To find the quadratic function that best fits the given data points, we can denote the quadratic function in the standard form as:
\[ f(x) = ax^2 + bx + c \]
Given the data points:
\[ \begin{align*} (0, 0), \ (2, 397), \ (4, 1602), \ (6, 3604), \ (8, 6401), \ (10, 9996) \end{align*} \]
We can set up a system of equations using the quadratic function for several \(x\) values. Let's use three points to create a system of equations. We can choose \( (0, 0) \), \( (2, 397) \), and \( (4, 1602) \):
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For \( (0, 0) \): \[ 0 = a(0^2) + b(0) + c \Rightarrow c = 0 \]
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For \( (2, 397) \): \[ 397 = a(2^2) + b(2) + c \Rightarrow 397 = 4a + 2b \]
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For \( (4, 1602) \): \[ 1602 = a(4^2) + b(4) + c \Rightarrow 1602 = 16a + 4b \]
Now we have a system with the equations:
\[ \begin{align*}
- \quad & 4a + 2b = 397 \quad \text{(1)} \
- \quad & 16a + 4b = 1602 \quad \text{(2)} \end{align*} \]
We can solve these equations step by step:
From the first equation, solve for \(b\):
\[ 2b = 397 - 4a \quad \Rightarrow \quad b = \frac{397 - 4a}{2} \]
Substituting this expression for \(b\) into the second equation:
\[ 16a + 4\left(\frac{397 - 4a}{2}\right) = 1602 \]
This simplifies to:
\[ 16a + 2(397 - 4a) = 1602 \]
Expanding gives:
\[ 16a + 794 - 8a = 1602 \]
Combining like terms results in:
\[ 8a + 794 = 1602 \]
Subtracting 794 from both sides:
\[ 8a = 808 \quad \Rightarrow \quad a = 101 \]
Now substitute \(a = 101\) back into the equation for \(b\):
\[ b = \frac{397 - 4(101)}{2} = \frac{397 - 404}{2} = \frac{-7}{2} = -3.5 \]
Now we have:
- \(a = 101\)
- \(b = -3.5\)
- \(c = 0\)
Thus, the quadratic function that best fits the data is:
\[ f(x) = 101x^2 - 3.5x + 0 \]
In standard form, the quadratic function is:
\[ \boxed{101x^2 - 3.5x} \]
If you need the function where \(c\) is explicitly included, it can be represented as:
\[ f(x) = 101x^2 - 3.50x + 0 \]
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