A quadratic function is a polynomial of degree two, meaning the highest power of the variable is two. Its visual representation is a distinctive U-shaped curve known as a parabola. Determining the precise algebraic equation requires identifying specific coordinates on the graph and applying them to one of the function’s established algebraic forms. The choice of method depends on which features of the parabola are most clearly discernible from the visual data.
Identifying the Essential Features of the Graph
The foundation for determining the equation lies in several significant points on the parabola. The most defining feature is the vertex, denoted by $(h, k)$, which is the highest or lowest point on the curve. This vertex is located on the axis of symmetry, the vertical line $x=h$ that perfectly divides the parabola into two mirror-image halves.
Other important points are the intercepts, where the curve crosses the coordinate axes. The x-intercepts, also known as the roots, are the points $(p, 0)$ and $(q, 0)$ where the parabola intersects the horizontal axis. A parabola may have two, one, or zero x-intercepts. The y-intercept is the single point $(0, c)$ where the parabola crosses the vertical axis. The clarity of these points determines the most efficient algebraic method for reconstruction.
Method 1: Deriving the Function Using the Vertex Form
The vertex form of a quadratic function, $y = a(x-h)^2 + k$, is the most direct method when the vertex $(h, k)$ is clearly identifiable on the graph. This form explicitly incorporates the vertex coordinates. The parameter $a$ is the stretch factor, controlling the vertical compression or expansion and the direction of opening. A positive $a$ means the parabola opens upward, while a negative $a$ means it opens downward.
The first step is to read the coordinates $(h, k)$ directly from the graph and substitute them into the vertex form, leaving $a$ as the unknown. For example, if the vertex is at $(2, 1)$, the equation becomes $y = a(x-2)^2 + 1$. Next, identify one other distinct point $(x, y)$ that the parabola passes through, such as the y-intercept.
This second point is substituted into the partially completed equation to solve for $a$. If the parabola passes through $(0, 5)$, the substitution yields $5 = a(0-2)^2 + 1$. Simplifying results in $5 = 4a + 1$. Solving for $a$ gives $a = 1$. The final quadratic function is then written by substituting $a$ back into the vertex form: $y = (x-2)^2 + 1$.
Method 2: Deriving the Function Using the Intercept Form
The intercept form, $y = a(x-p)(x-q)$, is the preferred method when the two x-intercepts are clearly visible on the graph. Here, $p$ and $q$ are the x-coordinates of the intercepts, $(p, 0)$ and $(q, 0)$. This structure immediately reveals the roots of the equation.
The process begins by identifying $p$ and $q$ and substituting them into the intercept form. For example, if a parabola crosses the x-axis at $x=-1$ and $x=3$, the equation is $y = a(x+1)(x-3)$. The stretch factor $a$ remains unknown and must be determined using a third point from the graph.
A convenient point to use is often the y-intercept, or any other distinct point $(x, y)$. If the parabola passes through $(0, -6)$, substitute these coordinates to solve for $a$: $-6 = a(0+1)(0-3)$. This simplifies to $-6 = -3a$, revealing that $a = 2$. The final quadratic function is $y = 2(x+1)(x-3)$.
Method 3: Deriving the Function Using Three Arbitrary Points
When the vertex or x-intercepts are not easily read as precise integer coordinates, a more general approach using three arbitrary points is necessary. This method relies on the standard form of the quadratic function, $y = ax^2 + bx + c$. Since this form contains three unknown coefficients ($a$, $b$, and $c$), three distinct, non-collinear points are required to solve for them.
The procedure involves selecting three points, such as $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, and substituting each pair of coordinates into the standard form. Each substitution generates a linear equation involving $a$, $b$, and $c$. For instance, substituting $(1, 4)$ yields the equation $a + b + c = 4$.
Repeating this substitution for the other two points creates a system of three linear equations with three variables. Solving this system, typically through substitution or elimination, is algebraically intensive but yields the unique values for $a$, $b$, and $c$. Once these coefficients are determined, they are substituted back into the standard form to complete the algebraic representation of the parabola.