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    Here are the questions

    1a. Find the coordinates O,A and B, the point of intersection of the curve y=x^3-3x^2+2x with the x-axis. Hence, state the equations.

    My attempt at finding the A intercept.
    From my understanding if something intercepts with the x-axis, x=0

    So subs back into original equation to get y=0.

    So I know my 2 points are (0,0), differentiating to get the gradient I get this:

    \frac{dy}{dx}= 3x^2-4x+2
    Putting x back in gives me the gradient which is 2, this is then put into the formula and I get the equation of the line as:

    y=2x

    My confusion is, how would I find the other points since x=0 y=0


    2. Find the equation of the tangent to y=3x^3-4x^2+2x-10 at the point where the curve meets the y-axis.

    Any hints for this question

    As always rep provided.
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    1st.
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    (Original post by CTurbinado)
    1st.
    Beat me to it. I hate it how tsr is seen as a marking service.
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    Substitute into the original thing. ie

    2x = x^3 -3x^2+2x

    (Original post by abstraction98)
    Beat me to it. I hate it how tsr is seen as a marking service.
    Help discussion, revision and homework help from KS3 and GCSE to degree level including for engineering, health-related subjects and sports science.
    Thats one of the points of F38
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    (Original post by SimonM)
    Substitute into the original thing. ie

    2x = x^3 -3x^2+2x
    subs 2x will give me x, I get it, but when I need to find just the coordinates of the letter, I get A as (0,0) which is right, then I don't understand how to get the coordinates of the other points.


    (Original post by CTurbinado)
    1st.
    That doesn't really help.
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    (Original post by Diaz)
    subs 2x will give me x, I get it, but when I need to find just the coordinated of the letter, I get A as (0,0) which is right, then I don't understand how to get the coordinated of the other points.
    2x = x^3 -3x^2+2x \Rightarrow 0 = x^3-3x^2

    What else can x be other than 0?

    Edit

    Misread your post. The graph intercepts the x-axis when y=0
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    (Original post by Diaz)
    That doesn't really help.
    But it was the fastest reply :yep:
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    (Original post by Diaz)
    Here are the questions

    1a. Find the coordinates O,A and B, the point of intersection of the curve y=x^3-3x^2+2x with the x-axis. Hence, state the equations.

    My attempt at finding the A intercept.
    From my understanding if something intercepts with the x-axis, x=0

    So subs back into original equation to get y=0.

    So I know my 2 points are (0,0), differentiating to get the gradient I get this:

    \frac{dy}{dx}= 3x^2-4x+2
    Putting x back in gives me the gradient which is 2, this is then put into the formula and I get the equation of the line as:

    y=2x

    My confusion is, how would I find the other points since x=0 y=0


    2. Find the equation of the tangent to y=3x^3-4x^2+2x-10 at the point where the curve meets the y-axis.

    Any hints for this question

    As always rep provided.
    I will help anyway. Factorise your initial equation. This requires finding the factors in a term with x^3 so you use special division i have forgotten the name of. That gives your x values (you should have been taught this at school)

    By "get the equations", I assume you mean the equations of the tangents?

    Once you ahve done this NOW differentiate as you have done correctly. Put x values in to get the gradient of each line. Shove the x values in to to coordinate form. All([xvalue], 0). Now we wheel out the formula:

    y - b = m(x - a)

    Lob the numbers from your intercepts in to get the tangents..

    Hope that helped. Please be a bit more subtle in the future though
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    (Original post by SimonM)
    Substitute into the original thing. ie

    2x = x^3 -3x^2+2x





    Thats one of the points of F38
    Homework help - not giving the answers
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    (Original post by abstraction98)
    Homework help - not giving the answers
    I would say I am probably the most Nazi-ish person on the forum for not posting full solutions. Neither he or I asked/gave a full solution
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    Ban OP please for not giving rep for the fastest reply
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    I still have no idea what any of you are on about.

    1. Why do I need to factorise? How to I obtain just the coordinates

    2. Any hints?
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    (Original post by SimonM)
    I would say I am probably the most Nazi-ish person on the forum for not posting full solutions. Neither he or I asked/gave a full solution
    Stand corrected. The thread title implies it though
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    (Original post by Diaz)
    I still have no idea what any of you are on about.

    1. Why do I need to factorise? How to I obtain just the coordinates

    2. Any hints?
    Factorise you initial expression. The initial formula. This involves doing a special sort of division i have forgotten the name of. SImonM will know. You should have been taught it. This will tell you if the function divides nicely by (x + y) where y is a whole number (assuming this is simple).

    For instance, if (x + 3) is a factor, an x intercept would be (-3, 0) (solving for (x + 3) = 0)
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    1. It crosses the x axis when Y = 0. So, throw that in, and factorize the equation to find out when it does equal zero

    2. It crosses the Y axis when x = 0, so if you differentiate it you can find the gradient of the tangent at that point....
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    (Original post by Diaz)
    Here are the questions

    1a. Find the coordinates O,A and B, the point of intersection of the curve y=x^3-3x^2+2x with the x-axis. Hence, state the equations.

    My attempt at finding the A intercept.
    From my understanding if something intercepts with the x-axis, x=0

    So subs back into original equation to get y=0.

    So I know my 2 points are (0,0), differentiating to get the gradient I get this:

    \frac{dy}{dx}= 3x^2-4x+2
    Shouldn't it be -6x and not -4x?
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    (Original post by LakHani_Hasnain)
    Shouldn't it be -6x and not -4x?
    I haven't been going through the numerics but agree entirely.
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    (Original post by Diaz)
    I still have no idea what any of you are on about.

    1. Why do I need to factorise? How to I obtain just the coordinates

    2. Any hints?
    You should be able to factorise x^3 equations.

    Take out x as the common factor and then you get a quadratic...then solve. There is a simpler way, look on some of the websites...they'll be a little more insightful. Also i really cannot be bothered to write it all out for you. :yep:
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    I think the values of x are 3 and 0. If you got something else like a graph with the question I would think that would help alot. Post the whole question if possible.
 
 
 
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