Complete the square of y^2 + 10y to answer this question.
x=y^2 + 10y + 22 becomes x = y^2 + 10y + 25 - 25 + 22, or
x = (y+5)^2 - 3
Comparing this result to the generic form x = a(y-k)^2 + h, we see that
h = -5 and k = -3.
Then the vertex of this parabola is at (-5, -3).
