Just as a quadratic equation can map a parabola, the parabola's points can help write a corresponding quadratic equation. Parabolas have two equation forms – standard and vertex. In the vertex form, *y* = *a*(*x* - *h*)^{2} + *k*, the variables *h* and *k* are the coordinates of the parabola's vertex. In the standard form, y = *ax*^{2} + *bx* + *c*, a parabolic equation resembles a classic quadratic equation. With just two of the parabola's points, its vertex and one other, you can find a parabolic equation's vertex and standard forms and write the parabola algebraically.

## Substitute in Coordinates for the Vertex

## Substitute in Coordinates for the Point

## Solve for a

## Substitute a

## Convert to Standard Form

Set either form to zero and solve the equation to find the points where the parabola crosses the x-axis.

Substitute the vertex's coordinates for *h* and *k* in the vertex form. For an example, let the vertex be (2, 3). Substituting 2 for *h* and 3 for *k* into y = *a*(*x* - *h*)^{2} + *k* results in *y* = *a*(*x* - 2)^{2} + 3.

Substitute the point's coordinates for *x* and *y* in the equation. In this example, let the point be (3, 8). Substituting 3 for *x* and 8 for *y* in *y* = *a*(*x* - 2)^{2} + 3 results in 8 = *a*(3 - 2)^{2} + 3 or 8 = *a*(1)^{2} + 3, which is 8 = *a* + 3.

Solve the equation for *a*. In this example, solving for *a* results in 8 - 3 = *a* - 3, which becomes *a* = 5.

Substitute the value of *a* into the equation from Step 1. In this example, substituting *a* into *y* = *a*(*x* - 2)^{2} + 3 results in *y* = 5(*x* - 2)^{2} + 3.

Square the expression inside the parentheses, multiply the terms by *a*'s value and combine like terms to convert the equation to standard form. Concluding this example, squaring (*x* - 2) results in *x*^{2} - 4_x_ + 4, which multiplied by 5 results in 5_x_^{2} - 20_x_ + 20. The equation now reads as *y* = 5_x_^{2} - 20_x_ + 20 + 3, which becomes *y* = 5_x_^{2} - 20_x_ + 23 after combining like terms.