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]}

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