Peter H.

Peter H.

£30 /hr

Industrial Systems Manufacture and Management (Masters) - Cambridge University

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17 completed lessons

About me

Hi! My name is Peter. I am a mechanical engineering graduate from the Cambridge University. I love all things science and technology and have a passion for turning complex problems into simple solutions. I have great experience in face to face tutoring in the subjects of Maths, Chemistry and Physics with all of my students achieving their goals and improving their grades. I firmly believe that students can become experts of science based subjects if they get a strong grasp of the fundamentals and do not just learn off rules!


On a personal note, I love water-sports such as windsurfing and waterskiing and I also love to sing and play the piano. I hope to hear from you soon and help you greatly improve your mathematical and scientific abilities :)

Hi! My name is Peter. I am a mechanical engineering graduate from the Cambridge University. I love all things science and technology and have a passion for turning complex problems into simple solutions. I have great experience in face to face tutoring in the subjects of Maths, Chemistry and Physics with all of my students achieving their goals and improving their grades. I firmly believe that students can become experts of science based subjects if they get a strong grasp of the fundamentals and do not just learn off rules!


On a personal note, I love water-sports such as windsurfing and waterskiing and I also love to sing and play the piano. I hope to hear from you soon and help you greatly improve your mathematical and scientific abilities :)

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About my sessions

From my experience in tutoring and taking exams, I find the best way to tackle a section of a syllabus is to first go over the fundamental rules and background to the area. After this I like to jump straight into past exam questions and show how the fundamental principles can be used to solve exam questions. Through understanding the basics, you add the correct tools to your tool box, that if used correctly, can answer any exam question you might meet.


I will measure progress by assigning you past exam questions to complete, and measuring your performance in these questions. These questions are the ones you will be given in the final exam. The more practice you get, the better!

From my experience in tutoring and taking exams, I find the best way to tackle a section of a syllabus is to first go over the fundamental rules and background to the area. After this I like to jump straight into past exam questions and show how the fundamental principles can be used to solve exam questions. Through understanding the basics, you add the correct tools to your tool box, that if used correctly, can answer any exam question you might meet.


I will measure progress by assigning you past exam questions to complete, and measuring your performance in these questions. These questions are the ones you will be given in the final exam. The more practice you get, the better!

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Personally interviewed by MyTutor

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

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02/10/2018

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Qualifications

SubjectQualificationGrade
MathsIrish leaving certA+
ChemistryIrish leaving certA+
PhysicsIrish leaving certA+
Mechanical EngineeringDegree (Bachelors)1ST
MusicIrish leaving certA

General Availability

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Pre 12pm
12 - 5pm
After 5pm

Pre 12pm

12 - 5pm

After 5pm
Mon
Tue
Wed
Thu
Fri
Sat
Sun

Subjects offered

SubjectQualificationPrices
MathsA Level£30 /hr
ChemistryGCSE£30 /hr
MathsGCSE£30 /hr
PhysicsGCSE£30 /hr
ScienceGCSE£30 /hr

Questions Peter has answered

Find the max/min value of the function: f(x) = 5x^2 - 20x + 15

A quadratic function will either have a minimum or a maximum value depending on the shape of the function. A quadratic function with a positive value in front of the x^2 term has a U shape and therefore has a defined minimum value at the turning point of the graph. A function with a negative value in front of the x^2 term has an n shape and therefore has a defined maximum value at the turning point of the graph. The function in this question is therefore U shaped and has a defined negative value at the turning point.To find the minimum value of the quadratic, we must find at what point the gradient (or the slope) of the function is 0. At this point, the function is at its turning point (the minimum point on the graph). The gradient is found by differentiating the function. Therefore, the minimum value is found by differentiating the function (to find the gradient) and setting it equal to 0. Our function is f(x) = 5x^2 - 20x + 15. Applying simple differentiation rules (Multiplying the power by the coefficient and then reducing the power by 1) we get the follow term: f’(x) = 10x - 20. If we set that equal to zero we get: 10x - 20 = 0. Solving this equation gives us: x = 2. Therefore, at x = 2, the quadratic has a minimum value. To find the y value of the minimum, we simply sub x = 2 back into the original equation. y = 5(2)^2 - 20(2) +15 gives us a y value of -5. The minimum point is therefore at (2,-5) A quadratic function will either have a minimum or a maximum value depending on the shape of the function. A quadratic function with a positive value in front of the x^2 term has a U shape and therefore has a defined minimum value at the turning point of the graph. A function with a negative value in front of the x^2 term has an n shape and therefore has a defined maximum value at the turning point of the graph. The function in this question is therefore U shaped and has a defined negative value at the turning point.To find the minimum value of the quadratic, we must find at what point the gradient (or the slope) of the function is 0. At this point, the function is at its turning point (the minimum point on the graph). The gradient is found by differentiating the function. Therefore, the minimum value is found by differentiating the function (to find the gradient) and setting it equal to 0. Our function is f(x) = 5x^2 - 20x + 15. Applying simple differentiation rules (Multiplying the power by the coefficient and then reducing the power by 1) we get the follow term: f’(x) = 10x - 20. If we set that equal to zero we get: 10x - 20 = 0. Solving this equation gives us: x = 2. Therefore, at x = 2, the quadratic has a minimum value. To find the y value of the minimum, we simply sub x = 2 back into the original equation. y = 5(2)^2 - 20(2) +15 gives us a y value of -5. The minimum point is therefore at (2,-5) 

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3 months ago

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