How do you find the vertex and the intercepts for #y=2x^2 -4x +3#?

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Answer 1 · 2017-04-08T15:00:50

Vertex is #(1,1)# and intercept on #y#-axis is #3#. There is no intercept on #x#-axis.

Writing the equation #y=2x^2-4x+3# in vertex form

we have #y=2(x^2-2x+1-1)+3#

or #y=2((x-1)^2)+3-2#

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

Hence, vertex is #(1,1)#

As #(x-1)^2>=0#, #y>=1# and hence it does not touch #x#-axis (#y=0#) and hence we do not have any #x# intercept.

When #x=0#, #y=3# and hence #y# intercept is #3#.

graph{2x^2-4x+3 [-4.332, 5.67, -0.42, 4.58]}