Cartesian form of the parabola

[f1player]

I'm having a little trouble with this question

Find the equation of the parabola with its axis of symmetry parallel to the x-axis, and which passes through the points (3,3), (6,5) and (6,-3)

Any ideas??

 

[stapel]

If the parabola is sideways ("parallel to the x-axis"), then the equation is of the form "x = ay² + by + c". So plug in the three given points to get three equations in a, b, and c.

. . . . .9a + 3b + c = 3
. . . . .25a + 5b + c = 6
. . . . .9a - 3b + c = 6

Then solve the system for the values of a, b, and c. This will give you your equation.

Hope that helps a bit.

Eliz.

 

[f1player]

thanks for that!!

I've now got the answer, but i have done it a different way.
your 3 simulatenous equations have given me an idea and i've done it my way, but it still works and the answer that i got is right.

thanks a lot

 

[stapel]

Glad I could help. And "congrats" on coming up with your own solution method. Creativity and a willingness to "try stuff" will take you far! :wink:

Eliz.