How do you find the focus, directrix and sketch y=x^2+2x+1y=x2+2x+1?

1 Answer
Nallasivam V
Jul 5, 2018

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

Directrix y=-0.25y=−0.25

Explanation:

Given -

y=x^2+2x+1y=x2+2x+1 ------------(1)

Vertx

x=(-b)/(2a)=(-2)/(2xx1)=(-2)/2=-1x=−b2a=−22×1=−22=−1

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

Vertex (-1,0)(−1,0)

Its y-intercept

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

(0,1)(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−h)2=4a(y−k)

(x+1)^2=4a(y-0)(x+1)2=4a(y−0)

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

We have to find the value of aa

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

Plug the value in equation (2)

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

4a=14a=1

a=1/4a=14

Plug the value a=1/4a=14 in equation (2)

(x+1)^2=1/4 xx4 xxy(x+1)2=14×4×y

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

Its focus is (x, (y+a)); (-1, (0+1/4))(x,(y+a));(−1,(0+14))

Focus (-1, 0.25)(−1,0.25)

Directrix y=-0.25y=−0.25

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