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Further Maths question - differentiation

Topic is differentiation.

The curve with the equation y=x^3-4x^2+3x crosses the x-axis at the points A, B and C.
a) Find the coordinates of A, B and C.
b) Find the gradient of the curve at each of the points.

I know the answers:
a) x=0, x=3, x=1
b) 3, 6, -2

But how do I get to those answers?
So you know that points A, B and C are on the x-axis. This means that at these points, y=0. So make the equation:

0=x^3-4x^2+3x.

Then you can factor this by removing an x from the equation to get:

0= x(x^-4x+3) which is equal to 0=x(x-3)(x-1)

from this, you know that x=0, x=3, x=1. This is the answer to (a).

For (b), you have to differentiate. So,

dy/dx = 3x^2-8x+3

then substitute x=0 and you get an answer of 3 for the gradient, substitute x=3 and you get 6 and substitute x=1 and you get -2.


Hope that made sense!
Reply 2
Original post by metellaest
so you know that points a, b and c are on the x-axis. This means that at these points, y=0. So make the equation:

0=x^3-4x^2+3x.

Then you can factor this by removing an x from the equation to get:

0= x(x^-4x+3) which is equal to 0=x(x-3)(x-1)

from this, you know that x=0, x=3, x=1. This is the answer to (a).

For (b), you have to differentiate. So,

dy/dx = 3x^2-8x+3

then substitute x=0 and you get an answer of 3 for the gradient, substitute x=3 and you get 6 and substitute x=1 and you get -2.


Hope that made sense!


thank you so so much
Original post by mxgx
thank you so so much


No problem :smile:

Is 'Further Maths' the same as Additional Maths (which is what I'm doing), or are they different subjects?
Reply 4
Could anyone also help on b for this question?

A curve has the equation y=x^3+x^2-4x+1.

a) Find the gradient of the curve at the point P (-1,5)
^^ I've already done this and the answer is -3.

b) Given that the gradient at the point Q on the curve is the same as the gradient at the point P, find, as exact fractions, the coordinates of the point Q.
Reply 5
Original post by metellaest
No problem :smile:

Is 'Further Maths' the same as Additional Maths (which is what I'm doing), or are they different subjects?


I'm not sure? My exam is the AQA Level 2 Further Maths exam.
Reply 6
Original post by mxgx
Could anyone also help on b for this question?

A curve has the equation y=x^3+x^2-4x+1.

a) Find the gradient of the curve at the point P (-1,5)
^^ I've already done this and the answer is -3.

b) Given that the gradient at the point Q on the curve is the same as the gradient at the point P, find, as exact fractions, the coordinates of the point Q.


Find dy/dx and equate it to -3, solve for x (should be a quadratic). One of the solutions will be x=-1, the other solution is the point you want. Then plug this x-value back into the equation to get your coordinates.

PS: Moved to maths.
Reply 7
this is the extra maths which some people do beyond GCSE

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