Find the Equation of a Quadratic Function from Its Graph

Learn how to determine the exact equation of a quadratic function when given its graph. Depending on which features of the parabola are visible (the vertex, the x-intercepts, or three points), you can use one of three methods. Below is a step-by-step guide with worked examples, followed by a practice applet that generates random parabolas.

Key Formulas Review

Recall the three main forms of a quadratic function:

Note: Some textbooks call \( ax^2 + bx + c \) the "standard form" and others call \( a(x-h)^2 + k \) the "standard form." The names differ; the algebra does not.

Example Problem

Find the equation of the quadratic function \( f \) whose graph is shown below:

Parabola passing through x-intercepts (-3,0) and (-1,0), y-intercept (0,6), and vertex (-2,-2)

Note: not every parabola will have exactly these features.

Solution Methods

We solve the problem three different ways, based on which points are easiest to read from the graph. All three methods produce the same equation.

Method 1: Using X-Intercepts (Factored Form)

From the graph, the x-intercepts are clearly at \((-3, 0)\) and \((-1, 0)\), and the y-intercept is at \((0, 6)\).

Show Step-by-Step Solution
  1. Start with the factored form using the x-intercepts: \[ f(x) = a(x - r_1)(x - r_2) \] \[ f(x) = a(x - (-3))(x - (-1)) \] \[ f(x) = a(x + 3)(x + 1) \]
  2. Substitute the y-intercept \((0, 6)\) to find \(a\): \[ f(0) = 6 \] \[ 6 = a(0 + 3)(0 + 1) \]
  3. Solve for \(a\): \[ 6 = a(3)(1) \] \[ 6 = 3a \implies a = 2 \]
  4. Write the final equation and optionally expand: \[ f(x) = 2(x + 3)(x + 1) \] \[ f(x) = 2(x^2 + 4x + 3) \] \[ f(x) = 2x^2 + 8x + 6 \]

Factored form requires real x-intercepts. If the parabola does not cross the x-axis, use Method 2 or Method 3.

Method 2: Using the Vertex (Vertex Form)

From the graph, the vertex is at \((-2, -2)\) and the y-intercept is at \((0, 6)\).

Show Step-by-Step Solution
  1. Start with the vertex form and substitute \((h, k) = (-2, -2)\): \[ f(x) = a(x - h)^2 + k \] \[ f(x) = a(x - (-2))^2 + (-2) \] \[ f(x) = a(x + 2)^2 - 2 \]
  2. Use the y-intercept \((0, 6)\) to find \(a\): \[ 6 = a(0 + 2)^2 - 2 \]
  3. Solve for \(a\): \[ 6 = a(2)^2 - 2 \] \[ 6 = 4a - 2 \] \[ 8 = 4a \implies a = 2 \]
  4. Write the final equation and optionally expand: \[ f(x) = 2(x + 2)^2 - 2 \] \[ f(x) = 2(x^2 + 4x + 4) - 2 \] \[ f(x) = 2x^2 + 8x + 8 - 2 \] \[ f(x) = 2x^2 + 8x + 6 \]

Vertex form requires the vertex. The vertex is a minimum if \( a > 0 \) and a maximum if \( a < 0 \); here \( a = 2 > 0 \), so the vertex is the lowest point.

Method 3: Using a System of Equations (Standard Form)

If neither the vertex nor the x-intercepts are clear, pick any three distinct points on the graph. We use \((-3, 0)\), \((-1, 0)\), and \((0, 6)\).

Show Step-by-Step Solution
  1. Write the standard form: \[ f(x) = ax^2 + bx + c \]
  2. Substitute the y-intercept \((0, 6)\) to find \(c\): \[ f(0) = a(0)^2 + b(0) + c = 6 \implies c = 6 \] So \( f(x) = ax^2 + bx + 6 \).
  3. Substitute \((-3, 0)\) to create the first equation: \[ 9a - 3b + 6 = 0 \] Divide by 3: \[ 3a - b = -2 \quad \text{(Equation 1)} \]
  4. Substitute \((-1, 0)\) to create the second equation: \[ a - b + 6 = 0 \implies a - b = -6 \quad \text{(Equation 2)} \]
  5. Solve the system: \[ \begin{cases} 3a - b = -2 \\ a - b = -6 \end{cases} \] Subtract Equation 2 from Equation 1: \[ 2a = 4 \implies a = 2 \] Substitute \(a = 2\) into Equation 2: \[ 2 - b = -6 \implies b = 8 \]
  6. Write the final equation with \(a = 2\), \(b = 8\), \(c = 6\): \[ f(x) = 2x^2 + 8x + 6 \]

All three methods give the same equation, as expected: a parabola is uniquely determined by any three of its points.

Which Method Should You Use?

Warning: reading points from a graph assumes they lie on grid lines. If the vertex or intercepts fall between grid lines, the "exact" reading may be approximate; verify by substituting the final equation back into the given points.

Practice: Generate a Random Parabola

Now test yourself. The applet below draws a random quadratic \( f(x) = ax^2 + bx + c \). Use the vertex (red dot), the x-intercepts, and the y-intercept on the grid to determine \( a \), \( b \), and \( c \). Then click Show Answer to check.

Quick method. Read the vertex \( (h, k) \) and one other point off the grid. Substitute into vertex form \( y = a(x - h)^2 + k \) to solve for \( a \). Then expand: \( b = -2ah \), \( c = ah^2 + k \).

\( f(x) = ? \)

Frequently Asked Questions

Which method should I use to find a quadratic equation from a graph?

Use factored form when the x-intercepts are clearly readable, vertex form when the vertex is clearly readable, and a linear system when only three arbitrary points (including the y-intercept) are available. Method 3 always works when three points are known, but it requires solving a system of equations.

Can I find a quadratic equation if the graph has no x-intercepts?

Yes. Factored form requires real x-intercepts, but you can still use vertex form if the vertex and any other point are visible, or a linear system if three points are known.

Do all three methods give the same equation?

Yes. Each method determines the same unique parabola from a different set of known points. Factored form, vertex form, and standard form are algebraically equivalent ways of writing the same function.

What range of answers can the practice applet produce?

The leading coefficient \( a \) is a nonzero integer from \( -4 \) to \( 4 \), and the vertex coordinates \( h \) and \( k \) are integers from \( -4 \) to \( 5 \). The resulting equations have integer coefficients \( b = -2ah \) and \( c = ah^2 + k \).

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Last updated: October 20, 2026