# 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

Note: there is no linear

Divide both sides by

We now have the hyperbola in the form

- This hyperbola is "centered" at the point
#(h,k)=(1/2,0).# - Since it is in
#x^2-y^2# form, it opens left and right. - 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).# - 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]}