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Coordinate Geometry C4 HELP!

How can I solve this???

Many thanks in advance. :biggrin:
(edited 8 years ago)
Original post by Scopeowner
How can I solve this???

Thanks in advance


I think you forgot to upload a photo, I'd be happy to help though!
Reply 2
CG4.JPG

Sorry, my computer has been messing me around. Please find attached the photo... :biggrin:
(edited 8 years ago)
Original post by Scopeowner
CG4.JPG

Sorry, my computer has been messing me around. Please find attached the photo... :biggrin:


Given that this curve crosses the x-axis at a point X, we may deduce that when
y=0,x=X    (X,0)y = 0, x = X \implies (X,0)

likewise:
x=0,y=Y    (0,Y)x = 0, y = Y \implies (0,Y)

Then you want to find the values of the variable t for y = 0 and x = 0 respectively

If the t3+1=0t^3 + 1 = 0 is the part that is confusing you, let f(x)=t3+1f(x) = t^3 + 1 and notice that f(1)=(1)3+1=0f(-1) = (-1)^3 + 1 = 0 does this give you a clue how to factorise this expression?
(edited 8 years ago)
Reply 4
Original post by JakeThomasLee
Given that this curve crosses the x-axis at a point X, we may deduce that when
y=0,x=X    (X,0)y = 0, x = X \implies (X,0)

likewise:
x=0,y=Y    (0,Y)x = 0, y = Y \implies (0,Y)

Then you want to find the values of the variable t for y = 0 and x = 0 respectively

If the t3+1=0t^3 + 1 = 0 is the part that is confusing you, let f(x)=t3+1f(x) = t^3 + 1 and notice that f(1)=(1)3+1=0f(-1) = (-1)^3 + 1 = 0 does this give you a clue how to factorise this expression?


Hmm, I am still stuck. After working out the value of 't' when y = 0, I can't seem to work out the value of 'q' for which I keep on getting as 0. Any chance of you showing me some working? Because I've been stuck on this question for ages now. :frown:
Original post by Scopeowner
Hmm, I am still stuck. After working out the value of 't' when y = 0, I can't seem to work out the value of 'q' for which I keep on getting as 0. Any chance of you showing me some working? Because I've been stuck on this question for ages now. :frown:


Coordinate Geom Q1.jpeg

Do let me know if the answer is different or if I made a mistake though! (It's been a long day)
Reply 6
Original post by JakeThomasLee
Coordinate Geom Q1.jpeg

Do let me know if the answer is different or if I made a mistake though! (It's been a long day)


No, there has been no mistake. Thank you very much for your all of your help, it makes a lot more sense now.

I have managed to attempt the question and I got the exact same answer as you which is '8/3'. I have verified this with the answer book and it is correct!

Once again, thank you so much for your kind help! :biggrin:
Original post by Scopeowner
No, there has been no mistake. Thank you very much for your all of your help, it makes a lot more sense now.

I have managed to attempt the question and I got the exact same answer as you which is '8/3'. I have verified this with the answer book and it is correct!

Once again, thank you so much for your kind help! :biggrin:


My pleasure

Have fun!

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