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Homework Help: Determine the range of x

  1. Dec 29, 2011 #1
    1. Plot the graph of thefunction f(x) = x^2 -2x -3 and determine the range of x for which the function is positive

    x^2 -2x - 3
    (x+1) (x-3) = 0
    x = -1 and x = 3
    Range -1, 3 <--- I am not sure am I right?

  2. jcsd
  3. Dec 29, 2011 #2


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    Staff: Mentor

    You found the zero crossing points, and there are two of them. One way to find out what the curve does in any region is to sketch it and see. Either use pencil and paper, or use one of the online graphing sites. Search google: online graphing
  4. Dec 29, 2011 #3
    Cheers! So the crossing points are x = -1 and x =3 , now I have to find how to graph in pencil because during the test I cant use graphic calculator ..
  5. Dec 29, 2011 #4


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    Well you know what parabolas generally look like, and can determine whether they are shaped more like a u or an n by checking whether the coefficient (number in front) of the highest power (x2 in this case) is positive or negative (it can't be zero because then there would be no x2 term and it wouldn't be a parabola then). In this case the coefficient of x2 is 1, which is positive, so the parabola is shaped like a u.

    Also keep in mind that a parabola is always symmetrical about its turning point (lowest or highest point) so since the zeroes are -1 and 3, so the parabola will have its lowest point directly in between the zeroes, at x=1.

    And you can also plot points to help aid you! Try x=0 to find where it cuts the y-axis for example, and maybe even x=1 to know where the lowest point is located.
  6. Dec 29, 2011 #5
    From what i see: it'll be decreasing from ]-∞,1] and increasing from [1,+∞[
  7. Dec 29, 2011 #6


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    Where the function is increasing of decreasing is not directly relevant to where the function is positive.

    Better: [itex]x^2- 2x- 3= (x+ 1)(x- 3)[/itex] and the product of two number is positive only if the two numbers have the same sign. So: (x+1< 0 and x- 3< 0) or (x+1> 0 and (x- 3> 0). x+ 1< 0 for x< -1 and x- 3< 0 for x< 3. Those are both true for x< -1. x+1> 0 for x> -1 and x- 3> 0 for x> 3. Those are both true for x> 3.

    More generally, any continuous function (and every polynomial is continuous) can change from "< 0" to "> 0" only where it "= 0". So we only need to check one point in each interval [itex](-\infty, -1)[/itex], [itex](-1, 3)[/itex], [itex](3, \infty)[/itex]. If x= -2, which is less than -1, [itex]f(-2)= (-2)^2- 2(-2)- 3= 4+ 4- 3= 5> 0[/itex] so f(x) is positive for all x< -3. If x= 0, which is between -1 and 3, [itex]f(0)= 0^2- 2(0)- 3= -3[/itex] so f(x) is negative for all x between -1 and 3. Finally, if x= 4, which is larger than 3, [itex]f(4)= 4^2- 2(4)- 3= 16- 8- 3= 5> 0[/itex].
  8. Dec 30, 2011 #7


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    You draw two columns, heading one 'x' and the other 'y'. Divide the x interval -1 ... 3 into, say, 5, and write these x values in the x column. Calculate the y value for each of those x-values, knowing that y=x^2 -2x -3.

    Finally, plot each x,y point on a set of axes.
  9. Dec 30, 2011 #8
    Since we know the graph intercepts the x-axis at x=-1 and x=3, we can pick x values that are less-than -1, in between -1 and 3, and greater then 3, and solve for each point. I like the points -2, 0, and 4. At x=-2 we get 5, at x=0 we get -3, at x=4 we get 5. With this data we clearly can determine where the function is positive. I usually prefer to do this using a number line so I can visually see and keep track of the functions behavior.
  10. Jan 4, 2012 #9
    i see...
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