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    Hi,
    I would appreciate it if someone could explain a few of the steps involved.

    y= (x+3)(x-3)(x-4)
    The above, is a factorised cubic equation.
    - What do the 3 brackets represent?
    - How does one obtain the value, at which the curve crosses the y - axis?
    -How do you derive the 3 possible solutions of x?
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    (Original post by AKRYL)
    Hi,
    I would appreciate it if someone could explain the a few of the steps involved.

    y= (x+3)(x-3)(x-4)
    The above, is a factorised cubic equation.
    - What do the 3 brackets represent?
    - How does one obtain the value, at which the curve crosses the y - axis?
    -How do you derive the 3 possible solutions of x?
    The 'solutions' are when y = 0.

    What do you know about the co-ordinates of the place where the curve crosses the y axis? (sketch a graph and think about the y axis)

    And when would y=0?
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    (Original post by AKRYL)
    Hi,I would appreciate it if someone could explain the a few of the steps involved.y= (x+3)(x-3)(x-4)The above, is a factorised cubic equation.- What do the 3 brackets represent?- How does one obtain the value, at which the curve crosses the y - axis?-How do you derive the 3 possible solutions of x?
    the three brackets represent the values of x for which they are factors.
    curve crosses y-axis when x is 0. therefore set each x to 0 and multiply each bracket by the other.
    curve crosses the x-axis when y is 0. therefore you will need to set y to 0 so that each bracket is set to 0 so the 'solutions' can be found. e.g. if (x-a) was a factor you'd do x-a = 0 => x = a.
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    to find the y intercept you evaluate the function with x = 0
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    You may wish to find the stationary points of the curve (where dy/dx=0) which will allow you to produce a more accurate sketch.
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    (Original post by AKRYL)
    Hi,
    I would appreciate it if someone could explain the a few of the steps involved.

    y= (x+3)(x-3)(x-4)
    The above, is a factorised cubic equation.
    - What do the 3 brackets represent?
    - How does one obtain the value, at which the curve crosses the y - axis?
    -How do you derive the 3 possible solutions of x?
    You know the general shape. The x intercept is when the brackets is equal to zero. I'll do the first one. One x value when y is zero is -3, as in the first bracket if x=-3, then (-3+3)=0
    So, x=-3 is one solution. Do the other two, and sketch the cubic graph going through these x intercepts.
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    To find the intercept on the y-axis, x must be 0 (imagine a ruled out graph). We can the substitute in 0 for where x appears:
    y= (0+3)(0-3)(0-4)
    Multiply the brackets to get your y value.

    To get your three possible solutions, you put each bracket equal to 0, e.g. x+3=0, then solve for x. You do this with each of the brackets. These are x intercept points (where it crosses the x axis).
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    (Original post by ozmo19)
    To find the intercept on the y-axis, x must be 0 (imagine a ruled out graph). We can the substitute in 0 for where x appears:
    y= (0+3)(0-3)(0-4)
    Multiply the brackets to get your y value.

    To get your three possible solutions, you put each bracket equal to 0, e.g. x+3=0, then solve for x. You do this with each of the brackets. These are x intercept points (where it crosses the x axis).
    ...
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    (Original post by AKRYL)
    x
    In between each bracket is a multiply sign. It would be like saying expand (x+1)(x+2), when you expand you multiply and you get a quadratic, having it in brackets is it factorised.
    The same applies with the brackets - the equation, for example, may be ax^3+bx^2+cx. This equation is factorised to get you those brackets.

    So, when x=0 you get 0+3 etc and you end up with y=(3)(-3)(-4). Multiplying those out gets you an integer that is your y value.
 
 
 
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