How do you identify the vertex, focus, directrix and the length of the latus rectum and graph #y=x^2-12x+20#?

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Answer 1 · 2018-02-19T13:02:28

Vertex is at #(6,-16)#, directrix is # y= -16.25#.
Focus is at #(6,-15.75)# and length of latus rectum is #1.0#

#y=x^2-12x+20=0# or

#y=(x^2-12x+36)-36+20 =0# or

#y=(x-6)^2-16 =0# The vertex form of equation of parabola is

#y=a(x-h)^2+k ; (h.k) ;# being vertex.

# h=6 , k = -16 ,a=1# .Therefore vertex is at #(6,-16)#

Vertex is at midway between focus and directrix . Since #a#

is positive , the parabola opens upward and directrix is below the

vertex. # d = 1/(4|a|)=1/4=0.25 ; d# is the distance of directrix

from vertex.Directrix is #y=-16-0.25 or y= -16.25#.

Focus is at #(6, (-16+0.25)) or (6,-15.75)#

The length of latus rectum is #4d=4*0.25=1.0#

graph{x^2-12x+20 [-40, 40, -20, 20]}