Question

Use differences to find the degree of a polynomial fitting the data. Then use a system

Round to two decimal places where necessary.


P(x) -20 -39 -66-101-144-195 -254 -321

A linear model fits the given data best, and its equation is:

A quadratic model fits the given data best, and its equation is:

A cubic model fits the given data best, and its equation is:

O A quartic model fits the given data best, and its equation is:

Answer 1

B) A quadratic model fits the given data best, and its equation is:

`\large \boxed{2}\:x^2 +\boxed{3}\:x+\boxed{13}`

Explanation:

The given data is:

`\begin{array}c\cline{1-9}\vphantom{\frac12}x&1&2&3&4&5&6&7&8\\\cline{1-9}\vphantom{\frac12}P(x)&18&27&40&57&78&103&132&165\\\cline{1-9}\end{array}`

To determine the degree of the polynomial fitting the given data, we can calculate the differences between consecutive values of P(x) until we reach a constant difference, which indicates the degree of the polynomial.

First-order differences:

`18 \underset{+9}{\longrightarrow}27 \underset{+13}{\longrightarrow}40 \underset{+17}{\longrightarrow}57 \underset{+21}{\longrightarrow}78 \underset{+25}{\longrightarrow}103 \underset{+29}{\longrightarrow}132 \underset{+33}{\longrightarrow}165`

Second-order differences:

`9 \underset{+4}{\longrightarrow}13\underset{+4}{\longrightarrow}17\underset{+4}{\longrightarrow}21\underset{+4}{\longrightarrow}25\underset{+4}{\longrightarrow}29\underset{+4}{\longrightarrow}33`

The second-order differences are all constant at +4, which means the degree of the polynomial is 2, indicating a quadratic model fits the data best.

The coefficient of x² is always half of the second difference. So, the coefficient of x² is 2.

Therefore, we can represent the quadratic model as:

`P(x) = 2x^2 + bx + c`

To find the values of b and c, we can use the data points (x, P(x)) to create two equations:

`\begin{aligned}P(1) = 2(1)^2 + b(1) + c&=18\\2+b+c&=18\\b+c&=16\end{aligned}`

`\begin{aligned}P(2) = 2(2)^2 + b(2) + c&=27\\8+2b+c&=27\\2b+c&=19\end{aligned}`

Solve this system of equations by rearranging the first equation to isolate c:

`\begin{aligned}b+c&=16\\b+c-c&=16-b\\c&=16-b\end{aligned}`

Substitute the equation for c into the second equation and solve for b:

`\begin{aligned}2b+(16-b)&=19\\2b+16-b&=19\\b+16&=19\\b&=3\end{aligned}`

Substitute the found value of b into the equation for c:

`\begin{aligned}c&=16-3\\c&=13\end{aligned}`

So, the quadratic model for the given data is:

`\large\boxed{\boxed{f(x) = 2x^2 +3x +13}}`

Therefore, a quadratic model fits the given data best, and its equation is:

`\large \boxed{2}\:x^2 +\boxed{3}\:x+\boxed{13}`.