Axis of symmetry of y = x^2 + 3x — Algebra (find x = -3/2)

Step-by-step solution to find the axis of symmetry for the parabola y = x^2 + 3x. Uses x = -b/(2a) and completing the square to show the axis is x = -3/2.

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For a parabola y = ax^2 + bx + c, the axis of symmetry is given by x = -b/(2a).

  1. Identify coefficients: a = 1, b = 3.
  2. Compute x = -b/(2a) = -3/(2·1) = -3/2.

So the equation of the axis of symmetry is x = -3/2.

Check by completing the square:
y = x^2 + 3x = (x^2 + 3x + 9/4) - 9/4 = (x + 3/2)^2 - 9/4.
The vertex is at (-3/2, -9/4), confirming the axis is x = -3/2.

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