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

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Answer 1 · 2018-06-29T16:09:32

Vertex= $(- 2, 0)$ $(-2,0)$
Its directrix is $y = - \frac{1}{4}$ $y=-1/4$
it's focus is $(- 2, \frac{1}{4})$ $(-2,1/4)$

$y = {(x + 2)}^{2} - 4 + 4$ $y=color(green)((x+2)^2-4)+4$

$y = {(x + 2)}^{2}$ $y=(x+2)^2$

the parabola is opened upwards

If a parabola is opened upwards then its equation will be

$y - k = 4 a {(x - h)}^{2}$ $color(blue)(y-k=4a(x-h)^2$

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

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

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

so applying this for the following equation

$y = {(x + 2)}^{2}$ $y=(x+2)^2$

$4 a = 1 \to a = \frac{1}{4}$ $4a=1rarra=1/4$

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

it's directrix is $y = 0 - \frac{1}{4} = - \frac{1}{4}$ $y=0-1/4=-1/4$

it's focus is $(- 2, 0 + \frac{1}{4}) = (- 2, \frac{1}{4})$ $(-2,0+1/4)=(-2,1/4)$

What are the vertex, focus and directrix of y=x^2+4x+4 y=x2+4x+4 y=x^2+4x+4 ?