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.

Enhanced DBS Check

31/12/20145from 45 customer reviews

Kumar (Parent from Maidstone)

November 16 2017

Kind, answers any questions you have

Louise (Parent from Christchurch)

October 16 2017

Another excellent session. Thank you!

Louise (Parent from Christchurch)

October 9 2017

Another great Session. Thank you Laura!

Kumar (Parent from Maidstone)

October 1 2017

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

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 b 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

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 b 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