Question

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

Answer 1

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]}