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How to find the vertex of G(t)=1/2(t^2-4t)

  1. Feb 17, 2015 #1
    1. The problem statement, all variables and given/known data
    G(t)=1/2(t^2-4t)

    3. The attempt at a solution
    Since this is a polynomial it's all real numbers for domain
    y-int:
    G(0)=1/2(0^2-4(0)
    y-int=(0,0)

    x-int:
    0=1/2(t^2-4t)
    0=(1/2)t(t-4)
    Not sure...

    And not sure about the vertex. I know the formula -b/2a but how do i find those two points for the graph thanks
     
  2. jcsd
  3. Feb 17, 2015 #2

    BvU

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    What exactly are you asking ?
    And what do you mean with x-int ? intercept on the x -axis ? well, with your y-int you already found (0,0) which is one of the roots.
    And if you can write 0=(1/2)t(t-4) , you can see that this is the case if x = 0 (already found) ##\vee## x = 4.

    You know the formula -b/(2a). It has to do with the equation ax2 + bx + c = 0 and represents the sum average of the roots. In your case 4.

    Could the vertex you mention be the extreme of G ?

    [edit] not sum but average
     
  4. Feb 17, 2015 #3

    SteamKing

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    Instead of being fixated on some formula for the vertex, what is characteristic about the vertex of the parabola?
     
  5. Feb 17, 2015 #4
    Would I have to complete the square to find the vertex? like so

    G(t)=1/2(t^2-4t)
    G(t)=1/2(t^2-4t+4) -4
    G(t)=1/2(t-2)^2 -4
    Vertex form: y=a(x-h)^2 +k
    and vertex is (h,k)
    so (2,-4)
     
  6. Feb 17, 2015 #5

    BvU

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    You've done it ! Well done. Can you make the link with the -b/(2a) formula ? If you can, you've really mastered the subject.
     
  7. Feb 17, 2015 #6
    I think. The vertex is actually (2,-2) I double checked it. Is that because when I have the form G(t)=1/2(t-2)^2 -4 I have to take the 1/2 out and times that with -4 to output -2? so then (2,-2). Is -b/(2a) connection between taking the form G(t)=1/2(t^2-4t+4) so A=1,B=-4,C=4. So I plug in 4/(2(1) = 2 that's the X and then the y-intercept is G(0)=1/2(0^2-4(0)+4) so then 1/2*4/1=(0,2) and put the 2 in for the x and keep the 2 for the Y?
     
  8. Feb 17, 2015 #7

    BvU

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    Sorry I forgot about the 1/2. You're right.

    In writing G in 'vertex form' you found the vertex directly.

    ax2 + bx + c in 'vertex form' would be someting like ##a (x - {-b\over 2a} )^2 - \left (({-b\over 2a})^2 - c\right )## (*)

    To form G, the 1/2 can serve as a. Then b is -2 and c is zero . So -b/(2a) is 2, as you found.


    (*) One step further you have this infamous

    'if b^2 > 4ac this can be written as
    $$
    a \left ( \left ( x - {-b\over 2a} \right )^2 - \left ( \sqrt { \left (\left ({-b\over 2a}\right )^2 - {c\over a} \ \right )}\ \right )^2\ \ \right ) = \\

    a\ \left ( x + {b\over 2a} + {\sqrt {b^2 - 4ac} \over 2a } \right ) \; \left ( x + {b\over 2a} - {\sqrt {b^2 - 4ac} \over 2a } \right ) = \\
    a\ \left ( x + {b + \sqrt {b^2 - 4ac} \over 2a } \right ) \; \left ( x + {b - \sqrt {b^2 - 4ac} \over 2a } \right )
    $$
    that will solve all quadratic equations in any career :smile:
     
  9. Feb 17, 2015 #8
    I just need to confirm this here since I have G(t)=1/2(t-2)^2 -4 why does the -4 become -2? does it multiply with 1/2?
    Thank you
     
  10. Feb 17, 2015 #9

    BvU

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    G(t)=1/2(t^2-4t)
    G(t)=1/2 ##\Bigl (## (t^2-4t+4) -4 ##\Bigr )##
    G(t)=1/2(t-2)^2 -2
     
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