Laura P. A Level Maths tutor, GCSE Maths tutor

Laura P.

Currently unavailable: for new students

Studying: Aerospace Engineering (Masters) - Bristol University

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1 review| 23 completed tutorials

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About me

Hi! I'm Laura, a fourth year Aerospace Engineering student. As someone who has always been passionate about Maths and Science, and always at the top of my class, I am keen to show others how understandable and even interesting, it can all be! 

Throughout High School and Sixth Form, I coached friends through their revision and past papers, as well as my twin brother, taking slightly different exams to me - very rewarding work.

I believe the most effective way of learning is firstly through a step-by-step explanation of a topic, followed by (most importantly I think), lots of practice questions to ensure the topic has been fully understood.

I am incredibly enthusiastic about my subjects - hoping to inspire you a little!

Hi! I'm Laura, a fourth year Aerospace Engineering student. As someone who has always been passionate about Maths and Science, and always at the top of my class, I am keen to show others how understandable and even interesting, it can all be! 

Throughout High School and Sixth Form, I coached friends through their revision and past papers, as well as my twin brother, taking slightly different exams to me - very rewarding work.

I believe the most effective way of learning is firstly through a step-by-step explanation of a topic, followed by (most importantly I think), lots of practice questions to ensure the topic has been fully understood.

I am incredibly enthusiastic about my subjects - hoping to inspire you a little!

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Ratings & Reviews

5from 1 customer review
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RACHEL (Student)

May 22 2016

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Qualifications

SubjectQualificationGrade
MathsA-level (A2)A*
PhysicsA-level (A2)A
ChemistryA-level (A2)A
Further Maths (AS)A-level (A2)A

Subjects offered

SubjectQualificationPrices
MathsA Level£20 /hr
MathsGCSE£18 /hr

Questions Laura has answered

How do I plot y=x^2-1?

To plot any graph, you need to find out where the curve cuts the x and y-axis, and where it's stationary points are, aswell as what type of stationary point you have.

Here are the steps:

1) Let x=0; find where the curve passes the y axis.

   Here, y =-1. This means the curve passes the y axis at (0,-1)

2) Let y=0; find where the curve passes the x axis.

   Here we get 0=x2-1 

   So  x2=1 

   The square roots of 1 are: 1 and -1. (1x1=1, -1x-1=1) 

  This means the curve crosses the x axis as (-1, 0) and (1,0).

Remember the order of this equation (the highest power to which a x is raised) gives the number of times the curve crosses the x axis, here it is two. (This is the same for the y axis. Here, there order of y is 1, so the curve passes through the y axis 1 time.

 

3) Differentiate the equation with respect to x and make this equal to 0; find the stationary points.

   dy/dx =2x=0

   This tells us there is a stationary point at x=0. We know from step 1 that this corresponds to the coordinate (0, -1). 

For the curve to pass through all of these points, it might be clear that the graph is a "sad face", passing through (0, -1), (1, 0) and (-1, 0) with the MINIMUM located at (0,-1)

However, if this is not clear to you, difference the equation again with respect to x. If this gives you a postive answer, you have a minimum, a negative answer means you have a maximum.

4) Differentiate again with respect to x to find the second derivative; does this give you a postive answer?

Here, d2y/dx2= 2. 

Obviously this is a postive number meaning you have a minimum.

 

To plot any graph, you need to find out where the curve cuts the x and y-axis, and where it's stationary points are, aswell as what type of stationary point you have.

Here are the steps:

1) Let x=0; find where the curve passes the y axis.

   Here, y =-1. This means the curve passes the y axis at (0,-1)

2) Let y=0; find where the curve passes the x axis.

   Here we get 0=x2-1 

   So  x2=1 

   The square roots of 1 are: 1 and -1. (1x1=1, -1x-1=1) 

  This means the curve crosses the x axis as (-1, 0) and (1,0).

Remember the order of this equation (the highest power to which a x is raised) gives the number of times the curve crosses the x axis, here it is two. (This is the same for the y axis. Here, there order of y is 1, so the curve passes through the y axis 1 time.

 

3) Differentiate the equation with respect to x and make this equal to 0; find the stationary points.

   dy/dx =2x=0

   This tells us there is a stationary point at x=0. We know from step 1 that this corresponds to the coordinate (0, -1). 

For the curve to pass through all of these points, it might be clear that the graph is a "sad face", passing through (0, -1), (1, 0) and (-1, 0) with the MINIMUM located at (0,-1)

However, if this is not clear to you, difference the equation again with respect to x. If this gives you a postive answer, you have a minimum, a negative answer means you have a maximum.

4) Differentiate again with respect to x to find the second derivative; does this give you a postive answer?

Here, d2y/dx2= 2. 

Obviously this is a postive number meaning you have a minimum.

 

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3 years ago

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