Currently unavailable: for new students
Degree: Chemistry and Maths (Bachelors) - Leeds University
Me, Myself & I
I am studying Chemistry and Maths at the University of Leeds and I am about to go into the second year of my degree. From quite an early age I realised Maths and Science are where my interests lie.
I volunteered as a Young Leader in Scouts and Beavers for 2 years so I have a lot of experience of working with children, some as young as 5. Having had to organise a large group of 5-7 year olds taught me the importance in pre planning so every session we have, I will be fully prepared to answer any questions you may have.
During the tutoring sessions you will have complete control over what we cover, whether its revision over a large topic or something very specific, preperation for a test or help with a difficult piece of homework, no matter the problem by the end of the hour we will have the problem solved and hopefully you will have a good enough understanding of the problem to then teach someone else.
If you have any questions regarding anything you have just read or how the whole tutoring process works you can contact me on here via a book the tutor session which is completely free and gives you a chance to decided if I will be able to help you out.
Hope to hear from you,
|Chemistry||A Level||£22 /hr|
|Maths||A Level||£22 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
Paul (Parent) November 17 2016
Dylan (Student) October 19 2016
Stephane (Parent) November 25 2016
Paul (Parent) December 1 2016
A tangent to a curve at a specific point along the line will have excatly the same gradient as the curve at that point. For example if the point was (0,0) the tangent would just be a horizontal line along the x-axis. To work out the gradient we simply differentiate the curve, set this to equal 0 and solve with the given value of x.
Here dy/dx = 3x2-8x
Setting this to 0 and solving gives us 3(3)2-8(3)=3 and so our gradient at the point (3,-7) is 3
We are now able to use the equation y-y(1)=m(x-x(1)) where m is the gradient and x(1) & y(1) are the coordinates we've been given.
Rearranging we get y=3(x-3)-7
y=3x-16 and this is the equation of the tangent to the curve y=x3-4x2+2 at the point (3,-7)see more