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 $(-2,0+1/4)=(-2,1/4)$

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