Parametric equation question showing minimum value of d^2 for the equation $d^2 = (1-a)^2t^2 + 18(1-a)t +117$
Show that when $a = 2$, the minimum value of $d^2$ is attained when
$t=9$.
I set $a=2$ to get $d^2 = t^2 - 18t + 117$
should i now just run it with $t = 9$ ?
this gets an answer of $d^2 = 36$ $d=6$  
Is this correct?
 A: Since they ask you for the minimum value of $d^2$, and you have a quadratic equation for $d^2(t)=t^2-18t+117$ you have to calculate the derivative of $d^2$ with respect to $t$. That is, $\frac{d(d^2)}{dt}=2t-18$ and find the value of $t$ for which $\frac{d(d^2)}{dt}=2t-18=0$. Of course, this implies $t=9$. 
Now, if you are not into derivatives, since the equation for $d^2$ is a parabola, and since the coefficient multiplying the $t^2$ term is $1$, positive, you therefore have a "happy face" parabola. And you can easily calculate de vertex of the parabola by setting $t_v=\frac{-b}{2a}$ and $y_v = d^2(x_v)$ where $t_v,y_v$ are the coordinates of the vertex, $-b=18$, $a=1$. You can easily see again that $t_v$ which is the value for which $d^2$ is minimum is, $t_v=9$
A: We need to show that $d^2$ is a minimum when $t=9$.  To that end we have
$$\begin{align}
d^2&=t^2-18t+117\\\\
&=(t-9)^2+36\\\\
&\le 36
\end{align}$$
for which it is easy to see that the minimum is 

$$d^2=36\,\,\,\text{when}\,\,t=9$$


Alternatively, we can take the derivative and set it equal to zero to find the critical point.  Proceeding, we have
$$\frac{d(d^2)}{dt}=2t-18=0\implies t=9$$
And we recover the expected result!
