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

Dec 3, 2016

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 $y$:

$4 {x}^{2} - {y}^{2} - 4 x - 3 = 0$

$\implies 4 {x}^{2} - 4 x - {y}^{2} = 3$

$\implies \textcolor{red}{4} \left({x}^{2} - x + \textcolor{red}{\frac{1}{4}}\right) - \textcolor{red}{4 \left(\frac{1}{4}\right)} - {y}^{2} = 3$

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

$\implies 4 {\left(x - \frac{1}{2}\right)}^{2} - 1 - {y}^{2} = 3$

$\implies 4 {\left(x - \frac{1}{2}\right)}^{2} - {y}^{2} = 4$

Divide both sides by $4$:

$\implies {\left(x - \frac{1}{2}\right)}^{2} - {y}^{2} / 4 = 1$

$\implies {\left(x - \frac{1}{2}\right)}^{2} / {1}^{2} - {\left(y - 0\right)}^{2} / {2}^{2} = 1$

We now have the hyperbola in the form ${\left(x - h\right)}^{2} / {a}^{2} - {\left(y - k\right)}^{2} / {b}^{2} = 1.$

1. This hyperbola is "centered" at the point $\left(h , k\right) = \left(\frac{1}{2} , 0\right) .$
2. Since it is in ${x}^{2} - {y}^{2}$ form, it opens left and right.
3. a) The left vertex is at $\left(h - a , \text{ "k)=(1/2-1," } 0\right)$
$= \left(- \frac{1}{2} , \text{ } 0\right) .$
b) The right vertex is at $\left(h + a , \text{ "k)=(1/2+1," } 0\right)$
$= \left(\frac{3}{2} , \text{ } 0\right) .$
4. The equations for the asymptotes are:
$y - k = \pm \frac{b}{a} \left(x - h\right)$
$\iff$
$y - 0 = \pm \frac{2}{1} \left(x - \frac{1}{2}\right)$
$\iff$
$y = \pm 2 \left(x - \frac{1}{2}\right)$

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]}