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

Jun 23, 2016

$\left(\frac{3}{4} , \frac{1}{2}\right)$

#### Explanation:

Note that for any Real value of $x$:

${\left(4 x - 3\right)}^{2} \ge 0$

and is only equal to zero when:

$4 x - 3 = 0$

That is when $x = \frac{3}{4}$

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 $- \frac{1}{2} {\left(4 x - 3\right)}^{2} = 0$, leaving $y = \frac{1}{2}$

So the vertex of the parabola is $\left(\frac{3}{4} , \frac{1}{2}\right)$

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]}