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

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    (Original post by Vinni0001)
    Heres the question, its (c)

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    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 +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.
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    (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 +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
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    (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
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    (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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    um, ok thanks
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    (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?
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    (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.
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    (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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    (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
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    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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    (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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