How to get quartic polynomial function given graph, 4 points, and 1 root

Question

I'm doing a math assignment right now, which is due in more or less four and a half hours. What I need to find is a polynomial function given this graph this graph and the points on it. Here are the points:

0,15

7,-1

22,-7

26,0

28,14

The other two points marked on the graph were just marked for another question; I'm not exactly sure if they are x intercepts because I can see that they are a few pixels above or below. Thanks in advance :)

Edit: So far, using the point 7,-1 and the x- and y-intercepts, I tried substituting and transposing the terms in the polynomial, so from y=ax^4+bx^3+cx^2+dx+e, I got 2401a + 343b + 49c + 7d + 16 = 0, and I'm stumped from there. I've also tried doing it in factored form, so y = a(x-26)(x-b)(x-c)(x-d), and I don't really know what to do next.

Answer 1

We can set up a general quartic equation like this:

$$ y = ax^4 + bx^3 + cx^2 + dx + e . $$

We can then set up a system of equations by plugging in each of the points individually:

$$ \begin{cases} 15 = 0a + 0b + 0c + 0d + 1e \\ -1 = 7^4a + 7^3b + 7^2c + 7d + e \\ -7 = 22^4a + 22^3b + 22^2c + 22d + e \\ 0 = 26^4a + 26^3b + 26^2c + 26d + e \\ 14 = 28^4a + 28^3b + 28^2c + 28d + e \\ \end{cases} .$$

This is a system of linear equations with five equations and five unknowns: $a, b, c, d, e$. It's a little different from the systems of equations you might be used to, where you know the equation and not the points - in that case, you have the coefficients but not $x$ and $y$. In this case, you know the points but not the function, so it's the reverse: coefficients are unknown while $x$ and $y$ are known.

If you're familiar with matrices, this is becomes very easy to solve. The system can be rewritten as

$$\begin{bmatrix} 15 \\ -1 \\ -7 \\ 0 \\ 14 \\ \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 & 0 & 1 \\ 7^4 & 7^3 & 7^2 & 7 & 1 \\ 22^4 & 22^3 & 22^2 & 22 & 1 \\ 26^4 & 26^3 & 26^2 & 26 & 1 \\ 28^4 & 28^3 & 28^2 & 28 & 1 \\ \end{bmatrix} \begin{bmatrix} a \\ b \\ c \\ d \\ e \\ \end{bmatrix} $$

$$\Rightarrow \begin{bmatrix} a \\ b \\ c \\ d \\ e \\ \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 & 0 & 1 \\ 7^4 & 7^3 & 7^2 & 7 & 1 \\ 22^4 & 22^3 & 22^2 & 22 & 1 \\ 26^4 & 26^3 & 26^2 & 26 & 1 \\ 28^4 & 28^3 & 28^2 & 28 & 1 \\ \end{bmatrix} ^{-1} \begin{bmatrix} 15 \\ -1 \\ -7 \\ 0 \\ 14 \\ \end{bmatrix} $$

This gives us the following solution:

$$ y = \frac{87}{69160}x^4 - \frac{31}{455}x^3 + \frac{82807}{69160}x^2 - \frac{268341}{34580}x + 15 $$

Here's the matrix multiplication using WolframAlpha.

If you're not familiar with matrices, you'll need to use substitution and elimination, and it'll take forever. But at least the first equation gets you $e=15$, so that's something you can start your substitution with.

Answer 2

Given a set $S=\{x_1,...,x_{n+1} \}$ of $n+1$ distinct values:

For $1\le j\le n+1$ let $S_j=S\setminus \{x_j\}.$

The polynomial $$P_j(x)=\frac {\prod_{s\in S_j}(x-s)}{\prod_{s\in S_j}(x_j-s)}$$ is $1$ when $x=x_j$ and is $0$ when $x_j\ne x\in S.$

Given a sequence $(y_1,...,y_{n+1}),$ the polynomial $$Q(x)=\sum_{j=1}^{n+1}y_jP_j(x)$$ satisfies $Q(x_j)=y_j$ for all $j\in \{1,...,n+1\}.$