Find Equation of Quadratic Function from Its Graph
Learn how to determine the equation of a quadratic function when given its graph. This guide explains three different methods with step-by-step examples.
Example Problem
Find the equation of the quadratic function f whose graph is shown below:
Solution Methods
Method 1: Using X-Intercepts (Factored Form)
The graph shows x-intercepts at \((-3,0)\) and \((-1,0)\), and a y-intercept at \((0,6)\).
- Start with the factored form using x-intercepts:
\[ f(x) = a(x + 3)(x + 1) \]
- Use the y-intercept \(f(0) = 6\):
\[ 6 = a(0 + 3)(0 + 1) \]
- Solve for \(a\):
\[ 6 = a(3)(1) \]
\[ 6 = 3a \]
\[ a = 2 \]
- Write the final equation:
\[ f(x) = 2(x + 3)(x + 1) = 2x^2 + 8x + 6 \]
Method 2: Using Vertex Form
The parabola has vertex at \((-2, -2)\) and y-intercept at \((0,6)\).
- Start with vertex form:
\[ f(x) = a(x + 2)^2 - 2 \]
- Use the y-intercept \(f(0) = 6\):
\[ 6 = a(0 + 2)^2 - 2 \]
- Solve for \(a\):
\[ 6 = a(4) - 2 \]
\[ 8 = 4a \]
\[ a = 2 \]
- Write the final equation:
\[ f(x) = 2(x + 2)^2 - 2 = 2x^2 + 8x + 6 \]
Method 3: Using System of Equations (Standard Form)
Three points on the graph: \((-3, 0)\), \((-1, 0)\), and \((0, 6)\).
- Standard quadratic form:
\[ f(x) = ax^2 + bx + c \]
- Using point \((0, 6)\):
\[ f(0) = a(0)^2 + b(0) + c = 6 \Rightarrow c = 6 \]
- Using point \((-3, 0)\):
\[ f(-3) = a(-3)^2 + b(-3) + 6 = 0 \]
\[ 9a - 3b + 6 = 0 \]
- Using point \((-1, 0)\):
\[ f(-1) = a(-1)^2 + b(-1) + 6 = 0 \]
\[ a - b + 6 = 0 \]
- Solve the system:
\[
\begin{cases}
9a - 3b = -6 \\
a - b = -6
\end{cases}
\]
- Solution: \(a = 2\), \(b = 8\)
- Final equation:
\[ f(x) = 2x^2 + 8x + 6 \]
Key Formulas
- Standard Form: \(f(x) = ax^2 + bx + c\)
- Vertex Form: \(f(x) = a(x - h)^2 + k\) where \((h,k)\) is vertex
- Factored Form: \(f(x) = a(x - r_1)(x - r_2)\) where \(r_1, r_2\) are x-intercepts
Practice Resources
Use this interactive applet to generate practice problems and check your answers.
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