How to Find X Intercepts of a Parabola

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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 x-axis in 0, 1 or 2 places. You will first determine if the parabola intercepts the x-axis 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 x-axis at all. If it is zero, the parabola touches the x-axis 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 x-axis 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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