Answer :
To determine the quadratic equation that best fits the given table of values, we assume that the relationship between \(y\) and \(x\) can be described by a quadratic equation of the form:
[tex]\[ y = ax^2 + bx + c \][/tex]
We are given the following data points:
[tex]\[ \begin{array}{c|c} x & y \\ \hline 0 & 4 \\ 1 & 7 \\ 2 & 16 \\ 3 & 31 \\ 4 & 52 \\ \end{array} \][/tex]
We can substitute these points into the quadratic equation to formulate a system of linear equations.
### Step-by-Step Solution
1. Substitute \(x = 0\) and \(y = 4\) into the equation:
[tex]\[ 4 = a(0)^2 + b(0) + c \][/tex]
[tex]\[ 4 = c \][/tex]
So, \( c = 4 \).
2. Substitute \(x = 1\) and \(y = 7\) into the equation:
[tex]\[ 7 = a(1)^2 + b(1) + c \][/tex]
Since \( c = 4 \):
[tex]\[ 7 = a + b + 4 \][/tex]
[tex]\[ 7 - 4 = a + b \][/tex]
[tex]\[ 3 = a + b \][/tex]
3. Substitute \(x = 2\) and \(y = 16\) into the equation:
[tex]\[ 16 = a(2)^2 + b(2) + c \][/tex]
Since \( c = 4 \):
[tex]\[ 16 = 4a + 2b + 4 \][/tex]
[tex]\[ 16 - 4 = 4a + 2b \][/tex]
[tex]\[ 12 = 4a + 2b \][/tex]
[tex]\[ 6 = 2a + b \][/tex]
4. We now have two equations from steps 2 and 3:
[tex]\[ 3 = a + b \][/tex]
[tex]\[ 6 = 2a + b \][/tex]
To solve for \(a\) and \(b\), we can subtract the first equation from the second:
[tex]\[ (2a + b) - (a + b) = 6 - 3 \][/tex]
Simplifying, we get:
[tex]\[ a = 3 \][/tex]
Substituting \(a = 3\) back into the first equation:
[tex]\[ 3 = 3 + b \][/tex]
[tex]\[ b = 0 \][/tex]
### Conclusion
Thus, we have found the coefficients:
[tex]\[ a = 3 \][/tex]
[tex]\[ b = 0 \][/tex]
[tex]\[ c = 4 \][/tex]
The quadratic equation describing the relationship between \( y \) and \( x \) is:
[tex]\[ y = 3x^2 + 0x + 4 \][/tex]
[tex]\[ y = 3x^2 + 4 \][/tex]
This quadratic equation fits the given table of values perfectly.
[tex]\[ y = ax^2 + bx + c \][/tex]
We are given the following data points:
[tex]\[ \begin{array}{c|c} x & y \\ \hline 0 & 4 \\ 1 & 7 \\ 2 & 16 \\ 3 & 31 \\ 4 & 52 \\ \end{array} \][/tex]
We can substitute these points into the quadratic equation to formulate a system of linear equations.
### Step-by-Step Solution
1. Substitute \(x = 0\) and \(y = 4\) into the equation:
[tex]\[ 4 = a(0)^2 + b(0) + c \][/tex]
[tex]\[ 4 = c \][/tex]
So, \( c = 4 \).
2. Substitute \(x = 1\) and \(y = 7\) into the equation:
[tex]\[ 7 = a(1)^2 + b(1) + c \][/tex]
Since \( c = 4 \):
[tex]\[ 7 = a + b + 4 \][/tex]
[tex]\[ 7 - 4 = a + b \][/tex]
[tex]\[ 3 = a + b \][/tex]
3. Substitute \(x = 2\) and \(y = 16\) into the equation:
[tex]\[ 16 = a(2)^2 + b(2) + c \][/tex]
Since \( c = 4 \):
[tex]\[ 16 = 4a + 2b + 4 \][/tex]
[tex]\[ 16 - 4 = 4a + 2b \][/tex]
[tex]\[ 12 = 4a + 2b \][/tex]
[tex]\[ 6 = 2a + b \][/tex]
4. We now have two equations from steps 2 and 3:
[tex]\[ 3 = a + b \][/tex]
[tex]\[ 6 = 2a + b \][/tex]
To solve for \(a\) and \(b\), we can subtract the first equation from the second:
[tex]\[ (2a + b) - (a + b) = 6 - 3 \][/tex]
Simplifying, we get:
[tex]\[ a = 3 \][/tex]
Substituting \(a = 3\) back into the first equation:
[tex]\[ 3 = 3 + b \][/tex]
[tex]\[ b = 0 \][/tex]
### Conclusion
Thus, we have found the coefficients:
[tex]\[ a = 3 \][/tex]
[tex]\[ b = 0 \][/tex]
[tex]\[ c = 4 \][/tex]
The quadratic equation describing the relationship between \( y \) and \( x \) is:
[tex]\[ y = 3x^2 + 0x + 4 \][/tex]
[tex]\[ y = 3x^2 + 4 \][/tex]
This quadratic equation fits the given table of values perfectly.
