Completing the Square Calculator

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The leading coefficient. Cannot be zero - that would make the expression linear, not quadratic.
Vertex form rewrites the quadratic; solving for roots sets it equal to zero and finds x.
When the discriminant is negative, roots involve the imaginary unit i. Turn this on to see them.
Vertex formTwo real roots
(x - 3)2 - 4
Vertex (h, k)(3, -4)
Vertex x (h)3
Vertex y (k)-4
Discriminant (b2 - 4ac)16
Root typeTwo distinct real roots
Root x15
Root x21
Axis of symmetryx = 3

The parabola opens upward, with a minimum value of -4 at x = 3.

  • The vertex sits at (3, -4); the axis of symmetry is x = 3.
  • Because a = 1 is positive, k = -4 is the minimum value the function can reach.
  • Discriminant = 16 > 0, so the parabola crosses the x-axis at two distinct points: x1 = 5, x2 = 1.

Next stepUse the roots to factor the quadratic as a(x - x1)(x - x2) for a complete algebraic picture.

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