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 = ax2 + 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 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 in Coordinates for the Point
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.
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Solve for a
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.
Convert to Standard Form
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 x2 - 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.
TL;DR (Too Long; Didn't Read)
Set either form to zero and solve the equation to find the points where the parabola crosses the x-axis.