How do find the vertex and axis of symmetry for a quadratic equation #y =x^2+2x-3#?

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Answer 1 · 2015-06-18T22:52:31

Rearrange the equation into vertex form by completing the square to get:

#y = (x-(-1))^2 + (-4)#

Hence, the vertex is at #(-1, -4)#

and the axis of symmetry is #x=-1#

Vertex form of the equation of a parabola with vertical axis is:

#y = a(x-x_0)^2 + y_0#

Starting with #y=x^2+2x-3#, rearrange as follows:

#y = x^2+2x-3#

#= x^2+2x+1-1-3#

#= (x+1)^2-4#

#= 1*(x-(-1))^2+(-4)#

This is vertex form, with #a=1#, #x_0=-1# and #y_0=-4#

So the vertex is at #(-1, -4)# and the axis of symmetry is #x=-1#