Answer
$x^2-y^2-4x+3 = 0$
$(x-2)^2-y^2 = 1$
This is the equation of a hyperbola, where $a = 1, b=1, h = 2,$ and $k=0$
Work Step by Step
$x^2-y^2-4x+3 = 0$
$x^2-4x-y^2+3 = 0$
$x^2-4x+4-4-y^2+3 = 0$
$(x-2)^2-y^2 = 1$
We can write the general form of a hyperbola:
$\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2} = 1$
The equation in this question is the equation of a hyperbola, where $a = 1, b=1, h = 2,$ and $k=0$
The equations of the asymptotes are $~~y = x-2~~$ and $~~y = -(x-2) = 2-x~~$
Note that these two lines intersect at the point $(2,0)$
The x-intercepts are $~~1~~$ and $~~3$
The y-intercepts are $~~\pm \sqrt{3}$
We can sketch a graph of this hyperbola:
