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### Graph based on y value as the input, and x as the output?

I need to identify the vertex of the following parabola: 3x-7 = y^2-5y

According to the book's author, since this contains a y^2 term rather than an x^2 term; I must obtain an end result with the standard form x = a(y - k)^2 + h

The answer, i.e. vertex, is (1/4, 5/2)

I did the following in sage 5.9, and as you can see the vertex point isn't located at the parabola, since the x in the 1st statement's equation, should be y. But sage says y is undefined:

p1 = plot( (1/3)*((x - 5/2)^2) + 1/4, (x,-15, 15), ymin =-5 , ymax = 25)
p3 = point((1/4, 5/2), size = 25)
p0 = p1 + p3
show(p0)


Is it possible for Sage to graph the equation based f(y), as opposed to the standard f(x), so i can see the vertex match up with the parabola.