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Finding vertex of quadratic equation?

  1. Apr 16, 2012 #1
    I have to find the vertex, axis, domain, & range of this quadratic equation:

    f(x)= -1/2(x+1)^2 -3

    I can't remember how to do the vertex. Is it y= -b/2a?

    Also, I tried factoring the equation, but I think I messed up. I ended up with:

    f(x)= .25x^2 +.5x-2.75



    Thanks
     
  2. jcsd
  3. Apr 16, 2012 #2

    LCKurtz

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    I assume your equation is$$
    f(x) = -\frac 1 2(x+1)^2-3$$If so, it is already in the form you need to locate the vertex. Note that the ##(x+1)^2## factor is always greater than or equal zero. Its contribution will be least when it is 0, which happens when ##x=-1##. So your vertex is at ##(-1,-3)##. No need to memorize formulas; just complete the square like in this problem.
     
  4. Apr 16, 2012 #3
    Thanks for the help.

    I'm trying to find the vertex of:

    f(x)= -3x^2 +24x-46

    and I ended up with:

    -3x+144-sqrtof 190
     
  5. Apr 16, 2012 #4
    What are your a and b terms?

    The x coordinate of the vertex is [itex]\frac{-b}{2a}[/itex], once you have that, you need another coordinate right?
     
  6. Apr 16, 2012 #5

    LCKurtz

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    Complete the square on that. Then do what I showed you in my previous post. You do know how to complete the square, right?
     
  7. Apr 16, 2012 #6
    I tried completing the square for:

    f(x)= -3x^2 +24x-46

    That's how I ended up with that.
    I divided 24/2 and added it to both sides and got

    -3x+144-sqrtof 190
     
  8. Apr 16, 2012 #7

    LCKurtz

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    No. You have to factor out the -3 to start, complete the square on the inside, and balance it out:
    ##f(x) = -3(x^2 -8x + ?) -46 + ?##
     
  9. Apr 17, 2012 #8
    Alright, I've gotten to:

    -3(x-4)(x-4)=62
    or
    -3(x-4)^2=62
    or
    -3(x-4)^2 -62=0

    What do I do from here?
     
  10. Apr 17, 2012 #9
    I've figured it out.
     
  11. Apr 17, 2012 #10

    LCKurtz

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    You don't set it equal to anything. What you should have written is$$
    f(x) = -3(x^2-8x+16)-46+48 = -3(x-4)^2+2$$From there it is clear that the vertex is ##(4,2)##.
     
  12. Apr 17, 2012 #11
    Why did you add 48 to -46 and instead of adding 16?
     
  13. Apr 17, 2012 #12

    LCKurtz

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    I didn't just add 16. There is a -3 out in front so I really put in -48 and I have to take it back out if I'm not to change the equation.
     
  14. Apr 17, 2012 #13
    That's what I thought. Thanks!
     
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