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!

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.

No DBS Check

5from 1 customer review

RACHEL (Student)

May 22 2016

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=x^{2}-1

So x^{2}=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, d^{2}y/dx^{2}= 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=x^{2}-1

So x^{2}=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).

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.

Here, d^{2}y/dx^{2}= 2.

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