In the real world, parabolas describe the path of any thrown, kicked or fired object. They're also the shape used for satellite dishes, reflectors and the like, because they concentrate all rays that enter them into a single point inside the bell of the parabola, called the focus. In mathematical terms, a parabola is expressed by the equation f(x) = ax^2 + bx + c. Finding the midpoint between the parabola's two x-intercepts gives you the x-coordinate of the vertex, which you can then substitute into the equation to find the y-coordinate as well.

Use basic algebra to write the parabola's equation in the form f(x) = ax^2 + bx + c, if it is not in that form already.

Identify which numbers are represented by a, b and c in the parabola's equation. If b and c aren't present in the equation, it means they are equal to zero. The number represented by a, however, will never be equal to zero. For example, if your parabola's equation is f(x) = 2x^2 + 8x, then a = 2, b = 8 and c = 0.

To find the midpoint between the parabola's two x-intercepts, calculate -b/2a, or negative b divided by twice the value of a. This gives you the x-coordinate of the vertex. To continue the example above, the x-coordinate of the vertex would be -8/4, or -2.

Find the y-coordinate of the vertex by substituting the x-coordinate back into the original equation, then solving for f(x). Substituting x = -2 into the example equation would look like this: f(x) = 2(-2)^2 + 8(-2) = 2(-4) - 16 = 8 - 16 = -8. The solution, -8, is the y-coordinate. So the coordinates of the vertex for the example parabola are (-2, -8).

#### Tip

If you can put the parabola's equation into the form f(x) = a(x - h)^2 +k, also known as the vertex form, the numbers that take the place of h and k are the x- and y-coordinates, respectively, of the vertex. Keep in mind that if k is absent when the equation is in this format, k = 0. So if the equation is just f(x) = 2(x - 5)^2, the vertex coordinates are (5, 0). If the equation in vertex form is f(x) = 2(x - 5)^2 + 2, the coordinates of the vertex would be (5, 2).

#### Warning

Pay close attention to negative signs when dealing with the x^2 term of the equation. Remember that when you square a negative number, the result is positive -- so x^2 on its own will always be positive. However the coefficient "a" may be positive or negative, so the ax^2 term as a whole may be either positive or negative.