Find the quadratic function that is the best fit for​ f(x) defined by the table below.

x
0
2
4
6
8
10

​f(x)
0
397
1602
3602
6402
9999
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Part 1
The quadratic function is yequalsnothing x squared plus nothing x plus left parenthesis nothing right parenthesis
  
  
  
.
​(Type an integer or decimal rounded to two decimal places as​ needed.)

1 answer

To find the quadratic function that best fits the given data, we will assume the function has the form:

\[ f(x) = ax^2 + bx + c \]

We have the data points:

\[ \begin{align*} (0, 0) \ (2, 397) \ (4, 1602) \ (6, 3602) \ (8, 6402) \ (10, 9999) \ \end{align*} \]

We'll create a system of equations based on these points to solve for \(a\), \(b\), and \(c\).

Step 1: Set Up the System of Equations

Using the points, we can substitute into the equation \(f(x) = ax^2 + bx + c\):

  1. For \(x = 0\), \(f(0) = 0\): \[ c = 0 \]

  2. For \(x = 2\), \(f(2) = 397\): \[ 4a + 2b + c = 397 \ 4a + 2b + 0 = 397 \quad (c = 0) \ 4a + 2b = 397 \quad \text{(Eq. 1)} \]

  3. For \(x = 4\), \(f(4) = 1602\): \[ 16a + 4b + c = 1602 \ 16a + 4b + 0 = 1602 \quad (c = 0) \ 16a + 4b = 1602 \quad \text{(Eq. 2)} \]

  4. For \(x = 6\), \(f(6) = 3602\): \[ 36a + 6b + c = 3602 \ 36a + 6b + 0 = 3602 \quad (c = 0) \ 36a + 6b = 3602 \quad \text{(Eq. 3)} \]

Step 2: Solve the System

First, we can eliminate \(b\) by solving Eq. 1 and Eq. 2:

From Eq. 1: \[ 2b = 397 - 4a \ b = \frac{397 - 4a}{2} \]

Substitute \(b\) into Eq. 2: \[ 16a + 4\left(\frac{397 - 4a}{2}\right) = 1602 \ 16a + 2(397 - 4a) = 1602 \ 16a + 794 - 8a = 1602 \ 8a = 1602 - 794 \ 8a = 808 \ a = 101 \]

Now substitute \(a\) back into Eq. 1 to find \(b\): \[ 4(101) + 2b = 397 \ 404 + 2b = 397 \ 2b = 397 - 404 \ 2b = -7 \ b = -3.5 \]

Summary of the Coefficients

  • \(a = 101\)
  • \(b = -3.5\)
  • \(c = 0\)

Step 3: Write the Quadratic Function

Now we can write the quadratic function: \[ f(x) = 101x^2 - 3.5x + 0 \]

So, the best-fit quadratic function is:

\[ f(x) = 101x^2 - 3.5x \]

Final Answer Format

The quadratic function in the requested format is:

\[ y = 101x^2 - 3.5x + 0 \]

The values to fill in are:

  • \(a = 101\)
  • \(b = -3.5\)
  • \(c = 0\)

Write it as:

\[ y = 101x^2 - 3.5x + 0 \]