How do you graph and label the vertex and axis of symmetry of #y=(x+2)(x+3)#?

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Answer 1 · 2018-02-12T01:38:12

Vertex: #(-5/2,-1/4)#
Axis of symmetry: #x=-5/2#

You first convert to vertex form:

#y=a(x-h)^2+k# with #(h,k)# being the vertex. To get to this, you have to complete the square.

First, expand:

#y=x^2+5x+6#

#y=(x^2+5x+(5/2)^2-(5/2)^2)+6#

#y=(x+5/2)^2+6-25/4#

#y=(x+5/2)^2+24/4-25/4#

#y=(x+color(red)(5/2))^2color(blue)(-1/4)#

Since the vertex is #(h,k)#, the vertex here is

#(color(red)(-5/2),color(blue)(-1/4))#

The axis of symmetry is just the x-coordinate of the vertex or #x=-b/(2a)# in #y=ax^2+bx+c#:

The axis of symmetry is:

#x=-5/2#