What are the vertex, focus and directrix of # y=x^2+4x+4 #?

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Answer 1 · 2018-06-29T16:09:32

Vertex=#(-2,0)#
Its directrix is #y=-1/4#
it's focus is #(-2,1/4)#

#y=color(green)((x+2)^2-4)+4#

#y=(x+2)^2#

the parabola is opened upwards

If a parabola is opened upwards then its equation will be

#color(blue)(y-k=4a(x-h)^2#

where #color(blue)((h,k)# are it's vertex

it's directrix is #color(blue)(y=k-a#

and its focus is #color(blue)((h,k+a)##rarr##"Where a is positive real number"#

so applying this for the following equation

#y=(x+2)^2#

#4a=1rarra=1/4#

it's vertex is #(-2,0)#

it's directrix is #y=0-1/4=-1/4#

it's focus is #(-2,0+1/4)=(-2,1/4)#