Equation for a Function with Given Points

**suavrectangle** · April 14th 2010, 04:28 PM

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?

**pickslides** · April 14th 2010, 04:39 PM

Originally Posted by suavrectangle

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?

**Soroban** · April 14th 2010, 05:35 PM

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

**HallsofIvy** · April 14th 2010, 10:46 PM

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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