The Student Room Group

Solving cubic graphs

Hi, I'm currently doing the 2011 paper 1 for revision and I have came across a question, which I'm not sure about. I would really appreciate your help.

This is the paper
http://pastpapers.download.wjec.co.uk/a11-0185-09.pdfIt's 14c which I'm stuck with, I can't understand why the 'appropriate line' should be at y=10, but after that would be okay with solving it. Does anyone know why the line has to be drawn at y=10?Really appreciate it, thank you.
Reply 1
Original post by emily380
Hi, I'm currently doing the 2011 paper 1 for revision and I have came across a question, which I'm not sure about. I would really appreciate your help.

This is the paper
http://pastpapers.download.wjec.co.uk/a11-0185-09.pdfIt's 14c which I'm stuck with, I can't understand why the 'appropriate line' should be at y=10, but after that would be okay with solving it. Does anyone know why the line has to be drawn at y=10?Really appreciate it, thank you.


I'm getting an Error 404, check the link or post the question
Reply 2
Original post by Andy98
I'm getting an Error 404, check the link or post the question


Does this link work?

http://pastpapers.download.wjec.co.uk/a11-0185-09.pdf
Reply 3


Yes, I'll just find 14c then
Reply 4
Original post by Andy98
Yes, I'll just find 14c then


Thank you so much :smile:
I'm not sure if it's on my syllabus anymore but I guess I'd better see what I'm getting wrong just in case
Reply 5
Original post by emily380
Hi, I'm currently doing the 2011 paper 1 for revision and I have came across a question, which I'm not sure about. I would really appreciate your help.

This is the paper
http://pastpapers.download.wjec.co.uk/a11-0185-09.pdfIt's 14c which I'm stuck with, I can't understand why the 'appropriate line' should be at y=10, but after that would be okay with solving it. Does anyone know why the line has to be drawn at y=10?Really appreciate it, thank you.


Ahhh yes, that makes sense now. Look at the difference between the two equations. You'll notice x33x213x+5x^3 - 3x^2 - 13x + 5 is 10 less than the original expression. It could be written as x33x213x+15=10x^3 - 3x^2 - 13x + 15 = 10 which would represent the intersections between the graphs y=x33x213x+5y=x^3 - 3x^2 - 13x + 5 and y=10y=10
Reply 6
Original post by Andy98
Ahhh yes, that makes sense now. Look at the difference between the two equations. You'll notice x33x213x+5x^3 - 3x^2 - 13x + 5 is 10 less than the original expression. It could be written as x33x213x+15=10x^3 - 3x^2 - 13x + 15 = 10 which would represent the intersections between the graphs y=x33x213x+5y=x^3 - 3x^2 - 13x + 5 and y=10y=10


Thank you so much! I didn't notice that the equations were actually different! Makes complete sense now, thank you :smile: really appreciate it
Reply 7
Original post by emily380
Thank you so much! I didn't notice that the equations were actually different! Makes complete sense now, thank you :smile: really appreciate it


Yeah, it is a subtle difference.

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