Perfect squares always one root? I had an exam today and I was thinking about this task now, after the exam of course. 
$f(x)=a(x-b)^2 +c$  
Now, the point was to find C so that the function only has one root. Easy enough, I played with the calculator and found this. But I hate explanations like that, yes. You get a few points but far from full score. But overall I should still get an A, I hope. If $C=0$ then the expression is a perfect square and they only have one root? Is that far of?  
$a(x-b)^2= - c$  
$\frac{a(x-b)^2}{a}= - \frac{c}{a}$  
$(x-b)^2= - \frac{c}{a}$  
This also argues that c should be 0 for it to only be one root? 

An alternative way to think about it is geometrically. The graph of $y=x^2$ is a parabola that opens up with vertex at the origin. The graph of
$$y = (x-b)^2$$
is then a horizontal shift by $b$ units (so $b$ units to the right if $b\geq 0$, and $|b|$ units to the left if $b\lt 0$) of the same graph. There is still only one root: the vertex.
If $a\neq 0$, then 
$$y = a(x-b)^2$$
is a vertical stretch of this graph, possibly with a flip (if $a\lt 0$); it does not change the number of intersections with the $x$-axis.
Finally, $$y=a(x-b)^2 + c$$
is a vertical shift by $c$ units (up if $c\gt 0$, down if $c\lt 0$). 
If $y=a(x-b)^2$ is a parabola that opens "up" (if $a\gt 0$), then shifting it up ($c\gt 0$) will remove all intersections with the $x$-axis; and shifting it down ($c\lt 0$) will create two intersections with the $x$-axis as the vertex moves down. 
If $y=a(x-b)^2$ is a parabola that opens "down" (if $a\lt 0$), then the situation is reversed: $c\gt 0$ will create two intersections with the $x$-axis, and $c\lt 0$ will remove all intersections with the $x$-axis.
Either way, in order to maintain one and only one intersection, you need the vertex of the parabola to stay on the $x$-axis, so you need $c=0$.  Conversely, if $c=0$, you have a parabola with vertex on the $x$-axis, hence with a single intersection. 
