How do find the vertex and axis of symmetry, and intercepts for a quadratic equation ?

1 Answer
Jun 14, 2015

Complete the square using:

ax^2+bx+c = a(x+b/(2a))^2 + (c - b^2/(4a))

From this the vertex, axis of symmetry, etc is easy to derive.

Explanation:

Given quadratic equation ax^2+bx+c = 0

We can complete the square using the equation:

ax^2+bx+c = a(x+b/(2a))^2 + (c - b^2/(4a))

Then the vertex is at (-b/(2a), c - b^2/(4a))

Axis of symmetry: x=-b/(2a)

y intercept (0, c)

x intercepts ((-b +- sqrt(b^2-4ac))/(2a), 0)