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  1. #1
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    Post one variable

    Determine the quadratic polynomial in one variable whose graph passes thru the points (1,6), (2,3) and (3,2).
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by puneet View Post
    Determine the quadratic polynomial in one variable whose graph passes thru the points (1,6), (2,3) and (3,2).
    let f(x) be a curve. a point (a,b) is on the curve if f(a)=b.

    if you want a quadratic in one variable, say in x, then your it is of the form f(x)=ax^2 + bx + c

    you are given 3 points. then you can solve a,b,c by solving the system of equation

    f(1)=6
    f(2)=3
    f(3)=2
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    Is this right

    Q)Determine the quadratic polynomial in one variable whose graph passes thru the points (1,6), (2,3) and (3,2).

    ANS)Cont.
    Thus we have
    6 = a + b + c --------(1)
    3 = 4a + 2b + c --------(2)
    2 = 9a + 3b + c --------(3)
    Multiply (1) by 2 and subtract from (2)
    Multiply (1) by 3 and subtract from (3)
    We get
    2a-c= -9 c-2a=9 --------(4)
    And
    6a-2c = -16 3a-c=-8 --------(5)
    Adding (4) and (5) We get
    a=1
    Putting a in (4) We get
    c-2(1)=9 c-2=9 c=11
    putting a and c in (1)
    6=1+b+11 6=12+b b=-6

    Hence in order we get
    a=1,b=-6,c=11 --------(6)

    We have
    F(x) =ax2 + bx + c --------(7)

    Putting values of f and (6) in (7) we get
    x2 - 6x + 5 = 0
    x2 - 6x + 8 = 0
    x2 - 6x + 9 = 0

    Is This right I am getting three equations whats wrong.
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  5. #5
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by puneet View Post
    Q)Determine the quadratic polynomial in one variable whose graph passes thru the points (1,6), (2,3) and (3,2).

    ANS)Cont.
    Thus we have
    6 = a + b + c --------(1)
    3 = 4a + 2b + c --------(2)
    2 = 9a + 3b + c --------(3)
    Multiply (1) by 2 and subtract from (2)
    Multiply (1) by 3 and subtract from (3)
    We get
    2a-c= -9 c-2a=9 --------(4)
    And
    6a-2c = -16 3a-c=-8 --------(5)
    Adding (4) and (5) We get
    a=1
    Putting a in (4) We get
    c-2(1)=9 c-2=9 c=11
    putting a and c in (1)
    6=1+b+11 6=12+b b=-6

    Hence in order we get
    a=1,b=-6,c=11 --------(6)

    We have
    F(x) =ax2 + bx + c --------(7)

    Putting values of f and (6) in (7) we get
    x2 - 6x + 5 = 0
    x2 - 6x + 8 = 0
    x2 - 6x + 9 = 0

    Is This right I am getting three equations whats wrong.
    your equation is f(x)=x^2 -6x + 11.. you should not substitute it on the system..
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  6. #6
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    Hello, puneet!

    With this type of system of equations,
    . . there is a much simpler approach.


    . . \begin{array}{cccc}a + b + c &=& 6 & [1] \\<br />
4a + 2b + c &=& 3 & [2] \\<br />
9a + 3b + c &=& 2 & [3] \end{array}


    Eliminate c.
    . . \begin{array}{ccccc}\text{Subtract: [2] - [1]} & 3a + b &=& \text{-}3 & [4] \\<br />
\text{Subtract: [3] - [2]} & 5a + b &=& \text{-}1 & [5] \end{array}

    Eliminate b.
    . . \text{Subtract: [5] - [4] }\;\;2a\:=\: 2 \quad\Rightarrow\quad \boxed{a \:=\:1}

    \text{Substitute into [4]: }\;3 + b \:=\:\text{-}3 \quad\Rightarrow\quad\boxed{ b \:=\:\text{-}6}

    \text{Substitute into [1]: }\;1 - 6 + c \:=\:6 \quad\Rightarrow\quad\boxed{ c \:=\:11}


    \text{Therefore: }\;f(x) \;=\;x^2-6x + 11

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