Find Equation of Quadratic Function Calculator
Enter the coordinates of three distinct points that lie on the parabola to find the equation of the quadratic function (y = ax² + bx + c).
Results:
a(x₁)² + b(x₁) + c = y₁
a(x₂)² + b(x₂) + c = y₂
a(x₃)² + b(x₃) + c = y₃
Visualization and Data
Graph showing the three input points and the calculated quadratic function.
| Point | x-coordinate | y-coordinate |
|---|---|---|
| 1 | 1 | 3 |
| 2 | 2 | 8 |
| 3 | 3 | 15 |
Table of input points.
What is Finding the Equation of a Quadratic Function?
Finding the equation of a quadratic function means determining the specific values of the coefficients 'a', 'b', and 'c' in the standard form y = ax² + bx + c, such that the graph of this function (a parabola) passes through a given set of points. If you know three distinct points that lie on the parabola, and no two of these points are vertically aligned, you can uniquely determine the equation of the quadratic function. The find equation of quadratic function calculator automates this process.
This is useful in various fields like physics (to model projectile motion), engineering (to design parabolic reflectors), and finance (to model certain cost or revenue functions). Anyone working with data that appears to follow a parabolic trend might use this to find the underlying model. Our find equation of quadratic function calculator is a tool for students, teachers, engineers, and scientists.
A common misconception is that any three points will define a quadratic function. While true for non-collinear points that aren't vertically aligned, if the three points lie on a straight line, you won't get a quadratic function (the 'a' coefficient will be zero or undefined if x-values are the same). The find equation of quadratic function calculator handles cases where points might lead to issues.
Find Equation of Quadratic Function Calculator Formula and Mathematical Explanation
Given three points (x₁, y₁), (x₂, y₂), and (x₃, y₃), we substitute them into the standard quadratic equation y = ax² + bx + c:
- a(x₁)² + b(x₁) + c = y₁
- a(x₂)² + b(x₂) + c = y₂
- a(x₃)² + b(x₃) + c = y₃
This forms a system of three linear equations with three unknowns (a, b, c). We can write this in matrix form:
[ x₁² x₁ 1 ] [ a ] [ y₁ ]
[ x₂² x₂ 1 ] [ b ] = [ y₂ ]
[ x₃² x₃ 1 ] [ c ] [ y₃ ]
To solve for a, b, and c, we can use methods like substitution, elimination, or matrix methods (like Cramer's rule or finding the inverse of the coefficient matrix). Our find equation of quadratic function calculator uses these principles.
Let:
D = x₁²(x₂ – x₃) – x₁(x₂² – x₃²) + (x₂²x₃ – x₃²x₂)
Da = y₁(x₂ – x₃) – x₁(y₂ – y₃) + (y₂x₃ – y₃x₂)
Db = x₁²(y₂ – y₃) – y₁(x₂² – x₃²) + (x₂²y₃ – x₃²y₂)
Dc = x₁²(x₂y₃ – x₃y₂) – x₁(x₂²y₃ – x₃²y₂) + y₁(x₂²x₃ – x₃²x₂)
If D is not zero, then:
a = Da / D
b = Db / D
c = Dc / D
The find equation of quadratic function calculator performs these calculations.
Variables Table
| Variable | Meaning | Unit | Typical Range |
|---|---|---|---|
| x₁, y₁ | Coordinates of the first point | None (or units of the problem) | Real numbers |
| x₂, y₂ | Coordinates of the second point | None (or units of the problem) | Real numbers |
| x₃, y₃ | Coordinates of the third point | None (or units of the problem) | Real numbers |
| a, b, c | Coefficients of the quadratic equation y = ax² + bx + c | Varies based on x and y units | Real numbers |
Practical Examples (Real-World Use Cases)
Example 1: Projectile Motion
An object is thrown, and its height (y) at different horizontal distances (x) is recorded: (1, 5), (2, 8), (3, 9). We want to find the parabolic path y = ax² + bx + c.
Using the find equation of quadratic function calculator with x₁=1, y₁=5, x₂=2, y₂=8, x₃=3, y₃=9, we get:
a = -1, b = 6, c = 0. So, the equation is y = -x² + 6x.
