What is the vertex of # y = -1/2(4x - 3)^2 + 1/2 #?

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Answer 1 · 2016-06-23T06:24:24

#(3/4, 1/2)#

Note that for any Real value of #x#:

#(4x-3)^2 >= 0#

and is only equal to zero when:

#4x-3 = 0#

That is when #x = 3/4#

So this is the #x# value of the vertex of the parabola.

Substituting this value of #x# into the equation will make the first expression #-1/2(4x-3)^2 = 0#, leaving #y = 1/2#

So the vertex of the parabola is #(3/4, 1/2)#

graph{(y-(-1/2(4x-3)^2+1/2))((x-3/4)^2+(y-1/2)^2-0.001) = 0 [-2.063, 2.937, -1.07, 1.43]}