Mathematics

Question

What is y= 6x^2 + 12x -10 in vertex form?

2 Answer

  • vertex for  is a(x - b)^2 + c

    first divide 6x^2 + 12x by 6  so we have

    6(x^2 + 2x) - 10

    now complete the square  on the expression in the brackets:-
    = 6[ (x + 1)^2 - 1] -10
    =6(x + 1)^2 - 6 -10
    = 6( x+ 1)^2 - 16
  • basically complete the square
    so
    conver tto y=a(x-h)²+k
    steps:
    group x terms
    factor out leading coefient
    take 1/2 of ilnear coefient and square it, add negative and positive inside
    factor perfect square
    expand/distribute





    y=6x²+12x-10

    group x
    y=(6x²+12x)-10

    factor out leading coefient
    y=6(x²+2x)-10

    take 1/2 of linear coefient and square it
    2/2=1, 1²=1
    add negativ and positivve inside
    y=6(x²+2x+1-1)-10

    factor perfect square
    y=6((x+1)²-1)-10

    distribute
    y=6(x+1)²-6-10
    y=6(x+1)²-16
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