Laura M. A Level Maths tutor, A Level Chemistry tutor, GCSE Chemistry...

Laura M.

£22 - £24 /hr

Currently unavailable: for new students

Studying: Natural Sciences (Other) - Durham University

5.0
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34 reviews| 95 completed tutorials

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About me

Hi, I'm Laura and I'm a 3rd year undergraduate studying Natural Sciences (Chemistry and Maths) at the University of Durham. I have taken part in tutoring schemes both at sixth form and university and through this am CRB enhanced checked. I have always enjoyed helping others with concepts they may find difficult or challenging, both as a tutor and with my peers through-out my education. I believe that the real goal of education should be to understand the topic and questions rather than just learning the answer. In my tutoring sessions I will aim to answer any particular questions a student may have and then present them with a similar problem but in a different style to help them achieve an all-round understanding, which is more benficial and longer lasting. I am approachable, friendly and patient which have all proved invaluable traits in my previous tutoring experiences. My A-levels are in Maths (AQA, A*), Further Maths (AQA,A), Chemistry (OCR A, A*) and Physics (OCR B: Advancing Physics, A*), although I am willing to do further research into content required for other courses if specified. I also achieved a Distinction in the Mathematics AEA, which can be a requirement for many university Mathematics course, and would be happy to give any help or guidance with this too.

Hi, I'm Laura and I'm a 3rd year undergraduate studying Natural Sciences (Chemistry and Maths) at the University of Durham. I have taken part in tutoring schemes both at sixth form and university and through this am CRB enhanced checked. I have always enjoyed helping others with concepts they may find difficult or challenging, both as a tutor and with my peers through-out my education. I believe that the real goal of education should be to understand the topic and questions rather than just learning the answer. In my tutoring sessions I will aim to answer any particular questions a student may have and then present them with a similar problem but in a different style to help them achieve an all-round understanding, which is more benficial and longer lasting. I am approachable, friendly and patient which have all proved invaluable traits in my previous tutoring experiences. My A-levels are in Maths (AQA, A*), Further Maths (AQA,A), Chemistry (OCR A, A*) and Physics (OCR B: Advancing Physics, A*), although I am willing to do further research into content required for other courses if specified. I also achieved a Distinction in the Mathematics AEA, which can be a requirement for many university Mathematics course, and would be happy to give any help or guidance with this too.

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Enhanced DBS Check

31/12/2014

Ratings & Reviews

5from 34 customer reviews
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Louise (Parent)

October 9 2017

Another great Session. Thank you Laura!

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Kumar (Parent)

October 1 2017

Really kind and goes in depth about the specific topic you want to learn

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Louise (Parent)

September 18 2017

Laura is a fantastic tutor and her lessons are always well prepared. Everything covered is catered to my daughter’s need and clearly explained. Couldn't ask for a better tutor.

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Louise (Parent)

September 10 2017

Laura is extremely knowledgeable, friendly and helpful. Very lucky to have found such a great tutor and would highly recommend Laura to any students.

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Qualifications

SubjectQualificationGrade
MathematicsA-level (A2)A*
Further MathematicsA-level (A2)A
ChemistryA-level (A2)A*
PhysicsA-level (A2)A*
Natural Sciences (Chemistry and Mathematics)Degree (Bachelors)1ST CLASS (HONS)

Subjects offered

SubjectQualificationPrices
ChemistryA Level£24 /hr
Further Mathematics A Level£24 /hr
MathsA Level£24 /hr
ChemistryGCSE£22 /hr
MathsGCSE£22 /hr
PhysicsGCSE£22 /hr
ScienceGCSE£22 /hr

Questions Laura has answered

How to translate a function of form y = f(x)

In a translation the graph of the function is moved in either the x or y direction. 

To perform a translation by amount a in the positive x direction (to the right on a graph) we replace every x in the equation with (x-a), so the function y = f(x) becomes y = f(x-a). A translation of a in the negative x direction can be thought of as a translation of -a in the positive direction, so this time y = f(x) becomes y = f(x-(-a)) = f(x+a).

For translations in the y direction a similar rule applies but this time we substitute y. A translation of in the positive y direction (upwards) will change y = f(x) into (y-b) = f(x), which is often written as y = f(x) + b. Similarly a translation by b in the negative y direction would give (y+b) = f(x) which may be written y = f(x) - b.

Example: Translate the function y = x^2 + 3x + 5 by vector (4,-1).

Solution: The vector given means that we want to translate the function by 4 units in the positive x direction and 1 units in the negative y direction. This transforms the original equation into

(y+1) = (x-4)^2 + 3(x-4) + 5
which expands to
y +1 = x^2 - 8x + 16 + 3x - 12 + 5
and simplifies to
y = x^2 - 5x + 8

 

In a translation the graph of the function is moved in either the x or y direction. 

To perform a translation by amount a in the positive x direction (to the right on a graph) we replace every x in the equation with (x-a), so the function y = f(x) becomes y = f(x-a). A translation of a in the negative x direction can be thought of as a translation of -a in the positive direction, so this time y = f(x) becomes y = f(x-(-a)) = f(x+a).

For translations in the y direction a similar rule applies but this time we substitute y. A translation of in the positive y direction (upwards) will change y = f(x) into (y-b) = f(x), which is often written as y = f(x) + b. Similarly a translation by b in the negative y direction would give (y+b) = f(x) which may be written y = f(x) - b.

Example: Translate the function y = x^2 + 3x + 5 by vector (4,-1).

Solution: The vector given means that we want to translate the function by 4 units in the positive x direction and 1 units in the negative y direction. This transforms the original equation into

(y+1) = (x-4)^2 + 3(x-4) + 5
which expands to
y +1 = x^2 - 8x + 16 + 3x - 12 + 5
and simplifies to
y = x^2 - 5x + 8

 

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3 years ago

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