Finding the Quadratic Function From a Graph

There are three common starting points, and each one matches a different piece of visual information:

• Vertex form, \(y = a(x-h)^2+k\), is the natural choice when the vertex \((h,k)\) is marked on the graph. See characteristics of quadratic functions for a refresher on why \(h\) and \(k\) sit where they do.
• Intercept form, \(y = a(x-p)(x-q)\), is the natural choice when the parabola crosses the x-axis at two clear points \(p\) and \(q\).
• General form, \(y = ax^2+bx+c\), is the fallback when you only have coordinate points and no obvious vertex or intercepts. This is the same layout covered in quadratic function in general form.

In every case, the graph will show one extra point beyond the vertex or roots. That extra point is what lets you solve for \(a\), the stretch factor that controls how wide or narrow the parabola is.

If the graph clearly marks the vertex, write \(y = a(x-h)^2+k\) using that vertex, then substitute any other point on the curve to solve for \(a\).

Example: A parabola has vertex \((1,-4)\) and passes through \((3,4)\). Find the quadratic function.

Start with \(y = a(x-1)^2-4\). Substitute \((3,4)\):

\(4 = a(3-1)^2 - 4\)

\(4 = 4a - 4\)

\(8 = 4a\), so \(a = 2\).

The function is \(f(x) = 2(x-1)^2-4\), which expands to \(f(x) = 2x^2 - 4x - 2\).

If a graph only gives you the general form but not a clean vertex, you can locate the vertex first using the -b/2a vertex formula before switching to this method.

If the parabola visibly crosses the x-axis at two points, use intercept form \(y = a(x-p)(x-q)\), where \(p\) and \(q\) are the roots. Then plug in one more point, often the y-intercept, to solve for \(a\).

Example: A parabola crosses the x-axis at \(x=-2\) and \(x=4\), and passes through \((0,-8)\). Find the quadratic function.

Start with \(y = a(x+2)(x-4)\). Substitute \((0,-8)\):

\(-8 = a(0+2)(0-4)\)

\(-8 = -8a\), so \(a = 1\).

The function is \(f(x) = (x+2)(x-4)\), which expands to \(f(x) = x^2 - 2x - 8\).

When the graph does not show a clean vertex or clean intercepts, but you can read off three coordinate points, use general form \(y = ax^2+bx+c\) and build a system of three equations.

Example: A parabola passes through \((0,3)\), \((1,0)\), and \((2,-1)\). Find the quadratic function.

Since \((0,3)\) is on the graph, \(c = 3\) right away.

Substitute \((1,0)\): \(a+b+3 = 0\), so \(a+b = -3\).

Substitute \((2,-1)\): \(4a+2b+3 = -1\), so \(4a+2b = -4\), which simplifies to \(2a+b = -2\).

Subtracting the first simplified equation from the second gives \(a = 1\), so \(b = -4\).

The function is \(f(x) = x^2 - 4x + 3\).

This general-form equation can always be rewritten in vertex form afterward if you need to see the vertex clearly — see converting from general to vertex form for that process.

Once you have a candidate function, test it against every labeled point on the graph, not just the ones you used to solve for \(a\), \(b\), or \(c\). If the graph gives you four or more marked points, one of them should be left over purely as a check — if your equation does not satisfy it, recheck your algebra before moving on.

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