Math Problem Statement

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)

Solution

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

y=ax2+bx+cy = 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 xx and yy into the quadratic form.

  1. For the point (0,2)(0, 2): 2=a(0)2+b(0)+cc=22 = a(0)^2 + b(0) + c \quad \Rightarrow \quad c = 2

  2. For the point (1,1)(1, -1): 1=a(1)2+b(1)+21=a+b+2a+b=3-1 = 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)(3, 4): 4=a(3)2+b(3)+24=9a+3b+29a+3b=24 = 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:

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.

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Math Problem Analysis

Mathematical Concepts

Quadratic Approximation
Algebra
Systems of Equations

Formulas

Quadratic equation formula: y = ax^2 + bx + c
System of linear equations to solve for a, b, c

Theorems

Solving Systems of Equations
Quadratic Interpolation

Suitable Grade Level

Grades 9-12