Question 533476
What is the equation of a parabola that passes through the point at (3,-1), has its vertex at (2,-3) and opens to the left?
<pre>
First we plot those two points to see if it's possible:

{{{drawing(400,400,-10,10,-10,10, graph(400,400,-10,10,-10,10),

circle(2,-3,.1), locate(2+.5,-3+.5,"(2,-3)"), 
circle(3,-1,.1), locate(3+.5,-1,"(3,-1)")  )}}} 

No it's not possible, for if it opens to the left, and have its
vertex at (2,-3) it would have to look something like this:

{{{drawing(400,400,-10,10,-10,10, graph(400,400,-10,10,-10,10,-3+2sqrt(4-2x)),
graph(400,400,-10,10,-10,10,-3-2sqrt(4-2x)),

circle(2,-3,.1), locate(2+.5,-3+.5,"(2,-3)"), 
circle(3,-1,.1), locate(3+.5,-1,"(3,-1)")  )}}} 

But as you see it can't possibly go through (3,-1), for that's 
further to the right than the vertex, and the vertex has to be
the rightmost point if it opens to the left.-

Did you make a mistake and copy a sign wrong?

Or maybe the parabola was supposed to open to the right like this.

{{{drawing(400,400,-10,10,-10,10, graph(400,400,-10,10,-10,10,sqrt(4x-8)-3),
graph(400,400,-10,10,-10,10,-sqrt(4x-8)-3),

circle(2,-3,.1), locate(2+.5,-3+.5,"(2,-3)"), 
circle(3,-1,.1), locate(3+.5,-1,"(3,-1)")  )}}} 

So it would be possible for the parabola to open to the right
but not the left:

So I'll show you how to get the equation of this one:

Use the form

(y - k)² = 4p(x - h)

where the vertex is (h,k).  So we substitute (h,k) = (2,-3)

(y - (-3))² = 4p(x - 2)

(y + 3)² = 4p(x - 2)

Now we substitute (x,y) = (2,-3)

(-1 + 3)² = 4p(3 - 2)

       2² = 4p(1)

        4 = 4p

        1 = p

And substitute that in

(y + 3)² = 4p(x - 2)
 
(y + 3)² = 4(1)(x - 2)

(y + 3)² = 4(x - 2)

That's the equation of the parabola with vertex
at (2,-3) that goes through (3,-1) but that opens to
the right.  

Edwin</pre>