Parabolas are defined by quadratic equations involving an x^2 term, an optional linear term and an optional constant term. As such, they have a "U" shape and, depending on the coefficients of the quadratic, linear and constant terms, a parabola will either intersect the xaxis in 0, 1 or 2 places. You will first determine if the parabola intercepts the xaxis at all and, if so, all the points of intersection.

Write the equation of the parabola in the form y = ax^2 + bx + c. For example, rewrite "y = 2 + x^2  3x" as "y = x^2  3x + 2" to get a = 1, b = 3 and c = 2. If your parabola is given as the product of two linear terms, "(ax + b)(cx + d)," your intercepts are already found for you: x = b/a and x = d/c.

Subtract the product of 4, a and c from b squared. This quantity "b^2  4ac" is known as the determinant of the parabola. If it is negative, the parabola does not intersect the xaxis at all. If it is zero, the parabola touches the xaxis only once. If it is positive, the parabola will have two intersections. For example, y = 4x^2 + 3 has no intersections because its determinant, 0^2  4 4 3 = 48, is less than 0. On the other hand, y = x^2 has exactly one intersection because its determinant, 0^2  4 1 0 = 0, is exactly 0.

Divide b by the product of 2 and a to get the one xaxis intersection if the determinant is 0. For example, y = x^2  2x + 1 has determinant 0, and has x intercept of x = (2) / (2 * 1) = 1.

Take the square root of the determinant you found in Step 2, and divide it by 2a, calling this quantity "d." Divide b by 2a and add d to get the first x intercept, and divide b by 2a and subtract d to get the second intercept. For example, y = 2x^2  5x + 2, d = sqrt(b^2  4 a c) = sqrt(25  16) = sqrt(9) = 3. Its intercepts are then (d  b) / 2a = (3 + 5) / 4 = 2 and (d + b) / 2a = (3  5) / 4 = 0.5.
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