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    Show that the curve y= (2x-1)^2 only meets the x axis at the point (0.5, 0)
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    b^2-4ac=0
    put 0.5 into the equation and you get y=0
    when y=0 you cross the x axis
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    differentiate it and you find that x=0.5, plug that value into original equation to get y=0
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    when crosses the x axis (y=0): therefore in the equation for curve y=(2x-1)^2

    --> 0 = (2x-1)^2
    --> 0=(2x-1) (square root bothe sides)
    -->2x = 1 therefore x = 0.5
    therefore crosses x axis at the point (0.5,0)
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    just think of it graphically. If you have x-intercept= 0.5, y must have a value of zero

    so sub in y as 0 and you get: 0 = (2x-1)(2x-1) << solve it and you can only get a value of 0.5, so the coordiantes are (0.5, 0)
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    Simple maths shows that...
    Maths is useful in everyday life.

    Complex maths shows that...
    Maths is useless in everyday life.
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    (Original post by asem93)
    Show that the curve y= (2x-1)^2 only meets the x axis at the point (0.5, 0)
    I know you still got an answer to your question but in future it might be better to post maths-related questions in the maths forum here

    Cheers.
 
 
 
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