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Math Help - intercept points

  1. #1
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    intercept points

    Find the points of intersection of the curve[/FONT]
    y = x2 – 2x – 29 and the line y + 5 – 3x = 0
    Sketch the line and the curve on the same grid.

    i take it y=x2-2x-29 would be x=0 y =-29 making it (0,29)
    iam just stuck on the 2nd one.
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  2. #2
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    Not quite understood

    What it wants is for you to find the points where the curve crosses the line, not where the curve and line cross the axes. The easiest way to do that is plot the curve and the line on the same graph and look at where they meet, but as it asks you to sketch the curve after, I would suggest it needs to be solved by simultaneous equations.
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  3. #3
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    Quote Originally Posted by Here to Help View Post
    What it wants is for you to find the points where the curve crosses the line, not where the curve and line cross the axes. The easiest way to do that is plot the curve and the line on the same graph and look at where they meet, but as it asks you to sketch the curve after, I would suggest it needs to be solved by simultaneous equations.
    i dont know were to start
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  4. #4
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    Getting started

    If it were me I would rearrange the second equation to give y = 3x - 5. Then you have y = something in both, and if y = x2 - 2x - 29 and y = 3x - 5, then x2 -2x-29 = 3x-5. Subtracting 3x and adding 5 to both sides gives x2 - 5x - 24 = 0, which wil solve to give two values of x. You can the substitute these into either equations (I would suggest the one without the sqaure is easier) to give two y values. Each pair of x and y values give one of the 2 coordinates where the curve and the line cross.
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  5. #5
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    do you have msn?
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