Math (Help This Will Be on My Test Tomorrow!)

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y= ax^2+bx-4. How would i find the equation of this parabola with the points (0,-4) (1,0) (2,2) (4,0) and 6,-10)?

  • Math (Help This Will Be on My Test Tomorrow!) -

    Use the first few given points
    for (0,-4) : -4 = 0 + 0 - 4 , nothing new learned here
    for (1,0) : 0 = a + b - 4 ----> a+b+4 or b = 4-a
    for (2,2) : 2 = 4a + 2b - 4 --- 2a + b = 3

    use substitution
    2a + (4-a) = 3
    a = -1
    then b = 4-(=1) = 5

    y = -x^2 + 5x - 4
    test if the other points satisfy, they do!

  • Math (Help This Will Be on My Test Tomorrow!) -

    You must write equation for any pair of points.


    y=ax^2+bx-4


    x=0 y= -4

    -4=a*0^2+b*0+c

    0+0+c= -4

    c= -4



    x=1 y=0

    0=a*1^2+b*1-4

    a*1+b*1-4=0

    a+b-4=0

    a=4-b



    x=2 y=2

    y=ax^2+bx-4

    2=(4-b)*2^2+b*2-4

    (4-b)*4+2b-4=2

    16-4b+2b=2+4

    -2b=6-16

    -2b= -1 Divide with -2

    b=5

    a=4-b

    a=4-5

    a= -1


    So:

    a= -1 b=5 c= -4

    y=ax^2+bx+c= -x^2+5x-4

    y=-x^2+5x-4

    Checking of results:

    For x=4

    y=ax^2+bx-4

    y= -4^2+5*4-4

    y= -16+20-4

    y=4-4

    y=0

    Correct value

    For x=6

    y=ax^2+bx-4

    y= -6^2+6*5-4

    y= -36+30-4

    y= -6-4

    y= -10

    Correct value

  • Math (Help This Will Be on My Test Tomorrow!) -

    I make one mistake in typig.

    -2b= -1 Divide with -2 is mistake


    -2b= -10 Divide with -2 is correct

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