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    mathsucks12's Avatar
    mathsucks12 Posts: 1, Reputation: 1
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    #1

    May 29, 2008, 03:40 PM
    How do I find the x-intercepts of a graph?
    :confused:
    galactus's Avatar
    galactus Posts: 2,271, Reputation: 282
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    #2

    May 29, 2008, 05:07 PM
    Set y=0 and solve for x.
    AJ54's Avatar
    AJ54 Posts: 6, Reputation: 1
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    #3

    May 30, 2008, 12:52 AM
    if it is a quadratic, factorise using adds to give b and multiplies to give ac. i.e. when y =ax^2 + bx + c. or just let y = 0.
    hope this kind of helped!
    kateuk's Avatar
    kateuk Posts: 20, Reputation: 2
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    #4

    Jun 4, 2008, 12:00 PM
    If it's a quadratic, then take the equation to be y = ax^2 + bx + c.
    (Remember - if it simply says x^2, then the value of a is 1, NOT 0)
    Then substitute into the following equation:


    This will give you two values of x, if you do it once taking the +/- as +, and the second time as a subtract sign.
    This only really works if it's simple maths or you have a calculator. If not factorize it.
    Then, the values of x are the numbers in the brackets multiplied by -1.
    Here's an example:
    x^2+6x+5 = (x+1)(x+5)
    1 x -1 = -1
    5 x -1 = -5

    Therefore the x values are -1 and -5.


    If it's a linear equation, for example, 2x+4=y
    Take y to be 0.
    So 2x+4=0
    Then apply the following steps:
    2x-4=0-4
    2x=-4
    2x/2=-4/2
    x=-2
    So, -2 is your intercept. With a quadratic equation there's 2 intercepts, with a linear only one. Does that all make sense? Eek, hope so!

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