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

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

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- Jul 12th 2010, 02:10 PMkenzie103109find vertex of parabola
find the vertex of the parabola and give its direction of opening.

x = 1/2y^2 + 3y-6 - Jul 12th 2010, 02:21 PMpickslides
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? - Jul 12th 2010, 02:25 PMkenzie103109
im not sure how to solve using those ways. i have to solve to go through by completing the square

- Jul 12th 2010, 02:35 PMpickslides
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?