Write Quadratic Equations from Parabola Graphs: Vertex Form to Standard Form

Algebra and Function Practice

A parabola graph gives two powerful clues: the vertex and any other labeled point. Use vertex form y = a(x - h)^2 + k, substitute the vertex (h, k), then use the extra point to solve for a. Finally expand to standard form y = ax^2 + bx + c. This lesson keeps arithmetic simple and shows how to verify each equation against the graph.

Practice materials: Printable downloads are not available for this lesson yet. Use the instructions and four exercises below.

Read the vertex and choose vertex form

Every parabola has a vertex, the highest or lowest point. On a coordinate grid, read its coordinates carefully and note whether the parabola opens upward or downward. Vertex form is y = a(x - h)^2 + k, where (h, k) is the vertex. The sign inside the parentheses is subtracted, so a vertex at (3, -2) becomes (x - 3)^2 - 2.

The value of a controls width and direction. If the graph opens upward, a is positive; if downward, a is negative. You cannot determine a from the vertex alone, which is why one additional labeled point is needed.

1. Identify the vertex (h, k) from the graph.
2. Write y = a(x - h)^2 + k with h and k substituted.
3. Check the opening direction to predict the sign of a.
4. Keep a as the unknown until the extra point is used.

Use the extra point to solve for a

Substitute the coordinates of the additional labeled point into the vertex-form equation. This produces a simple equation in a. Solve it by isolating a, often by dividing after subtracting k. Keep fractions exact unless the graph clearly supports a decimal.

For example, if the vertex is (1, 2) and the point (3, 10) lies on the graph, substitute: 10 = a(3 - 1)^2 + 2. Then 10 = 4a + 2, so 4a = 8 and a = 2. The vertex form is y = 2(x - 1)^2 + 2.

1. Substitute the extra point's x and y into the vertex-form equation.
2. Simplify the squared term.
3. Subtract k from both sides.
4. Divide by the squared term to find a.
5. Write the final vertex form with the found a.

Expand to standard form and verify

Expand the squared binomial and distribute a. For y = 2(x - 1)^2 + 2, expand (x - 1)^2 = x^2 - 2x + 1, multiply by 2 to get 2x^2 - 4x + 2, then add 2: y = 2x^2 - 4x + 4. Standard form is y = ax^2 + bx + c.

Verify by substituting the vertex and the extra point into the standard form. Both should satisfy the equation. Also confirm the sign of a matches the opening direction. If a point fails, recheck the sign inside the parentheses and the arithmetic.

1. Expand (x - h)^2.
2. Distribute a across all terms.
3. Add k to the constant term.
4. Substitute the vertex and extra point to verify.
5. Confirm the opening direction matches the sign of a.

Worked example

A parabola has vertex (-2, 3) and passes through (0, 7). Write its equation in vertex form and standard form.

Four practice exercises

Common mistakes

• Using (x + h) instead of (x - h) in vertex form. Vertex form subtracts h. For vertex (4, 1), write (x - 4)^2 + 1, not (x + 4)^2 + 1.
• Forgetting to subtract k before dividing to find a. Substitute the point, then isolate the squared term first: y - k = a(x - h)^2, then divide.
• Assuming a = 1 because the graph looks normal. Always solve for a using the extra point; width and direction depend on a.
• Expanding incorrectly by squaring each term separately. Expand (x - h)^2 as x^2 - 2hx + h^2, not x^2 - h^2.

Self-check rubric

• Vertex coordinates correctly identified and substituted with the correct sign.
• Extra point substituted correctly into vertex form.
• Value of a solved accurately, with sign matching the opening direction.
• Vertex form written clearly with parentheses and correct operations.
• Expansion to standard form is algebraically correct.
• Verification shows both the vertex and extra point satisfy the final equation.

Questions and answers