Question

What si the algebraic model of
X 4 7 12 19 28 39
Y 1 2 3. 4. 5. 6

Answer 1

To find an algebraic model for the given data points, we can analyze the values of \(X\) and \(Y\):

\[
\begin{align*}
X & : 4, 7, 12, 19, 28, 39 \\
Y & : 1, 2, 3, 4, 5, 6
\end{align*}
\]

First, we'll observe the relationship between \(X\) and \(Y\). As \(Y\) increases linearly from 1 to 6, we should see if \(X\) can be expressed as a function of \(Y\).

Next, let’s examine the differences in the \(X\) values:

\[
\begin{align*}
X_1 & = 4 \\
X_2 & = 7 \quad (X_2 - X_1 = 3) \\
X_3 & = 12 \quad (X_3 - X_2 = 5) \\
X_4 & = 19 \quad (X_4 - X_3 = 7) \\
X_5 & = 28 \quad (X_5 - X_4 = 9) \\
X_6 & = 39 \quad (X_6 - X_5 = 11)
\end{align*}
\]

The differences between consecutive \(X\) values are:

\[
3, 5, 7, 9, 11
\]

Now, let's calculate the second differences:

\[
\begin{align*}
5 - 3 & = 2 \\
7 - 5 & = 2 \\
9 - 7 & = 2 \\
11 - 9 & = 2
\end{align*}
\]

The second differences are constant (equal to 2), which suggests that the relationship between \(X\) and \(Y\) is quadratic. We can express this as:

\[
X = aY^2 + bY + c
\]

Next, we can use the points to establish a system of equations.

To find the coefficients \(a\), \(b\), and \(c\), we can use the first three values:

1. When \(Y = 1\), \(X = 4\):
\[
a(1)^2 + b(1) + c = 4 \quad \text{(1)}
\]

2. When \(Y = 2\), \(X = 7\):
\[
a(2)^2 + b(2) + c = 7 \quad \text{(2)}
\]

3. When \(Y = 3\), \(X = 12\):
\[
a(3)^2 + b(3) + c = 12 \quad \text{(3)}
\]

Substituting these equations, we get:

1. \(a + b + c = 4\) \quad (1)
2. \(4a + 2b + c = 7\) \quad (2)
3. \(9a + 3b + c = 12\) \quad (3)

Next, solve this system of equations. We can eliminate \(c\) by subtracting (1) from (2) and (2) from (3):

From (2) - (1):

\[
(4a + 2b + c) - (a + b + c) = 7 - 4
\]
\[
3a + b = 3 \quad (4)
\]

From (3) - (2):

\[
(9a + 3b + c) - (4a + 2b + c) = 12 - 7
\]
\[
5a + b = 5 \quad (5)
\]

Now, subtract (4) from (5):

\[
(5a + b) - (3a + b) = 5 - 3
\]
\[
2a = 2 \quad \Rightarrow \quad a = 1
\]

Substituting \(a = 1\) into (4):

\[
3(1) + b = 3 \quad \Rightarrow \quad b = 0
\]

Now substituting \(a\) and \(b\) back into (1):

\[
1 + 0 + c = 4 \quad \Rightarrow \quad c = 3
\]

Thus, we have:

\[
a = 1, \quad b = 0, \quad c = 3
\]

The algebraic model that describes the relationship is:

\[
X = Y^2 + 3
\]

So the algebraic model for the data is:

\[
X = Y^2 + 3
\]