find the vertex of the parabola and give its direction of opening.
x = 1/2y^2 + 3y-6
Completing the square is the same method as turning pt form.
$\displaystyle x = \frac{1}{2}y^2 + 3y-6$
First step is to tkae out half as a common factor
$\displaystyle x = \frac{1}{2}(y^2 + 6y-12)$
Now we need to change the constant term from inside the brackets so we can form a perfect square. To do this we take the co-efficient of the middle term, half it and square it. This gives +9 giving
$\displaystyle x = \frac{1}{2}((y^2 + 6y+9)-9-12)$
$\displaystyle x = \frac{1}{2}((y^2 + 6y+9)-21)$
Then making the perfect square complete.
$\displaystyle x = \frac{1}{2}((y+3)^2 -21)$
Can you take it from here?