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Cubic+ Quad Equations - y Axis

  1. Feb 14, 2007 #1
    Hi, I was wondering in a cubic equation such as

    x³ - 5x = 2x - 1
    can be rearranged to

    x³ - 7x + 1 = 0

    is that 0 the y value, and +1 where the line crosses the y axis? Is - 7x the gradient or is x³ - 7x the gradient?

  2. jcsd
  3. Feb 14, 2007 #2


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    "is that 0 the y value"?? What y value? An equation is not a function. If you are talking about y= x3- 7x+ 1, then the "y-intercept" is the value of y when x= 0: y= 1- the y coordinate when the curve (not a straight line) crosses the y-axis. Of course, we still have a value of y for every value of x. I'm not clear what you mean by "the gradient". Since this is NOT a straight line, it doesn't have a "slope" (sometimes called "gradient"). Sometimes the term "gradient" is used to mean "derivative" but, from the problems you have been posting here I would not expect you to be referring to that.
  4. Feb 14, 2007 #3

    Question 14 of a paper is:

    A graph -2< x < 3 with points

    x = -2, -1 , 0 , 1 , 2 ,3
    y = 1 , 4 , 1 ,-2 , 1 ,16

    I drew the graph. Now part b says

    By drawing suitable lines on you graph, find all the solutions to the following equations for -2 < x < 3. Give your solutions to 1 decimal place

    x³ - 5x = 2x - 1

    so x³ - 7x + 1 = 0

    dont know where to go from there...
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