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C1 Question. Help Please...

How would I sketch the curve with equation
y=(x-1)^3-4(x-1)
Please help, i'm so confused!
Reply 1
Heres the question, its (c)

Posted from TSR Mobile
Original post by Vinni0001
Heres the question, its (c)

Posted from TSR Mobile


Well you can find the points of intersection with the co-ordinate axis quite easily from factorising that cubic. Furthermore, you can differentiate it to find the maxima and minima points and mark those on. Since the leading term is +x3+x^3 you should have a general idea of what it should look like concerning directions when x goes to plus/minus infinity. Then you should have enough information to sketch it.
Reply 3
Original post by RDKGames
Well you can find the points of intersection with the co-ordinate axis quite easily from factorising that cubic. Furthermore, you can differentiate it to find the maxima and minima points and mark those on. Since the leading term is +x3+x^3 you should have a general idea of what it should look like concerning directions when x goes to plus/minus infinity. Then you should have enough information to sketch it.

I'm sorry, I still don't understand
Reply 4
Original post by Vinni0001
How would I sketch the curve with equation
y=(x-1)^3-4(x-1)
Please help, i'm so confused!


this is the equation i was referring to
Original post by Vinni0001
I'm sorry, I still don't understand


Step 1: Draw yourself the axis.
Step 2: Fully factorise the cubic - this should indicate where it crosses the x-axis. Mark these points down on your graph.
Step 3: Differentiate the cubic and set it equal to 0 - solve this quadratic. This will show you the turning points of the cubic, keep a hold of these x-coordinates and plug them through the original cubic to find the y-coordinates. Mark these coordinates on your graph.
Step 4: Join up the points and you should get something that looks like a cubic.
Step 5: Well done.
Reply 6
um, ok thanks
Reply 7
Original post by Vinni0001
um, ok thanks


If you look back at part b you've already drawn something similar. can you see that part c is a transformation of the graph from part b?
Reply 8
Original post by RDKGames
Step 1: Draw yourself the axis.
Step 2: Fully factorise the cubic - this should indicate where it crosses the x-axis. Mark these points down on your graph.
Step 3: Differentiate the cubic and set it equal to 0 - solve this quadratic. This will show you the turning points of the cubic, keep a hold of these x-coordinates and plug them through the original cubic to find the y-coordinates. Mark these coordinates on your graph.
Step 4: Join up the points and you should get something that looks like a cubic.
Step 5: Well done.


It's just a transformation from the previous part, differentiation isn't required here.
Original post by solC
It's just a transformation from the previous part, differentiation isn't required here.


Oh. Didn't even read part A lol. Love sideways images.
Reply 10
Original post by solC
If you look back at part b you've already drawn something similar. can you see that part c is a transformation of the graph from part b?

This makes much more sense, thank you very much. i get it now
for future reference - this guy teaches all the stuff, if you don't get something watch his theory videos maybe it will help https://www.youtube.com/watch?v=ZnVwKerOgsY&list=PLg2tfDG3Ww4vUYMrfblduaM9fOeiQ6BGU
Reply 12
Original post by Rowanstruggling
for future reference - this guy teaches all the stuff, if you don't get something watch his theory videos maybe it will help https://www.youtube.com/watch?v=ZnVwKerOgsY&list=PLg2tfDG3Ww4vUYMrfblduaM9fOeiQ6BGU


coolio

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