(x2-1)/(x2-2x-8)=(x-1){(x+1)/[(x-4)(x+2)]}. There are no cancelling factors.
We can split part into partial fractions:
(x+1)/[(x-4)(x+2)]=A/(x-4)+B/(x+2).
Ax+2A+Bx-4B=x+1⇒2A-4B=1; A+B=1, so B=1-A.
2A-4(1-A)=1, 2A-4+4A=1, 6A=5, A=⅚, B=⅙.
(x+1)/[(x-4)(x+2)]=⅚/(x-4)+⅙/(x+2).
The original expression can be written:
⅙(x-1)[5/(x-4)+1/(x+2)].
If the expression were to be graphed there would be vertical asymptotes x=4 and x=-2, and x-intercepts at -1 and 1. The y-intercept is y=⅛. The curve would be split into 3 parts, 2 wholly above the x-axis and the third between the other two parts mostly below the x-axis except for where the curve intersects the x-axis (at the x-intercepts) and the y-axis (at the y-intercept).
When x is very large (positive or negative) the expression approaches y=1, which is the horizontal asymptote for the two upper parts. However, the left upper part intersects y=1 at x=-3.5 (see below) and as x becomes more negative the curve approaches y=1 from below. The right upper part approaches y=1 from above, and the lower part is enclosed by the vertical asymptotes (an inverted U-shape).
y=1 when x2-1=x2-2x-8, -1=-2x-8, 2x=-7, x=-7/2. So the curve intersects y=1 at x=-3.5.