Answer
See below
Work Step by Step
Given: $y^2-10y-8x+1=0$
Because $A=0,B=0$, and $C=1$, the discriminant is $B^2-4AC=0$, so the conic is a parabola. Complete the square to write the equation in standard form.
$y^2-10y-8x+1=0\\y^2-10y+25-25-8x+1=0\\(y-5)^2=8(x+3)$
From the equation, we can see the vertex is at $(-3,5)$. The graph is shown below.
