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

edit

Try using implicit_plot so you don't have to solve the equation for y in terms of x.

f = lambda x,y:y^2-3*x-5*y+7
p1=implicit_plot(f, (-4, 4), (-2, 6))
p2= point((1/4, 5/2), size = 45, color='red')
p0=p1+p2
show(p0)

This will do it... you can add your own features.

Just as a note here, you don't need to use a lambda function here to get implicit plots to work. You will need to define y as a variable first, though. So, the code would be:

var('y')
f = y^2-3*x-5*y+7
Yax = x
Xax = y
SYMAXIS = y-5/2
p1=implicit_plot(f, (x,-4, 4), (y,-2, 6))
p2= point((1/4, 5/2), size = 45, color='red')
p3=implicit_plot(Yax,  (x,-4, 4), (y,-2, 6),color='black')
p4=implicit_plot(Xax,  (x,-4, 4), (y,-2, 6),color='black')
p5=implicit_plot(SYMAXIS,  (x,-4, 4), (y,-2, 6),color='green')
p0=p1+p2+p3+p4+p5 
show(p0)

This definition of your function f will also allow you to do the substitutions that you want.

Here I've displayed the X and Y axes as well as the axis of symmetry. I'm not sure what you are trying to do using "substitution".

f = lambda x,y:y^2-3*x-5*y+7
Yax = lambda x,y:x
Xax = lambda x,y:y
SYMAXIS = lambda x,y:y-5/2
p1=implicit_plot(f, (-4, 4), (-2, 6))
p2= point((1/4, 5/2), size = 45, color='red')
p3=implicit_plot(Yax, (-4, 4), (-2, 6),color='black')
p4=implicit_plot(Xax, (-4, 4), (-2, 6),color='black')
p5=implicit_plot(SYMAXIS, (-4, 4), (-2, 6),color='green')
p0=p1+p2+p3+p4+p5 
show(p0)