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A parabola: symmetry, intercepts and coefficients

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      Functions · Coordinate geometry

      Question

      The parabola y = ax² + bx + 5 intersects the x-axis at A and B, with AB = 4, and the y-axis at C. Its axis of symmetry x = 3 meets the line y = kx − 1 through A at D. Find the equations of AD and the parabola. This lesson covers part (1) of the original problem.

      Parabola and line in the coordinate planeThe axis of symmetry is x = 3. Intercepts are A(1,0), B(5,0) and C(0,5). Line AD passes through A(1,0) and D(3,2).OxyAxis x = 3C(0, 5)A(1, 0)B(5, 0)D(3, 2)
      The axis x = 3 bisects AB. Since AB = 4, A = (1, 0), B = (5, 0), and C = (0, 5).

      Step-by-step solution

      1. Use symmetry to locate A and B

      Each intercept is 4 ÷ 2 = 2 units from the axis x = 3. Thus A = (3 − 2, 0) = (1, 0) and B = (3 + 2, 0) = (5, 0).

      2. Find the equation of AD

      Substitute A = (1, 0) into y = kx − 1: 0 = k − 1, so k = 1. Hence AD has equation y = x − 1.

      3. Write the parabola in intercept form

      Its x-intercepts are 1 and 5, so write y = a(x − 1)(x − 5). The constant term of the original equation gives C = (0, 5).

      4. Determine a and expand

      Substitute C: 5 = a(0 − 1)(0 − 5) = 5a, hence a = 1. Expanding gives y = (x − 1)(x − 5) = x² − 6x + 5.

      The equations are AD: y = x − 1; parabola: y = x² − 6x + 5.

      Which method does this problem use?

      Use symmetry to locate the intercepts. Once both roots are known, intercept form leaves only one coefficient to determine.