Vertex Form Calculator
Calculate vertex form instantly with our math tool. Shows detailed work, formulas used, and multiple solution methods. Free to use with no signup required.
Reviewed by Manoj Kumar, Mathematics Educator
Formula
y = a(x - h)^2 + k
Where (h, k) is the vertex of the parabola, a determines the direction and width, h = -b/(2a) gives the x-coordinate of the vertex from standard form, and k = c - b^2/(4a) gives the y-coordinate.
Worked Examples
Example 1: Standard Form to Vertex Form
Problem:Convert y = 2x^2 - 12x + 14 to vertex form and find the vertex, axis of symmetry, and x-intercepts.
Solution:a = 2, b = -12, c = 14\nh = -b/(2a) = -(-12)/(2*2) = 12/4 = 3\nk = c - b^2/(4a) = 14 - 144/8 = 14 - 18 = -4\nVertex form: y = 2(x - 3)^2 - 4\nVertex: (3, -4)\nAxis of symmetry: x = 3\nDiscriminant: 144 - 112 = 32 > 0, so two x-intercepts\nx = (12 +/- sqrt(32)) / 4 = (12 +/- 5.657) / 4\nx1 = 4.414, x2 = 1.586
Result:Vertex form: y = 2(x - 3)^2 - 4 | Vertex: (3, -4) | Two x-intercepts
Example 2: Vertex Form to Standard Form
Problem:Convert y = -0.5(x + 2)^2 + 8 to standard form. Find the maximum value and y-intercept.
Solution:a = -0.5, h = -2, k = 8\nExpand: y = -0.5(x^2 + 4x + 4) + 8\ny = -0.5x^2 - 2x - 2 + 8\ny = -0.5x^2 - 2x + 6\nMaximum value = k = 8 (at x = -2, since a < 0)\nY-intercept: y(0) = -0.5(0) - 2(0) + 6 = 6
Result:Standard form: y = -0.5x^2 - 2x + 6 | Maximum: 8 at x = -2 | Y-intercept: 6
Frequently Asked Questions
What is vertex form and how does it differ from standard form?
Vertex form of a quadratic equation is written as y = a(x - h)^2 + k, where (h, k) is the vertex of the parabola and a determines its width and direction. Standard form is written as y = ax^2 + bx + c, which directly shows the y-intercept but hides the vertex location. The key advantage of vertex form is that you can immediately read off the vertex coordinates without any calculation. Converting between the two forms involves completing the square, a technique where you manipulate the standard form to create a perfect square trinomial. Both forms represent the same parabola, just emphasizing different geometric properties.
How do you convert from standard form to vertex form?
To convert from standard form y = ax^2 + bx + c to vertex form, you use the completing the square method or the vertex formula. The quickest approach uses h = -b/(2a) to find the x-coordinate of the vertex, then substitutes h back into the original equation to find k = c - b^2/(4a). For example, converting y = 2x^2 - 8x + 5: first find h = -(-8)/(2*2) = 2, then k = 5 - (-8)^2/(4*2) = 5 - 8 = -3, giving vertex form y = 2(x - 2)^2 - 3. The completing the square method factors out a from the first two terms, adds and subtracts the appropriate constant to form a perfect square, then simplifies.
How do you find the x-intercepts from vertex form?
To find x-intercepts from vertex form y = a(x - h)^2 + k, set y = 0 and solve for x. This gives a(x - h)^2 = -k, then (x - h)^2 = -k/a, and finally x = h plus or minus the square root of (-k/a). For real x-intercepts to exist, -k/a must be non-negative. If -k/a > 0, there are two distinct x-intercepts. If -k/a = 0, the vertex itself is the only x-intercept (the parabola just touches the x-axis). If -k/a < 0, there are no real x-intercepts, meaning the parabola floats entirely above or below the x-axis. This connects to the discriminant: b^2 - 4ac determines the same conditions.
What is the axis of symmetry and how is it related to the vertex?
The axis of symmetry is a vertical line that passes through the vertex of the parabola, dividing it into two mirror-image halves. Its equation is simply x = h, where h is the x-coordinate of the vertex. Every point on the parabola has a corresponding mirror point on the opposite side of this line at the same height. In standard form, the axis of symmetry is x = -b/(2a), which is also the formula for finding the vertex x-coordinate. This symmetry property is extremely useful for graphing: once you know a few points on one side of the axis, you can reflect them to find points on the other side. The axis of symmetry always passes through both the vertex and the focus of the parabola.
References
Reviewed by Manoj Kumar, Mathematics Educator · Editorial policy