Example 2: Cost Function
A company finds its cost (y) to produce x units is (10, 500), (20, 800), (30, 1300). Assuming a quadratic cost model:
Using the find equation of quadratic function calculator with x₁=10, y₁=500, x₂=20, y₂=800, x₃=30, y₃=1300, we get:
a = 1, b = 10, c = 300. So, the cost equation is y = x² + 10x + 300.
How to Use This Find Equation of Quadratic Function Calculator
- Enter Point 1: Input the x-coordinate (x₁) and y-coordinate (y₁) of the first point.
- Enter Point 2: Input the x-coordinate (x₂) and y-coordinate (y₂) of the second point.
- Enter Point 3: Input the x-coordinate (x₃) and y-coordinate (y₃) of the third point.
- Calculate: Click the "Calculate" button or simply change input values. The find equation of quadratic function calculator will automatically update the results.
- Read Results: The primary result shows the equation
y = ax² + bx + cwith the calculated values of a, b, and c. Intermediate results show the individual values of a, b, and c. - Visualize: The graph shows the three points you entered and the parabola that passes through them. The table below the graph also reflects your input points.
- Reset: Click "Reset" to return to the default example values.
- Copy: Click "Copy Results" to copy the equation and coefficients.
Ensure the three points are distinct and don't have the same x-values if you expect a unique quadratic function. The find equation of quadratic function calculator will warn if the points are collinear or vertically aligned, making 'a' zero or the system unsolvable in the standard quadratic form.
Key Factors That Affect Find Equation of Quadratic Function Calculator Results
- Coordinates of the Points (x₁, y₁), (x₂, y₂), (x₃, y₃): These are the direct inputs. Small changes in these values can significantly alter the coefficients a, b, and c, and thus the shape and position of the parabola.
- Distinctness of x-values: If any two x-values are the same (x₁=x₂, x₁=x₃, or x₂=x₃), you cannot define a unique *function* y=f(x) that is quadratic passing through them (it would imply a vertical line, not a function, or the points are identical). The find equation of quadratic function calculator expects distinct x-values for a standard quadratic function result.
- Collinearity of the Points: If the three points lie on a straight line, the coefficient 'a' will be zero, meaning the equation is linear (y = bx + c), not quadratic. The calculator might indicate this or show a=0.
- Magnitude of Coordinates: Very large or very small coordinate values can lead to very large or very small coefficients 'a', 'b', or 'c', potentially affecting the precision visible in the results.
- Relative Positions of Points: The arrangement of the points determines the concavity (opening upwards or downwards, controlled by 'a') and the vertex of the parabola.
- Precision of Input: The accuracy of the calculated coefficients a, b, and c depends on the precision of the input coordinates. Using more decimal places in the inputs, if available, leads to more precise results from the find equation of quadratic function calculator.
Frequently Asked Questions (FAQ)
A quadratic function is a polynomial function of degree 2, generally expressed as f(x) = ax² + bx + c, where a, b, and c are constants and 'a' is not zero. Its graph is a parabola.
A quadratic function has three coefficients (a, b, c). To uniquely solve for these three unknowns, you need a system of three independent equations, which can be obtained by substituting the coordinates of three distinct points into the equation.
If the three points are collinear, the 'a' coefficient will be zero, and the equation will simplify to y = bx + c, which is a linear equation, not quadratic. The find equation of quadratic function calculator will likely show a=0 or indicate the points are collinear.
If two points have the same x-coordinate but different y-coordinates, they form a vertical line. A function cannot pass through two such points. If the y-coordinates are also the same, the points are identical, and you effectively have only two distinct points, which are not enough to uniquely define a quadratic function.
If 'a' > 0, the parabola opens upwards. If 'a' < 0, the parabola opens downwards. The magnitude of 'a' affects how wide or narrow the parabola is.
It sets up a system of three linear equations using the coordinates of the three points and solves it using algebraic methods, often involving determinants or matrix inversion, as described in the formula section.
Yes, once you have the equation y = ax² + bx + c, the x-coordinate of the vertex is given by -b / (2a). You can then substitute this x-value back into the equation to find the y-coordinate of the vertex.
This can happen depending on the scale and position of your input points. It doesn't necessarily mean an error, but it reflects the sensitivity of the coefficients to the point locations.