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anonymous

  • 5 years ago

verify that the x-coordinate of the vertex of a parabola of the form y= ax^2 +bx +c is -b/2a.

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  1. anonymous
    • 5 years ago
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    differentiate

  2. anonymous
    • 5 years ago
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    set it equal to zero and solve for x

  3. anonymous
    • 5 years ago
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    but i have to solve it using 'completing the squares'

  4. anonymous
    • 5 years ago
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    so no calc?

  5. anonymous
    • 5 years ago
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    wat do u mean by that?

  6. myininaya
    • 5 years ago
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    f(x)=ax^2+bx+c=a(x^2+bx/a)+c= a(x^2+bx/a+(b/[2a])^2)-a(b/a)^2 =a(x+b/(2a))^2-b/a so the vertex is (b/[2a],-b/a)

  7. myininaya
    • 5 years ago
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    oops i left something out

  8. myininaya
    • 5 years ago
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    f(x)=ax^2+bx+c =a(x^2+bx/a)+c =a(x^2+bx/a+(b/[2a])^2)-a(b/[2a])^2 =a(x+b/[2a])^2-b/[2a] The vertex is (-b/[2a],-b/[2a])

  9. myininaya
    • 5 years ago
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    lol oops that 4th step should be =a(x+b/[2a])^2-b^2/[4a] the vertex is (-b/[2a],-b^2/[4a])

  10. amistre64
    • 5 years ago
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    this one?

  11. anonymous
    • 5 years ago
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    ya i dont get it

  12. amistre64
    • 5 years ago
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    "verify that the x-coordinate of the vertex of a parabola of the form y= ax^2 +bx +c is -b/2a." Ok. this is really a part of the quadratic formula. Do you know the quad formula?

  13. anonymous
    • 5 years ago
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    nop

  14. anonymous
    • 5 years ago
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    we are learning 'cpmpleting the sqaures'

  15. amistre64
    • 5 years ago
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    lets step thru it then by "completeing the square"......cue thunder clap........what?!?...no thunder?..........oh well, moving along.

  16. amistre64
    • 5 years ago
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    ax^2 + bx +c = 0 lets solve for any quadratic equation by going thru this setup..

  17. anonymous
    • 5 years ago
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    k

  18. amistre64
    • 5 years ago
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    a little history tho.... a "complete" square looks like this: (x+3)^2 is a complete square...... when we expand it we get something that looks like this: x^2 +6x +9.. right?

  19. anonymous
    • 5 years ago
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    ya

  20. amistre64
    • 5 years ago
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    ok.... then if we can get an equation to be a "complete" square, then solving it is easier... take the equation: x^2 +6x +3. this looks similar, but it is not a "complete" square is it?

  21. anonymous
    • 5 years ago
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    hmm..we divide the middle term by 2 ...then square it and stuff...yup

  22. amistre64
    • 5 years ago
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    good good good.... :) but even on a more basic note.... if we add "0" to this equation do we change its value?

  23. anonymous
    • 5 years ago
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    i dont know

  24. amistre64
    • 5 years ago
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    I think you do :) tell me: does (x^2 +6x +3) = (x^2 +6x +3 +0) ??

  25. anonymous
    • 5 years ago
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    ya

  26. amistre64
    • 5 years ago
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    good :) then what we really need is a convenient form of "0" to apply to this equation. we know we want a +9 in there somewhere right? so what happens if we add: 9-9 to this equation, does the value change? remember: 9-9 = 0

  27. anonymous
    • 5 years ago
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    s

  28. amistre64
    • 5 years ago
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    s?

  29. anonymous
    • 5 years ago
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    i get wat u r saying

  30. amistre64
    • 5 years ago
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    cool... so we do this: (x^2 +6x +9) -9 +3 in order to "complete" the square right?

  31. anonymous
    • 5 years ago
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    ya

  32. anonymous
    • 5 years ago
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    so hw do we apply that to this q:

  33. amistre64
    • 5 years ago
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    good... and you already pointed out the (b/2)^2 is what we will need given any equation of this form correct? (6/2)^2 = 3^2 = 9 right

  34. anonymous
    • 5 years ago
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    yup

  35. amistre64
    • 5 years ago
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    then lets begin to see what this form of equation is hiding from us :) ax^2 +bx + c = 0 ; subtract c from both sides... ax^2 + bx = -c ; divide at all by "a" x^2 +(b/a)x = (-c/a) ; complete the square x^2 +(b/a)x + (b/2a)^2 = (-c/a) + (b/2a)^2

  36. amistre64
    • 5 years ago
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    i hope your still with me on this :) (x +(b/2a))^2 = (-c/a) + (b^2/4a^2) ; lets add the right side together with the rules for add fractions.... (x +(b/2a))^2 = (b^2 -4ac)/2a ; now we squareroot both sides. x+ (b/2a) = +- sqrt(b^2 -4ac)/2a ; and to solve for "x" we subtract that (b/2a) and get... x = -b/2a +- sqrt(b^2 -4ac)/2a we good up to this point? cause this is where the question gets answered ;)

  37. amistre64
    • 5 years ago
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    now... do you see the answer or can i do my magic and show you whats right before your eyes :)

  38. anonymous
    • 5 years ago
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    ya

  39. amistre64
    • 5 years ago
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    that right side part...that +- sqrt(...)/2a is just a number..... lets go ahead and make it equal to, i dunno, 6 x = -b/2a (+or-) "6" what does this tell ya? <.......-6.........-b/2a..........+6...........> -b/2a sits right smack between this 6's right? its can be nothing else BUT the vertex

  40. amistre64
    • 5 years ago
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    put on your -b/2a shirt stand up and look to your left for 6 feet and look to the right for 6 feet.... are you in the middle?

  41. amistre64
    • 5 years ago
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    does that make sense for you?

  42. amistre64
    • 5 years ago
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    a parabola that is sagging beneath the x axis has its vertex halfway between its x intercepts..... its a giant "U".

  43. amistre64
    • 5 years ago
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    -b/2a is halfway between the intercept on the left and the intercept on the right...... that makes it the vertex.

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