SOLUTION: 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?

First we plot those two points to see if it's possible:

 

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:

 

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.

 

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