To find the best fit quadratic function for the data given, we can use the general form of a quadratic function:
\[ f(x) = ax^2 + bx + c \]
We'll need to determine the coefficients \( a \), \( b \), and \( c \) using methods such as polynomial regression. However, I can guide you through the process of setting up the equations based on the data provided to find these coefficients.
Given your data at the points:
\[ \begin{array}{|c|c|} \hline x & f(x) \ \hline 0 & 0 \ 2 & 397 \ 4 & 1602 \ 6 & 3601 \ 8 & 6401 \ 10 & 9998 \ \hline \end{array} \]
We can set up a system of equations by plugging the \( x \) values into the quadratic function form. From the first three data points, we can derive the following equations:
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When \( x = 0 \): \[ c = 0 \]
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When \( x = 2 \): \[ 4a + 2b + c = 397 \]
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When \( x = 4 \): \[ 16a + 4b + c = 1602 \]
Using \( c = 0 \), we can simplify these to:
- \( 4a + 2b = 397 \) (Equation 1)
- \( 16a + 4b = 1602 \) (Equation 2)
Now, we can solve Equation 1 for \( b \):
\[ b = \frac{397 - 4a}{2} \]
Substituting this expression for \( b \) into Equation 2:
\[ 16a + 4\left(\frac{397 - 4a}{2}\right) = 1602 \] \[ 16a + 2(397 - 4a) = 1602 \] \[ 16a + 794 - 8a = 1602 \] \[ 8a + 794 = 1602 \] \[ 8a = 808 \] \[ a = 101 \]
Now substituting \( a \) back into Equation 1 to find \( b \):
\[ 4(101) + 2b = 397 \] \[ 404 + 2b = 397 \] \[ 2b = 397 - 404 \] \[ 2b = -7 \] \[ b = -3.5 \]
Now we have \( a \), \( b \), and \( c \):
- \( a = 101 \)
- \( b = -3.5 \)
- \( c = 0 \)
Thus, the quadratic function that best fits the data is:
\[ f(x) = 101x^2 - 3.5x + 0 \]
In the format requested, this can be expressed as:
\[ f(x) = 101x^2 - 3.50x + 0 \]
So, the final answer is:
- \( a = 101 \)
- \( b = -3.50 \)
- \( c = 0 \)
So the quadratic function is:
\[ f(x) = 101x^2 - 3.50x \]
1 answer