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Math Help - Equation for a Function with Given Points

  1. #1
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    Equation for a Function with Given Points

    I'm trying to get a cubic function from the points (0, 7), (2, 10), (4, 5), and (6, 0). I know the equation is y = ax^3 + bx^2 + cx + d, and
    d = 7

    After plugging in the other points I got 3 equations:
    8a + 4b + 2c = 3
    64a + 16b + 4c = -2
    216a + 36b + 6c = -7

    My substitution and elimination between three unknowns is more than shabby, could someone please help out?
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  2. #2
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    Quote Originally Posted by suavrectangle View Post
    I'm trying to get a cubic function from the points (0, 7), (2, 10), (4, 5), and (6, 0). I know the equation is y = ax^3 + bx^2 + cx + d, and
    d = 7

    After plugging in the other points I got 3 equations:
    8a + 4b + 2c = 3
    64a + 16b + 4c = -2
    216a + 36b + 6c = -7

    My substitution and elimination between three unknowns is more than shabby, could someone please help out?

    8a + 4b + 2c = 3\implies a = \frac{3-4b-2c}{8}

    Sub this into both

    64a + 16b + 4c = -2
    216a + 36b + 6c = -7

    Giving

    64\left(\frac{3-4b-2c}{8}\right) + 16b + 4c = -2
    216\left(\frac{3-4b-2c}{8}\right) + 36b + 6c = -7

    Now you have 2 equations with 2 variables, can you finish?
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  3. #3
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    Hello, suavrectangle!

    You found: . d \:=\:7



    \begin{array}{cccc}8a + 4b + 2c &=& 3 & (a) \\<br />
64a + 16b + 4c &=& \text{-}2 & (b) \\216a + 36b + 6c &=& \text{-}7 & (c) \end{array}

    \begin{array}{cccccc}\text{Divide (a) by 2:} & 4a + 2b + c &=& \frac{3}{2} & [1] \\ \\[-3mm] \text{Divide (b) by 4:} & 16a + 4b + c &=& \text{-}\frac{1}{2} & [2] \\ \\[-3mm] \text{Divide (c) by 6:} & 36a + 6b + c &=& \text{-}\frac{7}{6} & [3]\end{array}


    \begin{array}{ccccccc}\text{Subtract [2] - [1]:} & 12a + 2b &=& \text{-}2 & [4] \\ \text{Subtract [3] - [2]:} & 20a + 2b &=& \text{-}\frac{2}{3} & [5] \end{array}


    Subtract [5] - [4]: . 8a \:=\:\frac{4}{3} \quad\Rightarrow \quad a \:=\:\frac{1}{6}

    Substitute into [4]: . 12\left(\tfrac{1}{6}\right) + 2b \:=\:-2 \quad\Rightarrow\quad b \:=\:-2

    Substitute into [1]: . 4\left(\tfrac{1}{6}\right)+ 2(-2) + C \:=\:\frac{3}{2} \quad\Rightarrow\quad c \:=\:\frac{29}{6}


    Therefore: . y \;=\;\frac{1}{6}n^3 - 2n^2 + \frac{29}{6}n + 7

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  4. #4
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    Third way! If you don't like fractions, Multiply 8a+ 4b+ 2c= 3 by 2 to get 16a+ 8b+ 4c= 6 and subtract from the second equation, 64a+ 16b+ 4c= -2, to get 48a+ 8b= -8, eliminating c. Similarly, multiply 8a+ 4b+ 2c= 3 by 3 to get 24a+ 12b+ 6c= 9 and subtract from the third equation, 216a+ 36b+ 6c= -7 to get 192a+ 24b= -16, again eliminating c.

    Now we have 48a+ 8b= -8 and 192a+ 24b= -16. Multiply 48a+ 8b= -8 by 3 to get 144a+ 24b= -24 and subtract from 192a+ 24b= -16 to get 48a= 8.

    a= \frac{8}{48}= \frac{1}{6} as Soroban got.
    (Of course, since a is a fraction, you can't avoid a fraction here.)

    Now put that back into 48a+ 8b= -8, which is the same as 6a+ b= -1 to get 1+ b= -1 so b= -2.

    Put a= 1/6 and b= -2 into any of the original three equations to solve for c.
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