Question 807833
<pre>
Let's plot the vertex and the y-intercept and see:

{{{drawing(5600/23,800,-2,5,-20,3,
circle(2,-5,.05), locate(2,-4.5,"(2,-5)"),circle(2,-5,.1),
circle(0,-17,.05), locate(.2,-16.5,"(0,-17)"),circle(0,-17,.15),
circle(0,-17,.1),circle(0,-17,.125),circle(0,-17,.07),

graph(5600/23,800,-2,5,-20,3) )}}}

Now the upper point has to be the vertex, and common sense 
tells you that the only way that the y-intercept can be way down
there at -17, is for the parabola to open downward, like this:  

{{{drawing(5600/23,800,-2,5,-20,3,
circle(2,-5,.05), locate(2,-4.5,"(2,-5)"),circle(2,-5,.1),
circle(0,-17,.05), locate(.2,-16.5,"(0,-17)"),circle(0,-17,.15),
circle(0,-17,.1),circle(0,-17,.125),circle(0,-17,.07),

graph(5600/23,800,-2,5,-20,3,-3(x-2)^2-5) )}}}

So this problem doesn't require any algebra knowledge, just
common sense.

Edwin</pre>