1. ## Diffrentiation help needed!!

1. A curve C has equation y=x^3 - 5x^2 + 5x +2
a) find dy/dx in terms of x
b)The points P and Q lie on C. The gradient of C at both p and q is 2. The x coordiante of P is 3.
i)Find the x coordinate of Q
ii)Find an equation for the tangest to C at P, giving your answer in the form of y =mx + c
iii)If this tangest intersects the coordinates axes at the points R and S find the length of RS giving your answer as a surd.

2. Sigh im off in a hour...noone can every solve my questions xD
BTW i worked out part a) its 3x^2-10x+5

3. Originally Posted by cougar
Sigh im off in a hour...noone can every solve my questions xD
BTW i worked out part a) its 3x^2-10x+5
Heh, i don't think you should post here and people do your homework for you whilst you go off its a help forum not a 'do my homework for me forum'.

I take it your doing core 1 in AS levels? Its a simple question that always comes up in an exam and its always the same ...and their always about 7 marks! An easy 7 marks to bag ....so please don't think im having a moan , i mean it in the best intentions(i did this exam last year) that you need to be able to do it for da exam .

Also i'll recommend http://www.mathhelpforum.com/math-he...ths.co.uk....i signed up for it last January and since i have been using it got over 85% in all exams afterwards...just a suggestion...I managed to get both an A in maths and further maths last year and livemaths.co.uk played a really big part in helping me do it, especially because i had some hard family situations going on.

Your correct in that part a is
$
3x^2 - 10x + 5
$

Part i'll work out and post in a second but i'll give you a hint...or are you stuck on something specific?

b.) you need to find when the gradient is 2...so do

$3x^2 - 10x + 5 = 2$
and solve it. You already know one root is 3...because it tells you, so the other answer you get will be the x co-ord to Q.

c.) The tangent...you need to work out the co-ordinate of of P...you already know x is 3(so u can get the y value)...you know the gradient is 2...so use the formular
y - y1 = m(x - x1)
Substitute in your values and re-arrange to get y = mx + c

d.) Now you have tangent etc...draw it and all the points to help you work out what the question is asking.
You can get R and S by putting using the equation of the tangent...subbing in x as 0 (to get where it crosses the y axis) and by subbing in y is 0 (to get where it crosses the x axis)...you now have points s and r and can work out its length by Pythagoras

Give it a go...if you still cant get the answer let me know and ill post the working for you ...in addition feel free to pm me since i've done furthermaths AS already and should be able to help you with D1, C1, C2, M1, FP1, S1.