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

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Answer 1 · 2018-07-05T05:33:48

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

Directrix #y=-0.25#

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#

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