find the vertex, focus and directrix of y^2-6u+11=2x. can someone check my answers please?

Question

anonymous · Answer

its should say Y^2-6y+11=2x

mathteacher1729 · Answer

Would you know how to do it this problem if it was $$\large x^2-6x+11=2y$$ instead?

anonymous · Answer

I actually can.. but i put the original problem in (y-k)^2=4p(x-h) form.. my vertex was (-1,3) and my foci were (-3/4,3) and (-5/4,3)... my directrix is x=-3/4. I'm not even sure if this is horizontal!

anonymous · Answer

@mathteacher1729 how would i know if it's horizontal or not??

mathteacher1729 · Answer

The parabola is a $\cup $ or $\cap $ shape if it's in the form $y = ax^2+bx+c$. It's shaped like $\subset$ or $\supset$ if the equation is of the form $x = ay^2+by+c$. You can also graph it using http://www.desmos.com for free. :)

mathteacher1729 · Answer

Time for me to sign off. Paul's Online Math Notes breaks this stuff down really well. Purplemath also has some good tutorials.  Good luck. :)

anonymous · Answer

Thank you :)