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#1
How would I sketch the curve with equation
y=(x-1)^3-4(x-1)
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#2
Heres the question, its (c)

Posted from TSR Mobile
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4 years ago
#3
(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 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.
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#4
(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 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
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#5
(Original post by Vinni0001)
How would I sketch the curve with equation
y=(x-1)^3-4(x-1)
this is the equation i was referring to
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4 years ago
#6
(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.
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#7
um, ok thanks
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4 years ago
#8
(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?
1
4 years ago
#9
(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.
2
4 years ago
#10
(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.
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#11
(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
1
4 years ago
#12
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=ZnVw...duaM9fOeiQ6BGU
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#13
(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=ZnVw...duaM9fOeiQ6BGU
coolio
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