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    I would really appreciate any help here. Thanks!!

    The curve C has equation
    y= (x+3) (x-1)^2
    a. sketch C showing clearly coordinates of points where curve meets coordinate axes.
    I got points (-3,0) (1,0) (1,0) (0,3) (0,-1) are they right? If so, then how come the graph won't turn out properly if I try to connect them?? If they're not right, what are the right points?
    b. Show that equation of C can be written in the form
    y= x^3 + x^2-5x+k
    where k is a positive integer, and state the value of k.
    Got this right. k =3

    There are two points on C where gradient of the tangent to C is equal to 3.
    c. Find x coordinates of these two points.
    Help here?
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    (Original post by BrownieLover)
    I would really appreciate any help here. Thanks!!

    The curve C has equation
    y= (x+3) (x-1)^2
    a. sketch C showing clearly coordinates of points where curve meets coordinate axes.
    I got points (-3,0) (1,0) (1,0) (0,3) (0,-1) are they right? If so, then how come the graph won't turn out properly if I try to connect them?? If they're not right, what are the right points?
    b. Show that equation of C can be written in the form
    y= x^3 + x^2-5x+k
    where k is a positive integer, and state the value of k.
    Got this right. k =3

    There are two points on C where gradient of the tangent to C is equal to 3.
    c. Find x coordinates of these two points.
    Help here?
    Well this is a cubic function with a repeated root, hence it crosses the x-axis just twice. - (-3,0) and (1,0). The y-intercept is found by subbing X=0.. Which is (0,3).

    How do you find the gradient of a polynomial ?
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    (Original post by MathsLover28)
    Well this is a cubic function with a repeated root, hence it crosses the x-axis just twice. - (-3,0) and (1,0). The y-intercept is found by subbing X=0.. Which is (0,3).

    How do you find the gradient of a polynomial ?

    yeah but when we sub x=0 for (x-1) we're also gonna get (0, -1) so there are two points on the y axis aren't there?
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    (Original post by BrownieLover)
    yeah but when we sub x=0 for (x-1) we're also gonna get (0, -1) so there are two points on the y axis aren't there?
    Nooo....

    let's change the notation around a bit.

     f(x) = (x+3)(x-1)^2

    so what is f(0), well where ever there's an x we put 0, so :

     f(0) = (0+3)*(0-1)^2
     f(0) = 3*(-1)^2
     f(0) = 3*1 = 3

    The product of two brackets, implies there are two roots (when the function crosses the x-axis), not the y-intercept. Y is a function of x, not the other way around.
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    (Original post by BrownieLover)
    yeah but when we sub x=0 for (x-1) we're also gonna get (0, -1) so there are two points on the y axis aren't there?
    How can you get two values for y when you substitute one value for x? That's not a function, is it?
 
 
 
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