What is the axis of symmetry and vertex for the graph #y=2x^2+6x+4#?

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Answer 1 · 2017-02-26T06:23:30

Vertex is #(-1/2,-3/2)# and axis of symmetry is #x+3/2=0#

Let us convert the function to vertex form i.e. #y=a(x-h)^2+k#,

which gives vertex as #(h,k)# and axis of symmetry as #x=h#

As #y=2x^2+6x+4#, we first take out #2# and make complete square for #x#.

#y=2x^2+6x+4#

= #2(x^2+3x)+4#

= #2(x^2+2xx3/2xx x+(3/2)^2)-(3/2)^2xx2+4#

= #2(x+3/2)^2-9/2+4#

= #2(x-(-3/2))^2-1/2#

Hence, vertex is #(-1/2,-3/2)# and axis of symmetry is #x+3/2=0#
graph{2x^2+6x+4 [-7.08, 2.92, -1.58, 3.42]}