# Thread: find vertex of parabola

1. ## find vertex of parabola

find the vertex of the parabola and give its direction of opening.

x = 1/2y^2 + 3y-6

2. Two ways to find the vertex that I can think of are:

Putting the eqn into turning pt form

or

Finding halfway between the x-intercepts if they exist

Which way suits you?

3. im not sure how to solve using those ways. i have to solve to go through by completing the square

4. Originally Posted by kenzie103109
im not sure how to solve using those ways. i have to solve to go through by completing the square
Completing the square is the same method as turning pt form.

$x = \frac{1}{2}y^2 + 3y-6$

First step is to tkae out half as a common factor

$x = \frac{1}{2}(y^2 + 6y-12)$

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

$x = \frac{1}{2}((y^2 + 6y+9)-9-12)$

$x = \frac{1}{2}((y^2 + 6y+9)-21)$

Then making the perfect square complete.

$x = \frac{1}{2}((y+3)^2 -21)$

Can you take it from here?