Question 894040
Do you want a parabola?  Should it open upward, downward, left, or right?  


As a parabola, it could be parallel to the y-axis with a maximum for the vertex.  This would mean that for standard {{{y=a(x-h)^2+k}}}, {{{a<0}}}.


Solve for a.
{{{a(x-h)^2=y-k}}}
{{{a=(y-k)/(x-h)^2}}};
Using the vertex coordinates,
{{{a=(y-5)/(x-1)^2}}}
Using also the given point on the parabola (as the shape I chose),
{{{a=(1-5)/(2-1)^2}}}
{{{a=-4/1^2}}}
{{{a=-4}}}
-
The equation can be {{{highlight(y=-4(x-1)^2+5)}}}.
You can use y=0 and find the x-axis intercepts.


{{{graph(300,300,-6,6,-6,6,-4(x-1)^2+5)}}}