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IGCSE Maths Calculus

I had Paper 3 for IGCSE Maths today and it was a bit of a stinker..too many questions on circle theoreom and no simeltaneous [sp] equations!

Luckily, there wasn't any questions on calculus, so they must be all on the 2nd paper tomorrow..Can anyone help me with this problem?

A curve has equations y = x^3 - 6x^2 + 9x - 2

a) Find the coordinates of the point of this curve at which the tangent is parallel to the line y = -3x + 5

b) Find the coordinates of the two turning points on this curve

Any help is much appreciated! :biggrin:
Reply 1
Avatar for mikesgt2
mikesgt2
a) Find dy/dx first. If the tangent is parallel to the line y=-3x+5 it has the same gradient. The gradient of that line is -3, so you need to find the values of x for which dy/dx = -3.
b) The stationary points can be found by solving the equation dy/dx=0.

I hope this helps.
Reply 2
Avatar for Casper
Casper
rachelina*
A curve has equations y = x^3 - 6x^2 + 9x - 2

a) Find the coordinates of the point of this curve at which the tangent is parallel to the line y = -3x + 5

b) Find the coordinates of the two turning points on this curve


a) dy/dx = 3x^2 - 12x + 9

The tangent is parallel to the line y = -3x + 5. This line has a gradient of -3, therefore the tangent has a gradient of -3 as well.

The tangent has the same gradient as the point on the curve that it touches.

Hence, gradient of the curve at this point is -3, therefore:

-3 = 3x^2 - 12x + 9
3x^2 - 12x + 12 = 0
(3x - 6)(x - 2) = 0

At the point on the curve where the gradient is -3, x = 2.

y = x^3 - 6x^2 + 9x - 2 = 2^3 - 6(2^2) + 9(2) - 2 = 8 - 24 + 18 - 2 = 0

Hence, the coordinates of the point on the curve at which the tangent is parallel to the line y = -3x + 5, are: (2, 0)

b) At the two turning points, the gradient of the curve will be zero, hence:

0 = 3x^2 - 12x + 9
0 = (3x - 3)(x - 3)

x = 1 and x = 3 at the two turning points.

For the turning point whose x-coordinate is 1:
y = x^3 - 6x^2 + 9x - 2 = 1^3 - 6(1^2) + 9(1) - 2 = 1 - 6 + 9 - 2 = 2

For the turning point whose x - coordinate is 3:
y = x^3 - 6x^2 + 9x - 2 = 3^3 - 6(3^2) + 9(3) - 2 = 27 - 54 + 27 - 2 = - 2

Hence, the coordinates of the two turning points on the curve are:

(1, 2) and (3, -2)
Reply 3
Avatar for Casper
Casper
mikesgt2
a) Find dy/dx first. If the tangent is parallel to the line y=-3x+5 it has the same gradient. The gradient of that line is -3, so you need to find the values of x for which dy/dx = -3.
b) The stationary points can be found by solving the equation dy/dx=0.

I hope this helps.


IGCSE Maths must be harder than GCSE Maths in the UK, because we don't do calculus until P1.
Reply 4
Avatar for rachelina*
rachelina*
OP
Cheers! I'll have to read that about 10 times until it computes in my mind but thank you very much! :smile:

Nah, IGCSE isn't THAT bad. Hard calculus is the worst..We get calculus, sets and function notation added in but lots of nasty things like 3D Trig, Standard Deviation, Sampling [ok thats not so nasty, but I was never that keen on it] are taken out
Reply 5
Avatar for etomac
etomac
what? i did cambridge's IGCSE Maths and there's no calculus (the extended paper -2 and 4) and 1 and 3 are for core people...
Reply 6
Avatar for rachelina*
rachelina*
OP
Well I don't know how that happened! There was simple differentiation on Paper 3 yesterday..I take Paper 3 and 4, the higher tier papers.

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