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need help with this nasty MATHS question !!!!

this question has just totally confused me - pls help !

the points P, Q and R on the curve with equation y=1-x^3 have the x- coordinates 2, -1, and -2. the curve crosses the x axis at S. Find the gradient of the curve at P, Q, R and S.
you need to find the gradient function and then put in suitable x values.
Reply 2
Avatar for jsMath
jsMath
8
1. Substitute the x valuse for each component of the coordinate in your function, to find the y values of the points.
2. Use m=y2-y1/x2-x1

An alternative way to solve this, is to differentiate and plug in values...
(edited 8 years ago)
Yeah I'd differentiate and substitute in the x-values to find the answer. Using m=y2-y1/x2-x1 normally only works for straight lines.

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(edited 8 years ago)
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Reply 4
Avatar for FAAWAH
FAAWAH
OP
9
Original post by nikhiltikare
Yeah I'd differentiate and substitute in the x-values to find the answer. Using m=y2-y1/x2-x1 normally only works for straight lines.

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THANK YOU GUYS SO SO SO SO MUCH !!!!! really appreciate it, but would that mean that pqr and s all have different gradients and cannot have one ???
Yes they have different gradients, it's a x^3 curve so therefore it has 2 turning points. If it was a straight line (y = 1-x) then it'd only have 1 gradient.

:smile:
Reply 6
Avatar for FAAWAH
FAAWAH
OP
9
Original post by nikhiltikare
Yes they have different gradients, it's a x^3 curve so therefore it has 2 turning points. If it was a straight line (y = 1-x) then it'd only have 1 gradient.

:smile:



thank you so much!!
Yeah no worries :biggrin:

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