Basically, you need the #x#-intercepts (if any), the #y#-intercepts, and the vertex point. Knowing the concavity helps as well.
To get the #x#-intercept, let #y=0#.
# 0=x^2-8x+15#
#(x-5)(x-3)=0#
So #x# can be #5# or #3# #-># the #x#-intercepts
For the #y#-intercept, I just let #x=0#. This gives the constant, so the #y#-intercept is at the point #(0,15)#.
For the vertex point, I can use #-b/(2a)#, so
#-b/(2a) = 8/2 = 4#
I'm not finished, I need to sub #x=4# back in.
#y=-1# when #x# is #4#.
Now I have the vertex #(4,-1)#, the #x#-intercepts #(3, 0)# and #(5, 0)#, the #y#-intercept #(0,15)#, and the concavity (concave up as it is a positive coefficient of #x^2#).
Now I graph.
graph{x^2-8x+15 [-10, 10, -5, 5]}
