Equation of inverted, parametric parabola

Question

I want to find the equation of an inverted parabola such that its vertex is $(1/2, 2)$, its y-intercept is $(0, 1)$ and right end point is $(1, 1))$. I'm working with parametrics so this curve is defining $x$ as a function of $t$, where $0\leq t\leq 1$

First I tried to generalize such an equation with some real numbers $a, k$. $$x(t)=-a(t-k)^2+2$$ To find $k$ I searched for the critical point, which I see by inspection is at $t=1/2$, \begin{align*}\frac{dx}{dt}&=-2a(1/2-k) \\ 0&= 2ak-a/2 \\ k&= \frac14\end{align*} I found $a$ by using the fact that $x(0)=1$, \begin{align*} 0&=-a(t-1/4)^2+2 \\ 1-2&= -a/16 \\ a&=16\end{align*}

Which gave me $x(t)=-16(t-1/4)^2+2$. But actually graphing this almost gets me the curve I was looking for. I get an inverted parabola that indeed passes through $(0, 1)$, but its vertex is on $(1/4, 2)$ with right endpoint $(1/2, 1)$.

I think both my values for $a$ and $k$ must be wrong.

Answer 1

$$\frac {dx}{dt}=-2a\left(\frac 12-k\right)=0$$ means $k=\frac 12\neq \frac 14$. The incorrect value of $k$ results in an incorrect value for $a$ as well.

Answer 2

$$x(t)=a(t-p)^2+q$$

where $(p,q)$ iş turning point

$$x(t)=a(t-\frac{1}{2})^2+2$$

$$1=a(0-\frac{1}{2})^2+2$$

$$a=-4$$

$$x(t)=-4(t-\frac{1}{2})^2+2$$