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Homework Help: Curve Sketching

  1. Aug 8, 2008 #1
    1. The problem statement, all variables and given/known data

    A curve has equation (x^2 - 2x + 2)/(x^2 + 3x + 9). I need to prove that the range of values of y for this curv eis 2/27 =< y =< 2

    I know how it should be worked but I can't understand why it's done that way.

    My teacher simply arranged the equation in the form (x^2 + 3x + 9)y = x^2 - 2x + 2 which leads to the equation (y-1)x^2 + (3y+2)x + 9y - 2 = 0 and then said "For range, b^2 - 4ac >= 0"

    This is what I'm having trouble understanding. What does the range have to do with b^2 - 4ac >= 0. I want to know the reasoning behind this

    I'd appreciate any answers :)

  2. jcsd
  3. Aug 8, 2008 #2


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    You teacher used algebraic manipulation to write the equation as [itex](y-1)x^2 + (3y+2)x + 9y - 2 = 0[/itex], which you can think of as a quadratic equation for x, with the coefficients involving y.

    The "range" of a function is the set of all possible y values. Here that means values of y that correspond to real number values of x.

    Since that is written as a quadratic equation, use the quadratic formula to solve it.

    Remember that the solutions to the quadratic equation [itex]ax^2+ bx+ c= 0[/itex] are given by [itex](-b\pm \sqrt{b^2- 4ac})/2a[/itex]. In particular, those will be real numbers as long as the value inside the squareroot, [itex]b^2- 4ac[/itex] is not negative. Here a= y-1, b= 3y+ 2, and c= 9y- 2.
    [tex]b^2- 4ac= (3y+2)^2- 4(y-1)(9y-2)= 9y^2+ 12y+ 4- 4(9y^2- 11y+ 2)[/tex]
    [tex]= 9y^2+ 12y+ 4- 36y^2+ 44y- 8= -27y^2+ 56y- 4[/tex]
  4. Aug 11, 2008 #3
    thanks a lot for your help. i perfectly understand it now. thank you
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