Simple Ways to Find the Vertex Form

Standard Form Translation

• The standard form of the parabolic equation is y = ax^2 + bx + c. The vertex form is y = a(x-h)^2 + k. The values (h, k) are the x- and y-values of the vertex point. Let's say that your standard form equation is y = 3x^2 + 2x + 5. To find the x-value of the vertex, divide "b" by "2a" and then switch the sign. Here, that would result in -2/3. Plug the "x" value into the equation. y = 3(2/3)(2/3) + 2(2/3) + 5. The answer is 12/9 + 12/9 + 5, or 5 8/3, or 22/3. The vertex form would then be y = 3(x-(-2/3)) + 22/3 which can be rewritten as y = 3(x + 2/3) + 22/3.

Changing Problems to Proper Form

• Let's consider the problem y = -3x^2 + 5. Here, h = 0, a = -3 and k = 5. So, to write it in proper format, it would look like this: y = -3(x-0)^2) + 5.

Moving from Graph to Equation

• Having the graph of a parabola gives us enough information to write the formula in vertex form. Let's say we have a parabola that crosses the x-axis at x = -2 and at x = 2. It crosses the y-axis at y = 3 as well, by definition. So, y = a(x+2)(x-2), using the x-intercepts. To find a, we will also add in 0,3 as the y-intercept.

  3 = a (0+2)(0-2).

  3 = -4a.

  a = -4/3.

  y = -4/3(x+2)(x-2).

  y = -4/3(x^2-4).

  y = -4/3x^2 + 0x + 16/3 in standard form.

  h = -b/2a, or 0/-8/3, or 0.

  k = -4/3(0-2)(0+2).

  k = -4/3(-4) or 16/3.

  The vertex is at 0, 16/3.

  Vertex form: y = -4/3(x-0)^2 + 16/3

Completing the Square

• Let's take the standard form equation y = x^2 + 6x + 7. Then: y = (x^2 + 6x) + 7

  Complete the square inside the parentheses.

  y = (x^2 + 6x + 9) - 9 + 7.

  If you're going to add 9, you have to subract it, too, unless you want a 9 over there with the y, and we don't.

  So, it simplifies to: y = (x+3)^2 - 2.

  Here's the vertex form: a = 1, h = 3, and k = -2. Your vertex would be at (3, -2).