You can determine a quadratic function by finding the coefficients a, b, and c in the standard form f(x) = ax2 + bx + c. According to the provided reference, this is possible if you have three points on the graph of the quadratic function.
Finding the Quadratic Function Algebraically
The most common method involves solving a system of three equations with three unknowns. Here's how it works:
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Start with the General Form: Begin with the general form of a quadratic function: f(x) = ax2 + bx + c.
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Substitute the Points: Substitute the x and y values (or f(x)) of each of the three given points into the general form. This will create three equations. Let's say the points are (x1, y1), (x2, y2), and (x3, y3). This results in the following system of equations:
- y1 = ax12 + bx1 + c
- y2 = ax22 + bx2 + c
- y3 = ax32 + bx3 + c
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Solve the System of Equations: Solve the system of three equations for a, b, and c. You can use various methods:
- Substitution: Solve one equation for one variable and substitute that expression into the other equations.
- Elimination: Add or subtract multiples of the equations to eliminate one variable at a time.
- Matrices: Represent the system as a matrix equation and solve using matrix operations (e.g., Gaussian elimination, inverse matrix).
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Write the Quadratic Function: Once you have found the values of a, b, and c, substitute them back into the general form f(x) = ax2 + bx + c to obtain the specific quadratic function.
Example
Let's say we have the points (1, 2), (2, 7), and (-1, 2).
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Substitute:
- 2 = a(1)2 + b(1) + c => 2 = a + b + c
- 7 = a(2)2 + b(2) + c => 7 = 4a + 2b + c
- 2 = a(-1)2 + b(-1) + c => 2 = a - b + c
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Solve: Solving this system (using any method) yields a = 2, b = 1, and c = -1.
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Write the function: Therefore, the quadratic function is f(x) = 2x2 + x - 1.
Summary
Step | Description |
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1 | Start with f(x) = ax2 + bx + c |
2 | Substitute three known points into the equation |
3 | Solve the resulting system of three equations for a, b, and c |
4 | Substitute the values of a, b, and c back into the original equation |