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

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Answer 1 · 2015-06-14T21:54:01

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.

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)#