What is the vertex, axis of symmetry, the maximum or minimum value, and the range of parabola #y=4x^2-2x+2#?

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Answer 1 · 2015-05-17T17:12:34

Vertex #(1/4, 7/4)# Axis of symmetry x= #1/4#, Min 7/4, Max #oo#

Re arrange the equation as follows

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

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

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

=#4(x-1/4)^2# +7/4

The vertex is #(1/4,7/4)# Axis of symmetry is x=#1/4#

Minimum value is y=7/4 and maximum is #oo#

Answer 2 · 2015-05-17T17:54:54

In the general case, the coordinates of the vertex for a function of the 2nd degree #a x^2 + b x + c# are the following:

#x_v# #=# #-b / (2 a)#

#y_v# #=# #- Delta / (4a)#

(where #Delta# #=# #b^2 - 4 a c#)

In our particular case, the vertex will have the following coordinates:

#x_v# #=# #- (-2) / (2 * 4)# #=# #1 / 4#

#y_v# #=# #- ((-2)^2 - 4 * 4 * 2) / (4 * 4)# #=# #7 / 4#

The vertex is the point #V (1/4, 7/4)#

We can see that the function has a minimum, that is #y_v# #=# #7 / 4#

The axis of symmetry is a parallel line to the #Oy# axis passing through the vertex #V (1/4. 7/4)#, i.e. the constant function #y# #=# #1/4#

As #y# #>=# #7/4#, the range of our function is the interval #[7/4, oo) #.