How do you find the axis of symmetry and vertex point of the function: #f(x)=2x^2+x-3#?

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Answer 1 · 2015-10-22T13:09:31

See explanation.

Firstly, you change this equation in the form of #a(x-h)^2 +k# which is also called as complete square form
So,

taking common the coefficient of #x^2#

#2(x^2 + x/2 -3/2)#

#2[x^2 + 2*x/(4) + (1/4)^2 - (3/2) - (1/4)^2]#

#2[(x^2 + 1/4) ^2 -(1/16 + 3/2)]#

#2[(x^2 + 1/4) ^2 -25/16]#

Opening the bracket and multiplying by 2.

#2(x^2 + 1/4) ^2 - 2*25/16#

#2(x^2 + 1/4) ^2 - 25/8# is in the form of #a(x-h)^2 +k#

where #a=2# , #h= -1/4# and #k = -25/8#

The vertex is given by #(h,k)#

So the vertex is #( -1/4,-25/8)#

and the line of symmetry is given by #(x-h)=0#
So

#x-(-1/4) =0#

#x+1/4=0 implies x = -1/4#