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

Barnum S.

£22 - £26 /hr

Currently unavailable: for new students

Studying: Mathematics (Bachelors) - Cambridge University

4.8
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66 reviews| 159 completed tutorials

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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.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.

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Ratings & Reviews

4.8from 66 customer reviews
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Jahnavi (Student)

January 17 2017

Great lesson

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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!

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Jahnavi (Student)

June 23 2016

Good lesson

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Dom (Parent)

June 4 2016

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

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Qualifications

SubjectQualificationGrade
MathsA-level (A2)A*
Further MathsA-level (A2)A*
PhysicsA-level (A2)A*
ChemistryA-level (A2)A*
GermanA-level (A2)A*
STEPUni admission test1

Subjects offered

SubjectQualificationPrices
Further MathematicsA Level£24 /hr
MathsA Level£24 /hr
PhysicsA Level£24 /hr
MathsGCSE£22 /hr
ScienceGCSE£22 /hr
.STEP.Uni Admissions Test£26 /hr

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 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).

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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