How do you graph #4x^2-y^2-4x-3=0#?

1 Answer
Dec 3, 2016

Answer:

See below.

Explanation:

Since the equation involves the difference between the quadratic terms (rather than their sum), this is the equation of a hyperbola (rather than an ellipse.)

Complete the squares for both #x# and #y#:

#4x^2-y^2-4x-3=0#

#=>4x^2-4x-y^2=3#

#=>color(red)4(x^2-x+color(red)(1/4))-color(red)(4(1/4))-y^2=3#

Note: there is no linear #y# term, so #y^2# is already a complete square.

#=>4(x-1/2)^2-1-y^2=3#

#=>4(x-1/2)^2-y^2=4#

Divide both sides by #4#:

#=>(x-1/2)^2-y^2/4=1#

#=>(x-1/2)^2/1^2-(y-0)^2/2^2=1#

We now have the hyperbola in the form #(x-h)^2/a^2-(y-k)^2/b^2=1.#

  1. This hyperbola is "centered" at the point #(h,k)=(1/2,0).#
  2. Since it is in #x^2-y^2# form, it opens left and right.
  3. a) The left vertex is at #(h-a," "k)=(1/2-1,"  "0)#
    #=(-1/2,"  "0).#
    b) The right vertex is at #(h+a," "k)=(1/2+1,"  "0)#
    #=(3/2,"   "0).#
  4. The equations for the asymptotes are:
    #y-k=+-b/a(x-h)#
    #<=>#
    #y-0=+-2/1(x-1/2)#
    #<=>#
    #y=+-2(x-1/2)#

Draw the vertices and the asymptotes; the rest should fall into place.

graph{(4x^2-y^2-4x-3)(y+2x-1)(y-2x+1)=0 [-5.82, 6.67, -3.076, 3.17]}