basically im screwed because i have my C1 exam in january and im doing really bad...
basically
a) how do you suggest revising
b) there is this one question that i've been trying to do...im sure its really easy but...
the curve C has equation y=4x^2+((5-x)/x)) x cannot = 0
the point P is on C and has x-coordinate 1.
show that the value of f'(x) at P is 3
i've managed that one..
then
find equation of the tangent to C at P
then
this tangent meets the x axis at point (k,0)
find value of k
cheers
its a level by the way
Help with maths?
the curve C has equation y=4x^2+((5-x)/x)) x cannot = 0
the point P is on C and has x-coordinate 1.
show that the value of f'(x) at P is 3
i've managed that one..
How did you do the derivative?
Did you do it this way?
y=4x^2+((5-x)/x))
y = 4 x^2 + 5/x - 1
y = 4 x^2 + 5 x^(-1)
y' = 2 * 4 * x - 5 * x^(-2)
y' = 8 x - 5/ x^2
at x = 1, y' = 8 - 5 = 3.
then
find equation of the tangent to C at P
then
You have just shown that the derivative at x = 1 is 3.
The slope of the tangent line at a point is the derivative
at that point.
The slope of the tangent line is 3.
The equation of the tangent line is (y - f(1) ) = 3 * (x-1)
From
y = 4 x^2 + 5/x - 1
we get
f(x) = 4 x^2 + 5/x - 1
and
f(1) = 4 + 5 - 1 = 8
The equation of the tangent line is (y - 8) ) = 3 * (x-1)
or y = 8 + 3 (x - 1)
or y = 3 x + 5
When y = 0, what does x equal?
3 x + 5 = 0
3 x = - 5
x = -5/3
this tangent meets the x axis at point (k,0)
find value of k
k = -5/3
Reply:go on samlearning.com that will help you
Reply:i think you sould re/post this question about the28th (you may then get a result)
Reply:Looks like you are going to fail :S
Never mind though, i hear McDonald's are starting a recruiting campaign in the new year.
Reply:in the equation substitute the values for x,y----(k,0) and find the value of k...
Reply:You need a personal tutor!
Reply:As you've managed the first part of the question, you're not a lost cause. I think it might help if you draw a rough sketch. Remember that the tangent at the point where it meets the curve has exactly the same gradient as the curve, but it's a straight line of course so you need y=mx +c. You can find the graident of the curve by differentiation. Once you have the equation of the tangent, the last part is easy.
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