve the quadratic equation,

dy/dx = 6x2 + 6x - 12 = 0.

This quadratic can be most easily solved by factorisation (alternative methods are completing the square and using the quadratic formula). 

Firstly we can take 6 out as a common factor, which results in 6(x+ x - 2) = 0. We can then factorise the terms in brackets which results in 6(x+2)(x-1) = 0.

This gives the two solutions x = -2, x = 1.

3) Substitution: So far, we have only found the x coordinate of the stationary point, but to answer the question fully we need the y coordinate as well. We do this by substituting x = -2, and x = 1 into the original equation of the curve, y = 2x3 + 3x- 12x + 1

When x = -2, we can calcuclate that y = 21. Similarly when x = 1, y = -6.

Thus the coordinates of the stationary points are (-2, 21) and (1, -6).

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9 months ago

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