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How does the formula -b over 2a give the x value of the vertex given ax2 bx c?
Wiki User
∙ 13y ago
From the quadratic formula, the zeros (where y=0) is -b/(2a) +/- sqrt(b^2 -4ac)/(2a). The expression inside the square root is known as the descriminant. Being that the parabola is symmetric, one of the zeros will be at the vertex + sqrt(b^2 -4ac)/(2a) and the other zero is at vertex - sqrt(b^2 -4ac)/(2a). The vertex is halfway between these two values, so the x coordinate of the vertex is -b/(2a).
Or if you use Calculus, you can take the derivative of f(x) = ax^2 + bx + c.
f'(x) = 2ax + b. Set equal to zero and solve for x. x = -b/(2a).
Wiki User
∙ 13y ago
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