Suppose you need to estimate the value of y at x = 2. Use quadratic approximation
with the following points: (0,2) , (1,-1), (3,4)

To estimate the value of \( y \) at \( x = 2 \) using **quadratic approximation** from the points \( (0, 2) \), \( (1, -1) \), and \( (3, 4) \), we will find a quadratic function of the form:

\[
y = ax^2 + bx + c
\]

### Step 1: Set up the system of equations
Using the given points, we can create a system of equations by substituting the values of \( x \) and \( y \) into the quadratic form.

1. For the point \( (0, 2) \):
\[
2 = a(0)^2 + b(0) + c \quad \Rightarrow \quad c = 2
\]

2. For the point \( (1, -1) \):
\[
-1 = a(1)^2 + b(1) + 2 \quad \Rightarrow \quad -1 = a + b + 2 \quad \Rightarrow \quad a + b = -3
\]

3. For the point \( (3, 4) \):
\[
4 = a(3)^2 + b(3) + 2 \quad \Rightarrow \quad 4 = 9a + 3b + 2 \quad \Rightarrow \quad 9a + 3b = 2
\]

### Step 2: Solve the system of equations

We now have the system:
\[
\begin{aligned}
a + b &= -3 \\
9a + 3b &= 2
\end{aligned}
\]

First, solve for \( b \) from the first equation:
\[
b = -3 - a
\]

Substitute this into the second equation:
\[
9a + 3(-3 - a) = 2
\]
\[
9a - 9 - 3a = 2
\]
\[
6a - 9 = 2
\]
\[
6a = 11
\]
\[
a = \frac{11}{6}
\]

Now substitute \( a = \frac{11}{6} \) into the equation for \( b \):
\[
b = -3 - \frac{11}{6} = -\frac{18}{6} - \frac{11}{6} = -\frac{29}{6}
\]

Thus, the quadratic equation is:
\[
y = \frac{11}{6}x^2 - \frac{29}{6}x + 2
\]

### Step 3: Estimate \( y \) at \( x = 2 \)

Now, substitute \( x = 2 \) into the quadratic equation:
\[
y = \frac{11}{6}(2)^2 - \frac{29}{6}(2) + 2
\]
\[
y = \frac{11}{6}(4) - \frac{29}{6}(2) + 2
\]
\[
y = \frac{44}{6} - \frac{58}{6} + 2
\]
\[
y = \frac{44 - 58}{6} + 2 = -\frac{14}{6} + 2 = -\frac{7}{3} + 2 = -\frac{7}{3} + \frac{6}{3} = -\frac{1}{3}
\]

Thus, the estimate for \( y \) at \( x = 2 \) is:
\[
y \approx -\frac{1}{3}
\]

### Summary
The quadratic approximation estimates that \( y \) at \( x = 2 \) is \( -\frac{1}{3} \).

Would you like further details or clarifications on this process?

### 5 Related Questions
1. How is quadratic approximation different from linear approximation?
2. How would you use cubic approximation with additional points?
3. Can we apply this method to estimate values outside the given range of points?
4. How do you derive the general formula for a quadratic function given any three points?
5. How does the accuracy of approximation change with different point selections?

### Tip
When using quadratic approximation, always choose points that are close to the point of interest to increase the accuracy of your estimate.

Suppose you need to estimate the value of y at x = 2. Use quadratic approximation with the following points: (0,2), (1,-1), (3,4).

Learn how to estimate y at x = 2 using quadratic approximation with the points (0, 2), (1, -1), and (3, 4). This problem involves solving a system of equations to find the quadratic formula. Suitable for students in Grades 9-12.

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