Thread: find vertex of parabola

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

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

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?

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

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.



First step is to tkae out half as a common factor



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





Then making the perfect square complete.



Can you take it from here?