How do you identify the vertex, focus, directrix and the length of the latus rectum and graph #y=1/2x^2-3x+19/2#?

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Answer 1 · 2017-03-17T07:58:30

The vertex is #V=(3,5)#
The focus is #F=(3,11/2)#
The directrix is #y=9/2#
The length of the latus rectum is #LR=4#

Let's rearrange the equation by completing the squares

#y=x^2/2-3x+19/2#

#y-19/2=1/2(x^2-6x)#

#y-19/2+9/2=1/2(x^2-6x+9)#

#y-10/2=1/2(x-3)^2#

#2(y-5)=(x-3)^2#

#(x-3)^2=2(y-5)#

We compare this equation to the parabola

#(x-a)^2=2p(y-b)#

#p=1#

The vertex is #V=(a,b)=(3,5)#

The focus is #F=(a,b+p/2)=(3,11/2)#

The directix is #y=b-p/2#

#y=5-1/2=9/2#

The length of the latus rectum is #LR=2p=2#

graph{(y-x^2/2+3x-19/2)(y-9/2)((x-3)^2+(y-11/2)^2-0.01)=0 [-19.54, 26.1, -3.83, 18.98]}