Question

How do you find the focus, directrix and sketch `y=x^2+2x+1`?

Answer 1

`y=(x+1)^2`
Focus `(-1, 0.25)`

Directrix `y=-0.25`

Explanation:

Given -

`y=x^2+2x+1` ------------(1)

Vertx

`x=(-b)/(2a)=(-2)/(2xx1)=(-2)/2=-1`

At `x=-1; y=(-1)^2+(-1)2+1=1-2+1=0`

Vertex `(-1,0)`

Its y-intercept

At `x=0; y=(0)^2+2(0)+1=1`

`(0,1)`

Considering vertex and intercept the curve opens up.

Knowing the vertex, let us rewrite the equation in vertex form

`(x-h)^2=4a(y-k)`

`(x+1)^2=4a(y-0)`

`(x+1)^2=4ay` --------------(2)

We have to find the value of `a`

The parabola is passing through `(0,1)`

Plug the value in equation (2)

`(0+1)^2=4a(1)`

`4a=1`

`a=1/4`

Plug the value `a=1/4` in equation (2)

`(x+1)^2=1/4 xx4 xxy`

`(x+1)^2=y`

Its focus is `(x, (y+a)); (-1, (0+1/4))`

Focus `(-1, 0.25)`

Directrix `y=-0.25`