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Make y a funciton of x

  1. Apr 18, 2005 #1
    please help, how can i make y=8x-2x^2 go to x= something y

    same with y=4x-x^2 goes to x=

    thank you for help!!
     
  2. jcsd
  3. Apr 18, 2005 #2
    Always by completing the square for quadratic functions.
     
  4. Apr 18, 2005 #3
    The method is pretty straight forward. Just divide both sides so that x^2 has a coefficient of 1. Then add on a constant C to both sides so that the right hand side is a square (remember square quadratics have the form X^2 + 2*b + b^2 = (x+b)^2). Then you just need to square root both sides and then it becomes clear how to get the x alone. You may want to add a [tex]\pm[/tex] when you take the square root if you are doing that in this class.

    so for y = ax^2 +bx just divide both sides by a, add (b/(2a))^2. Then the right hand side can be written out as (x + b/(2a))^2. Then just square root and add stuff around. Remember nothing says a or b can't be negative. a just can't be 0.
     
  5. Apr 18, 2005 #4

    dextercioby

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    You can express "x" in terms of "y" only on a certain interval.That is to say that the function "y(x)" is invertible only on an interval,and not on R.

    Daniel.
     
  6. Apr 18, 2005 #5
    Yes, but I think that that won't be a good function, since it doesn't map -to one on any interval.

    EDIT:
    Ah, I see now that you mean interval of x. Sorry.
     
    Last edited: Apr 18, 2005
  7. Apr 18, 2005 #6
    I think that I can correctly derive (without testing) that

    [tex]x = \sqrt { \frac {y} {2} } + \frac {8} {y}[/tex]
     
    Last edited: Apr 18, 2005
  8. Apr 18, 2005 #7
    I don't understand. 8*1-2*1^2=6 but 6*1-1^2=5
    So the two functions differ at 1 right?
    The first case you should get [tex]x=2\pm 1/2\,\sqrt {16-2\,y}[/tex]
    and the second cased
    [tex]x=3/4\pm 1/4\,\sqrt {9-4\,y}[/tex]
    Two different answers for two different equations.
     
  9. Apr 19, 2005 #8
    y=8x-2x^2--> 8x-2x^2-y=0--> 2x^2-8x+y=0
    use quadratic formula!
     
  10. Apr 19, 2005 #9
    never mind! that was wrong
     
  11. Apr 19, 2005 #10
    no! it's not! it's right! (sorry for not being sure )
     
  12. Apr 19, 2005 #11

    dextercioby

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    'Twas not.It was correct.How else could this problem b solved?

    Daniel.
     
  13. Apr 22, 2005 #12
    u should be strongwilled baby garfield, the step is right, the use of the quadratic formula gives x in terms of y and that satisfies ur problem dexter
     
  14. Apr 22, 2005 #13
    [tex]y = 8x - 2x^2[/tex]

    [tex]\Rightarrow 2x^2 - 8x + y = 0[/tex]

    Apply to quadratic formula: [tex]x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

    a = 2, b = -8 and c = y

    [tex]x = \frac{ 8 \pm \sqrt{(-8^2) - (4 \times 2 \times y)}}{2 \times 2}[/tex]

    [tex]\Rightarrow x = \frac{ 8 \pm \sqrt{64 - 8y}}{4}[/tex]

    [tex]\Rightarrow x = 2 \pm \sqrt{4 - \frac{1}{2}y}[/tex]

    The Bob (2004 ©)
     
    Last edited: Apr 22, 2005
  15. Apr 22, 2005 #14
    [tex]y = 4x - x^2[/tex]

    [tex]\Rightarrow x^2 - 4x + y = 0[/tex]

    Apply to quadratic formula: [tex]x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

    a = 1, b = -4 and c = y

    [tex]x = \frac{ 4 \pm \sqrt{(-4^2) - (4 \times 1 \times y)}}{2 \times 1}[/tex]

    [tex]\Rightarrow x = \frac{ 8 \pm \sqrt{16 - 4y}}{2}[/tex]

    [tex]\Rightarrow x = 4 \pm \sqrt{4 - y}[/tex]

    The Bob (2004 ©)
     
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